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Honors Analysis Course Notes Math 55b, Harvard University Contents 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 The real numbers . . . . . . . . . . . . . . . . . . . . . . . . . 3 3 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 4 Real sequences and series . . . . . . . . . . . . . . . . . . . . . 17 5 Differentiation in one variable . . . . . . . . . . . . . . . . . . 21 6 Integration in one variable . . . . . . . . . . . . . . . . . . . . 24 7 Algebras of continuous functions . . . . . . . . . . . . . . . . . 30 8 Differentiation in several variables . . . . . . . . . . . . . . . . 37 9 Integration in several variables . . . . . . . . . . . . . . . . . . 43 10 Elementary complex analysis . . . . . . . . . . . . . . . . . . . 55 11 Analytic and harmonic functions . . . . . . . . . . . . . . . . 67 12 Zeros and poles . . . . . . . . . . . . . . . . . . . . . . . . . . 73 13 Residues: theory and applications . . . . . . . . . . . . . . . . 76 14 Geometric function theory . . . . . . . . . . . . . . . . . . . . 83 1 Introduction This course will provide a rigorous introduction to real and complex analysis. It assumes strong background in multivariable calculus, linear algebra, and basic set theory (including the theory of countable and uncountable sets). The main texts are Rudin, Principles of Mathematical Analysis, and Marsden and Hoffman, Basic Complex Analysis. A convenient reference for set theory is Halmos, Naive Set Theory. Real analysis. It is easy to show that there is no x ∈ Q such that x2 = 2. One of the motivations for the introduction of the real numbers is to give solutions for general algebraic equations. A more profound motivation comes from the general need to introduce limits to make sense of, for example,∑ 1/n2. Finally a geometric motivation is to construct a model for a line, which should be a continuous object and admit segments of arbitrary length (such as π). 1 At first blush real analysis seems to stand apart from abstract algebra, with the latter’s emphasis on axioms and categories (such as groups, vector spaces, and fields). However R is a field, and hence an additive group, and much of real analysis can be conceived as part of the representation theory of R acting by translation on various infinite-dimensional spaces such C(R), Ck(R) and L2(R). Fourier series and the Fourier transforms are instances of this perspective. Differentiation itself arises as the infinitesimal generator of the action of translation. Complex analysis. The complex numbers (including the ‘imaginary’ num- bers of questionable ontology) also arose historically in part from the simple need to solve polynomial equations. Imaginary numbers intervene even in the solution of cubic equations with integer coefficients — which always have at least one real root. A signal result in this regard is the fundamental theo- rem of algebra: every polynomial p(x) has a complex root, and hence can be factors into linear terms in C[x]. The complex numbers take on a geometric sense when we regard z = a+ib as a point in the plane with coordinates (a, b) = (Re z, Im z). The remarkable point here is that complex multiplication respects the Euclidean length or absolute value |z|2 = a2 + b2: we have |zw| = |z| · |w|. It follows that if T ⊂ C is a triangle, then zT is a similar triangle (if z 6= 0). Passing to polar coordinates r = |z|, θ = arg(z) ∈ R/2πZ, we find: arg(zw) = arg(z) + arg(w). This gives a geometric way to visualize multiplication. We also note that |zn| = |z|n, arg(zn) = n arg(z). All rational functions, and many transcendental functions such as ez, sin(z), Γ(z), etc. have natural extension to the complex plane. For example we can define ez = ∑ zn/n! and prove this power series converges for all z ∈ C. Alternatively one can define ez = lim(1 + z/n)n. It is then easy to see geometrically that eiθ = cos θ+ i sin θ. The main point is that arg(1 + iθ/n) = θ/n+O(1/n2), 2 Dedekind cuts. This Corollary motivates both a construction of R and a proof of its uniqueness. Namely one can construct a standard field (call it R) as the set of Dedekind cuts (A,B), where Q = A⊔B, A < B, A 6= ∅ 6= B and B has no least element. (The last point makes the cut for a rational number unique.) Then (A1, B1) + (A2, B2) = (A1 + A2, B1 + B2), and most importantly: sup{(Aα, Bα)} = (⋃ Aα, ⋂ Bα ) , so R is complete. (A slight correction may be needed if the limit is rational.) Then, for any other complete ordered field K, one shows that map f : K → R given by f(x) = (A(x), B(x)), where A(x) = {y ∈ Q : y ≥ x} and B(x) = {y ∈ Q : y > x}, is an isomorphism. Remark: ideals. Dedekind also invented the theory of ideals. The idea here is that if you have a suitable number ring (say A = Z[ √ 7]), you can describe a number n ∈ A by associating to it the ‘ideal’ I = (n) of all x ∈ A which are divisible by n. Then you can axiomatize the properties of I (basically A/I should be a ring), and the consider all ideals in A as an extension of the ‘numbers’ in A. It turns out that, even though A may not have unique factorization (e.g. 6 = 2 · 3 = ( √ 7 + 1)( √ 7 − 1), the ideals in I do factor uniquely into prime ideals. This good theory of factorization holds in all Dedekind domains (which include all integrally closed number rings). Models. Are the real numbers R really unique? What we have shown above is that in any fixed model M for set theory, any two complete ordered fields K1 and K2 are isomorphic. In particular, they have the same cardinality. Whether or not |Ki| = ℵ1 or not depends on the model, but the answer is the same for both values of i. Uses of completeness. Here is a sample use of completeness. Theorem 2.3 For any real number x > 0 and integer n > 0, there exists a unique y > 0 such that yn = x. Proof. It is convenient to assume x > 1 (which can be achieve by replacing x with knx, k >> 0). The main point is the existence of y which is established by setting y = supS = {z ∈ R : z > 0, zn < x}. This sup exists because 1 ∈ S and z < x for all z ∈ S. Suppose yn 6= x; e.g. yn < x. Then for 0 < ǫ < y we have (y + ǫ)n = yn + · · · < yn + 2nǫyn−1. 5 By choosing ǫ small enough, the second term is less than x − yn and so y + ǫ ∈ S, contradicting the definition of y. A similar argument applies if yn > x. By the same type of argument one can show more generally: Theorem 2.4 Any polynomial of odd degree has a real root. Limits and continuity. The order structure makes it possible to define limits of real numbers as follows: we say xn → y if for every integer m > 0 there exists an N such that |xn − y| < 1/m for all n ≥ N . We then say a function f : R → R is continuous if f(xn) → f(y) whenever xn → y. Similarly f : A→ R is continuous if whenever xn ∈ A converges to y ∈ A, then f(xn) → f(y). Example: the function f(x) = 1/(x− √ 2) is continuous on A = Q. Extended real numbers. It is often useful to extend the real numbers by adding ±∞. These correspond to the Dedekind cuts where A or B is empty. Then every subset of R has a least upper bound: sup R = +∞, sup ∅ = −∞. Infs. We define inf E = − sup(−E). It is the greatest lower bound for E. Cauchy sequences in R. The completeness of R shows an increasing sequence which is bounded above converges to a limit, namely its sup. More generally, xn ∈ R is a Cauchy sequence if it clusters: we have lim n→∞ sup i,j>n |xi − xj | = 0. Theorem 2.5 Every Cauchy sequence in R converges: there exists an x ∈ R such that xi → x. Proof. Let an = infi≥n xi and let bn = supi≥n xi. Then a1 ≤ a2 ≤ · · · ≤ b2 ≤ b1, so there exists an A and B such that ai → A and bi → B. Moreover, |an − bn| ≤ sup i,j>n |xi − xj | → 0, so A = B. Since an ≤ xn ≤ bn, we have xn → A as well. 6 Constructing roots. A decimal number is just a way of specifying a Cauchy sequence of the form xn = pn/10n. Here is a constructive definition of √ 2: it is the limit of xn where x1 = 1 and xn+1 = (xn + 2/xn)/2. Limits, liminf, limsup. Because of the order structure of R, in addition to the usual limit of a sequence xn ∈ R (which may or may not exist), we can also form: lim sup xn = lim n→∞ sup{xi : i > n} and lim inf xn = lim n→∞ inf{xi : i > n}. These are limits of increasing or decreasing sequences, so they always exist, if we allow ±∞ as the limit. Example: Let f(x) = exp(x) sin(1/x), and let x → 0. Then lim f(x) does not exist, but lim sup f(x) = 1 and lim inf f(x) = −1. 3 Metric spaces A pair (X, d) with d : X ×X → [0,∞) is a metric space if d(x, y) = d(y, x), d(x, y) = 0 ⇐⇒ x = y, and d(x, z) ≤ d(x, y) + d(y, z). We let B(x, r) = {y ∈ X : d(x, y) < r} denote the ball of radius r about x. Euclidean space. The vector space Rk with the distance function d(x, y) = |x− y| = (∑ (xi − yi) 2 )1/2 is a geometric model for the Euclidean space of dimension k. The underlying inner product 〈x, y〉 = ∑ xiyi satisfies 〈x, y〉 = |x||y| cos θ, where θ is the angle between the vectors x and y. In particular 〈x, x〉 = |x|2. Norms. When V is a vector space over R or C, many translation invariant metrics are given by norms. A norm is a function |x| ≥ 0 such that |x+ y| ≤ 7 6. The Cantor set K ⊂ [0, 1] consists of all points which can be expressed in base 3 without using the digit 1. This is an example of a perfect set with no interior. Trees and Snowflakes. The Cantor set arises naturally as the ends of a bifurcating tree. The tree just gives the base three expansion of each point. If you build a tent over each complementary interval to the Cantor set in [0, 1], you get the beginnings of the Koch snowflake curve (a fractal curve of dimension log 4/ log 3 > 1). Figure 1. The Koch snowflake curve. Completeness. A metric space is complete if every Cauchy sequence has a limit. For example, R is complete, as is Rk. A closed subset of a complete space is complete. Theorem 3.3 Any metric space X can be isometrically embedded as a dense subset of a complete metric space X. Proof. Take X to be the space of Cauchy sequences in X with d(x, y) = lim d(xi, yi), and points at distance zero identified. Example. The real numbers are the completion of Q. (Can we take this as the definition of R? It is potentially circular, since we have used limits in the definition of the metric on X, and we have used R in the definition of metrics.) Compactness. Theorem 3.4 Let (X, d) be a metric space. The following are equivalent: 10 1. Every sequence in X has a convergent subsequence. 2. Every infinite subset of X has a limit point. 3. For every nested sequence of nonempty closed sets F1 ⊃ F2 ⊃ · · · in X, ⋂ Fi 6= ∅. The proof is straightforward. When these conditions hold, we say X is compact. More generally, a subset K ⊂ X is compact if these conditions hold for the induced metric. Note that a compact subspace of X is automatically closed. Separability and open covers. A metric space is separable if it has a countable dense set. E.g. Rn is separable since Qn is dense. Proposition 3.5 A compact metric space is separable. Proof. Let En ⊂ X be a maximal collection of points separated by at least 1/n. Then En has no limit point, so it must be finite; and ⋃ En is dense. Proposition 3.6 In a separable metric space, every open cover has a count- able subcover. Proof. Suppose U covers X. Let (xi) be a countable dense set, and let Ui,n be an element of U containing B(xi, 1/n) if such an element exists. The list of such Ui,n is countable, and for every x ∈ X there is a U ∈ U and i, n such that x ∈ B(xi, 1/n) ⊂ U ; thus ⋃ Ui,n covers X. Theorem 3.7 A metric space X is compact iff every open cover of X has a finite subcover. Proof. Suppose X is compact, and let U be an open cover of X. By the preceding results, we can assume U is a countable set {U1, U2, . . .}. Then Vi = ⋃i 1 Uj is an increasing collection of open sets with ⋃ Vi = X. Thus Fi = X −Vi is a decreasing collection of closed sets with ⋂ Fi = ∅. It follows that Fi = ∅ for some i, and hence ⋃i 1 Uj = X. The converse is similar. 11 Non-example. The line R is not compact. Check that every property above is violated. Theorem 3.8 Any interval I = [a, b] ⊂ R is compact. Proof. Suppose E ⊂ I is an infinite set. Cut I into two equal subintervals. One of these, say I1, meets E in an infinite set. Repeating the process, we obtain a nested sequence I1 ⊃ I2 ⊃ I3 . . . such there is at least one point xi ∈ E ∩ Ii. Since |Ii| = 2−i|I|, (xi) is a Cauchy sequence, and hence it converges to a limit x ∈ I which is also a limit point of E. Theorem 3.9 A subset E ⊂ Rn is compact iff it is closed and bounded. Proof. A bounded set is contained in [a, b]n for some a, b and then the argu- ment above can be applied to each coordinate. Conversely, if E is unbounded then a sequence xn ∈ E with |xn| → ∞ has no convergent subsequence. Similar arguments show: Theorem 3.10 A nonempty, compact, perfect metric space X is uncount- able. In fact, it contains a bijective copy of 2N, so |X| = |R|. Remark: the continuum hypothesis. CH asserts that any uncoutnable set E ⊂ R satisfies |E| = |R|. Although this statement is undecidable, it can be proved for many classes of sets. In particular, it holds if E is closed: in this case either E is countable or it contains a closed, perfect set, and hence |E| = R. (To prove this statement, show the set of condensation points E ′ ⊂ E, i.e. points x such that |B(x, r) ∩ E is uncountable for all r > 0, is perfect.) Theorem 3.11 If (X, d) is compact, then it is also complete. Infinite cubes. Let X = [0, 1]N denote the infinite cube. Consider the metrics d1(x, y) = sup |xi − ui| and d2(x, y) = ∑ |xi − ui|/2i. 12 Corollary 3.20 If an → a and bn → b in R, then anbn → ab. Corollary 3.21 The polynomials R[x] are in C(R). Question. Why is exp(x) continuous? A good approach is to show it is a uniform limit of polynomials (see below). N.B. the function g : [0, 1] → R given by g(x) = lim n→∞ xn is also a limit of polynomials but not continuous! Continuous functions on a compact set. We wish to give an interesting example of a complete metric space besides a closed subset of Rk, and also show the difference between completeness and compactness. Let C[a, b] be the vector space of continuous functions f : [a, b] → R. Define a norm on this space by ‖f‖∞ = sup [a,b] |f(x)|, and a metric by d(f, g) = ‖f − g‖∞ = sup |f(x) − g(x)|. This metric is finite because any continuous function on a compact space is bounded. We will show: Theorem 3.22 The metric space (C[a, b], d) is complete. Bounded sets. First note that a closed ball in C[a, b] is not compact. For example, what should sin(nx) converge to? Or, note that we can find infinitely many points in B(0, 1) with d(fi, fj) = 1, i 6= j. Even worse, we can have fn ∈ C[a, b] such that fn(x) → g(x) for all x, but g is not continuous. Also, is it clear that d(f, g) is even finite? The main point will be to use compactness of [a, b]. In fact the whole development works just as well for any compact metric space K. We define C(K) just as before. In particular, we can make the space of all bounded functions B(K) into a metric space using the sup-norm as well, and we have C(K) ⊂ B(K). Uniform convergence. We say a sequence of functions fn, g : X → R converges uniformly if g−fn is bounded and ‖g−fn‖∞ → 0. If g (and hence fn) is bounded, this is the same as convergence in B(X). 15 Theorem 3.23 If fn : X → R is continuous for each n, and fn → g uni- formly, then g is continuous. Proof. We illustrate the use of lim sup. Suppose xi → x in X. Then for any n, we have |g(xi) − g(x)| ≤ |fn(xi) − fn(x)| + 2d(fn, g). Letting i→ ∞, we have lim sup |g(xi) − g(x)| ≤ 2d(fn, g). Since n is arbitrary and d(fn, g) → 0, this shows g(xi) → g(x). Corollary 3.24 If K is compact then C(K) is complete. Proof. Let fi be a Cauchy sequence in C(K). Then fi(x) is a Cauchy sequence in R for each x. Thus fi(x) → g(x) for each x. Moreover ‖g − fi‖∞ → 0 since d(fi, fj) → 0. Thus fi → g uniformly, and therefore g is continuous. Note that compactness was used only to get distances finite. The same argument shows that B(X) is complete for any metric space X, and C(X)∩ B(X) is closed, hence also complete. The quest for completeness: comparison with R. We obtained the complete space R by starting with Q and requiring that all reasonable limits exist. We obtained C[0, 1] by a different process: we obtained completeness ‘under limits’ by changing the definition of limit (from pointwise to uniform convergence). This begs the question: what happens if you take C[0, 1] and then pass to the small set of functions which is closed under pointwise limits? This question has an interesting and complex answer, addressed in courses on measure theory: the result is the space of Borel measurable functions. Monotone functions. One class of non-continuous functions that are very useful are the monotone functions f : R → R. These satisfy f(x) ≥ f(y) whenever x > y (if they are increasing) or whenever x < y (if they are decreasing). If the strict inequality f(x) < f(y) holds, we say f is strictly monotone. 16 Theorem 3.25 A map f : R → R is a homeomorphism iff it is strictly monotone and continuous. Theorem 3.26 A monotone function has at most a countable number of discontinuities. Note that limx→y− f(x) and limx→y+f(x) always exists. Example; probability theory. Let qn be an enumeration of Q with n = 1, 2, 3, . . . and let f(x) = ∑ qn<x 1/2n. Then f is monotone increasing and its points of discontinuity coincide with Q. Note that f increases from 0 to 1. Quite generally, if X is a random variable, then its distribution function is defined by F (x) = P (X < x). In the example above we can take X = qn with probability 1/2n. In fact, random variables correspond bijectively with monotone functions that increase from 0 to 1, with a suitable convention of right or left continuity. The random variable X attached to f(x) above gives a random rational number. This variable can be described as follows: flip a coin until it first comes up heads; if n flips are required, the value of X is qn. 4 Real sequences and series The binomial theorem. The binomial theorem, which is easily proved by induction, states that: (1 + y)n = 1 + ny + ( n 2 ) y2 + · · ·+ yn, where ( a b ) = a!/(b!(a − b)!). The coefficients form Pascal’s triangle. We will see that this algebra theorem can also be used to evaluate limits. Some basic sequences. Perhaps the most basic fact about sequences is that if |x| < 1 then xn → 0. But how would you prove this? Here are more. Assume p > 0 and n→ ∞. Then: 1. np → ∞, i.e. 1/np → 0. 2. p1/n → 1. 3. n1/n → 1. 17 Irrationality of e. It is well-known, and not too hard to prove, that e is transcendental. It is even easier to prove e is irrational: we have e = (Nq + ǫq)/q!, where 0 < ǫq < 1; while if e = p/q, then q!e − Nq = ǫq is an integer. Root and ratio test. The ratio test says that if lim sup |an+1/an| = r < 1, then ∑ an converges (absolutely). This proof is by comparison with a geometric series C ∑ sn. with r < s < 1. The root test gives the same conclusion, by the same comparison, if lim sup |an|1/n = r < 1. There are converse theorems if the limsup is a limit r > 1, but these are no more interesting than the fact that ∑ an diverges if |an| does not tend to zero (the nth term test). Power series. The real virtue of the root test is the following. Theorem 4.4 Given an ∈ C, let r = lim sup |an|1/n. Then ∑ anz n con- verges uniformly on compact sets for all z ∈ C with |z| < 1/r. For the proof just observe that ∑ fn converges uniformly if ∑ ‖fn‖∞ is finite. The conclusion is almost sharp: the series diverges if |z| > 1/r. Example. The function exp(z) = ∑ zn/n! is well-defined for all z ∈ C, because (ratio test) lim |z|/n→ 0 or because (root test) (n!)1/n ≥ (n/2)1/2 → ∞. Corollary 4.5 If f(x) = ∑ anx n converges for |x| < R, then ∫ f and f ′(x) are given in (−R,R) by ∑ anx n+1/(n+ 1) and ∑ nanx n−1. Summation by parts. It is worth noting that differentiation and integra- tion have analogues for sequences. These can be based on the definition ∆an = an − an−1, which satisfies N∑ 1 ∆an = aN − a0 as well as the Leibniz rule (∆ab)n = an∆bn + bn−1∆an. 20 Summing both sides we get the summation by parts formula: N∑ 1 an∆bn = aNbN − a0b0 − N∑ 1 bn−1∆an. (It is convenient to think of an, bn as being defined for all n, but for the equation above it is enough that they are defined for n ≥ 0; then ∆an and ∆bn are defined for n ≥ 1.) Example: S = N∑ 1 n2 = N∑ 1 n2(∆n) = N3 − N∑ 1 (n− 1)∆n2 = N3 − N∑ 1 (n− 1)(2n− 1) = N3 − 2S + N∑ 1 (3n− 1) = N3 − 2S + 3N(N + 1)/2 −N, which gives S = N(2N2 + 3N + 1) 6 · 5 Differentiation in one variable Differentiation. Let f : [a, b] → R. We say f is differentiable at x if lim y→x f(y) − f(x) y − x =: f ′(x) exists (note that if x = a or b, the limit is one-sided); equivalently, if f(y) = f(x) + f ′(x)(y − x) + o(y − x). If f ′(x) exists for all x ∈ [a, b], we say f is differentiable . Calculating derivatives. The usual procedures of calculus (for computing derivatives of sums, products, quotients and compositions) are readily verified for differentiable functions. (Less is needed, e.g. all but the chain rule work for functions just differentiable at x.) In particular we will later use: Theorem 5.1 The derivative of a polynomial ∑ anx n is given by ∑ nanx n−1. 21 Continuity. If f is differentiable then it is also continuous. However there exist functions which are continuous but nowhere differentiable. An example is f(x) = ∑∞ 1 sin(n!x)/n2; the plausibility is seen by differentiating term by term. Properties of differentiable functions. Theorem 5.2 Let f : [a, b] → R be differentiable. Then: 1. If f(a) = f(b) then f ′(c) = 0 for some c ∈ (a, b). 2. There is a c ∈ (a, b) such that f(b) − f(a) = f ′(c)(b− a). 3. If f ′(a) < y < f ′(b), then f ′(c) = y for some c ∈ (a, b). Proof. (1) Consider a point c where f achieves it maximum or minimum. (2) Apply (1) to g(x) = f(x) −Mx, where M = (f(b) − f(a))/(b − a). (3) Reduce to the case y = 0 and again consider an interior max or min of f . Jumps. To see (3) is interesting, it is important to know that there exist examples where f ′(x) is not continuous! By (3), if f ′(x) exists everywhere, it cannot have a jump discontinuity. Corollary 5.3 If f is differentiable and f ′ is monotone, then f ′ is continu- ous. Taylor’s theorem. This is a generalization of the mean-valued theorem. Theorem 5.4 If f : [a, b] → R is k-times differentiable, then there exists an x ∈ [a, b] such that f(b) = ( k−1∑ 0 f (j)(a) j! (b− a)j ) + f (k)(x) k! (b− a)k. Proof. Subtracting the Taylor polynomial from both sides, we can also assume that f and its first k − 1 derivatives vanish at 0. Suppose we also knew f(b) = f(a). Then there would be an x1 ∈ [a, b] such that f ′(x1) = 0, 22 Proof. We must show for each h > 0 there is an r > 0 such that d(x, y) < r =⇒ |f(x) − f(y)| < h. Cover R by open intervals (Ii) of length h. Then their preimages Ui give an open cover of K (which can be reduced to a finite subcovering). Let r > 0 be the Lebesgue number of this covering. If |x− y| < r then x, y ∈ Ui for some i, and hence f(x), f(y) ∈ Ii which implies |f(x) − f(y)| ≤ h. Sketch of the proof of Theorem 6.1. Although we are interested in con- tinuous functions, it is useful to first define the integral I(s) of step functions s(x). Then we check that I(s) satisfies the axioms above, with continuous functions replaced by step functions. Especially, I(s) is linear. Then we observe that any function Ib a(f) satisfying the axioms above also satisfies I(s) ≤ Ib a(f) ≤ I(S) whenever s < f < S. Therefore we define I−(f) = sup s<f I(s), I+(f) = inf f<S I(S). By uniform continuity we find s, S with s < f < S and |S − s| < ǫ, which shows I−(f) = I+(f) and shows the existence of such the linear map Ib a on continuous functions. A similar argument proves uniqueness. Corollary 6.3 The map I : C[a, b] → R is continuous; in fact, |I(f)| ≤ |a− b|‖f‖∞. Corollary 6.4 If fn is a sequence of continuous functions and fn → g uni- formly, then ∫ b a fn → ∫ b a g. Counterexample. It is important to know that we cannot conclude ∫ fn →∫ g just from pointwise convergence! See an example below, where fn → 0 but ∫ fn = 1. 2/n n n f 0 Fundamental theorem of calculus. We will make a more detailed study of integration later, but for the moment we prove from the axioms above: 25 Theorem 6.5 If f : [a, b] is continuous and F (x) = ∫ x a f(t) dt, then F ′(x) = f(x). Proof. We have (1/t)(F (x+ t) − F (x)) = (1/t) ∫ x+t x f(t) dt→ f(x). Corollary 6.6 If f ′(x) is continuous, then ∫ x a f ′(t) dt = f(x) + C. Note: if we just assume f(x) is differentiable, then f ′(x) might not even have an integral. Limits of differentiable functions, reprise. Integration gives an alter- nate proof of Theorem 5.5 above, which asserts if fn ∈ C1[a, b], fn → f uniformly and f ′ n → g uniformly, then f ′ = g. Proof. We have fn(x) − fn(a) = ∫ x a f ′ n; taking limits on both sides gives f(x) − f(a) = ∫ x a g and hence f ′(x) = g(x). Riemann integrability. Let f : [a, b] → R be an arbitrary function. We say f is Riemann integrable if sup {∫ g : g ∈ C[a, b], g ≤ f } = inf {∫ h : h ∈ C[a, b], h ≥ f } , and denote the common value of these quantities by ∫ b a f(x) dx. Theorem 6.7 Any piecewise continuous function is in R. If f ∈ R then f is bounded. Theorem 6.8 A bounded function f : [a, b] → R is Riemann integrable iff its points of discontinuity form a set of measure zero. 26 Here the measure of a set E is the infimum of ∑ |Ii| over all countable collections of open intervals such that E ⊂ ⋃ Ii. Examples. The indicator function of Q gives an example of a function that is not Riemann integrable. On the other hand, the function f(x) = 1/q if x = p/q and f(x) = 0 otherwise, is discontinuous just on the rationals, so it is integrable. Hölder’s inequality. It is useful to define for p ≥ 1 the norms ‖f‖p = (∫ |f |p )1/p . These satisfy the triangle inequality (exercise) and we have the important: Theorem 6.9 (Hölder’s inequality) If f, g ∈ C[a, b] then ∣∣∣∣ ∫ b a fg ∣∣∣∣ ≤ ‖f‖p ‖g‖q. whenever 1/p+ 1/q = 1. The case p = q = 2 gives Cauchy-Schwarz. Proof. First check Young’s inequality xy ≤ xp/p + yq/q. Then by homo- geneity we can assume ‖f‖p = ‖g‖q = 1, and deduce ∫ |fg| ≤ ∫ |f |p/p+ ∫ |g|q/q = 1. (Proof of Young’s inequality. Draw the curve yq = xp, which is the same as the curve y = xp−1 or x = yq−1 (since pq = p+q). Then the area inside the rectangle [0, a]× [0, b] is bounded above by the sum of ap/p, the area between the graph and [0, a], and bq/q, the area between the graph and [0, b].) (Brute force proof: minimize f(x, y) = xp/p + yq/q subject to g(x, y) = xy = a. Then by the method of Lagrange multipliers, we have (y, x) = λ(xp−1, yq−1). This gives xy = λxp = λyq and hence xp/p+yq/q = λxy. And it also implies x = λ2(xp−1)q−1 = λ2x, so λ = 1.) Lebesgue theory. The completions of the space C[a, b] with respect to the norms above are the Lebesgue spaces Lp[a, b]. Their elements consist of measurable functions, to be discussed in Math 114. 27 7 Algebras of continuous functions In this section we prove some deeper properties of the space C(X). Compactness: Arzela–Ascoli. LetK be a compact metric space as above. A family of functions F ⊂ C(K) is equicontinuous if they all have the same modulus of continuity h(r). Theorem 7.1 The closure of F is compact in C(K) iff F is bounded and equicontinuous. Proof. The condition that F is totally bounded in the metric space C(K) translates into equicontinuity. Example. The functions fn(x) = sin(nx) are not equicontinuous on C[0, 1], but the functions with |f ′ n(x)| ≤ 1 are. Thus any sequence of bounded functions with bounded derivatives has a uniformly convergent subsequence. Approximation: Stone–Weierstrass. Theorem 7.2 (Weierstrass) The polynomials R[x] are dense in C[a, b]. This result gives a nice occasion to introduce convolution and approxi- mations to the δ-function. First, the convolution is defined by (f ∗ g)(x) = ∫ s+t=x f(s)g(t) dt = ∫ f(x− t)g(t) dt = ∫ g(x− t)f(t) dt. To make sure it is well-defined, it is enough to require e.g. that both functions are continuous and one has compact support. Note that: ‖f ∗ g‖∞ ≤ ‖f‖infty‖g‖1. From this one can readily verify that f ∗ g(x) is continuous. (Note: use the fact that if g has compact support, and gt(x) = g(x+ t), then ‖gt − g‖ → 0 as t→ 0; this is a restatement of uniform continuity.) The convolution inherits the best properties of both functions; e.g. if f has a continuous derivative, then so does f ∗ g, and we have (f ∗ g)′(x) = (f ′ ∗ g), as can easily be seen from the formula above. Thus shows: If f is a polynomial of degree d, then so is f ∗ g(x). 30 This can also be seen directly, using the fact that: ∫ (t− x)ng(t) dt = ∑( n k ) (−x)k ∫ tkg(t) dt; or conceptually, by noting that the polynomials of degree d form a closed, translation invariant subspace of C(R). Approximate identities. We say a sequence of functions Kn(x) ≥ 0 is an approximate identity if ∫ Kn = 1 for all n and ∫ |x|>ǫ Kn → 0 for all ǫ > 0. Theorem 7.3 If f is a compactly supported continuous function, then f ∗ Kn → f uniformly. Proof. We have |f | ≤ M for some M , and f is uniformly continuous, say with modulus of continuity h(r). Suppose for example we wish to compare f(0) and (Kn ∗ f)(0). Since Kn ∗ 1 = 1, we may assume f(0) = 0. Choose r such that h(r) < ǫ and N such that ∫ |x|>r Kn < ǫ. Then splitting the integral into two pairs at |x| = r, we find |(Kn ∗ f)(0)| = ∣∣∣∣ ∫ Kn(x)f(−x) dx ∣∣∣∣ ≤ h(r) +Mǫ ≤ (1 + ǫ)M. Example. Let Kn(x) = an(1 − x2)n for |x| ≤ 1, and 0 elsewhere, where an is chosen so ∫ Kn = 1. Then Kn(x) is an approximate identity. To see this, we first note that (1−x2)n ≥ 1/2 if 1−x2 > (1/2)1/n ≈ 1−Cn, i.e. if |x| < C/ √ n; this shows ∫ (1 − x2)n ≥ C/ √ n and hence an = O( √ n). Now suppose |x| > r. Then Kn(x) = O( √ n(1 − r2)n) → 0, i.e. Kn → 0 uniformly on |x| > r. It follows that ∫ |x|<r Kn → 1. Proof of Weierstrass’s theorem. We may suppose [a, b] = [0, 1], and adjusting by a linear function, we can assume f(0) = f(1) = 0. Extend f to the rest of R by zero. As above, choose Kn(x) = an(1 − x2)n for |x| ≤ 1, and 0 elsewhere, where an is chosen so ∫ Kn = 1. Then pn = Kn ∗ f → f uniformly on R. By differentiating, one can see that pn|[0, 1] (like Kn|[−1, 1]) is a polynomial of degree 2n or less, because Kn ∗ f |[0, 1] is independent of whether we cut off the polynomial Kn or not. 31 Theorem 7.4 (Stone) Let A ⊂ C(X) be an algebra of real-valued functions that separates points. Then A is dense. Sketch of the proof. Replace A by its closure; then we must show A = C(X). Since A is an algebra, f ∈ A =⇒ P (f) ∈ A for any polynomial P (x). Since |x| is a limit of polynomials (by Weierstrass’s theorem), we find |f | is in A. This can used to show that f, g ∈ A =⇒ f ∧ g, f ∨ g are both in A. The proof is completed using separation of points and compactness. Complex algebras. This theorem does not hold as stated for complex- valued functions. A good example is the algebra A of polynomials in z in C(S1). These all satisfy ∫ zf |dz| = 0 and this property is closed under uniform limits, so g(z) = z is not in the closure. If, however we require that A is a ∗-algebra (it is closed under complex conjugation), then the Stone- Weierstrass theorem holds as stated for complex functions as well. Fourier series. Let S1 = R/2πZ. We make the space of continuous complex functions C(S1) into an inner product space by defining: 〈f, g〉 = 1 2π ∫ 2π 0 f(x)g(x) dx. In particular, ‖f‖2 2 = 〈f, f〉 = 1 2π ∫ |f |2. This inner product is chosen so that en = exp(inx) satisfies 〈ei, ej〉 = δij. The Fourier coefficients of f ∈ C(S1) are given by an = 〈f, en〉. If f = ∑N −N bnen, then an = bn. Convergence of Fourier series: Féjer and Dirichlet. One of the main concerns of analysts for 150 years has been the following problem: given a function f(x) on S1, in what sense is f represented by its Fourier series∑ an exp(inx)? Trig polynomials. Let An ⊂ C(S1) denote the finite-dimensional subspace spanned by (e−n, . . . , en) and let A = ⋃ An be the space of trigonometric polynomials. Clearly A is an algebra, closed under complex conjugation. Thus the Stone-Weierstrass theorem shows: 32 After this phenomenon was discovered, a common sentiment was that it was only a matter of time before a continuous function would be discovered whose Fourier series diverged everywhere. Thus it was even more remarkable when L. Carleson proved: Theorem 7.11 For any f ∈ L2(S1), the Fourier series of f converges to f pointwise almost everywhere. The proof is very difficult. -3 -2 -1 1 2 3 -5 5 10 15 20 -3 -2 -1 0 1 2 3 2 4 6 8 10 Figure 2. The Dirichlet and Fejér kernels. The Fejér kernel. However in the interim Fejér, at the age of 19, proved a very simple result that allows one to reconstruct the values of f from its Fourier series for any continuous function. Theorem 7.12 For any f ∈ C(S1), we have f(x) = lim S0(f) + · · · + SN−1(f) N uniformly on the circle. This expression is a special case of Césaro summation, where one replaces the sequence of partial sums by their averages. This procedure can be iter- ated. In the case at hand, it amounts to computing ∑∞ −∞ an as the limit of the sums 1 N N∑ i=−N (N − |i|)an. 35 Approximate identities. The proof of Fejér’s result again uses approxi- mate identities. If we let TN(f) = S0(f) + · · · + SN−1(f) N , then TN(f) = f ∗ FN where the Fejér kernel is given by FN = (D0 + · · ·DN−1)/N. Of course ∫ FN = 1 since ∫ Dn = 1. But in addition, FN is positive and concentrated near 0, i.e. it is an approximation to the identity. Indeed, we have: FN(x) = sin2(Nx/2) N sin2(x/2) · To see the positivity more directly, note for example that (2N + 1)F2N+1 = z−2N + 2z−2N+1 + · · ·+ (2N + 1) + · · · 2z2N−1 + z2N = (z−N + · · · zN )2 = D2 N , where z = exp(ix). For the concentration near zero, observe that if |x| > 1/(ǫ √ N), then |FN(x)| ≥ ǫ or so. Modern questions. Perhaps a better question is: given any (an) ∈ ℓ2, is there some function on S1 such an as its Fourier coefficients? The answer again leads into the Lebesgue theory. If we let L2(S1) de- note the metric completion of C(S1) in the L2 norm, it is then clear than ℓ2 ∼= L2(S1); so the only issue is, how to interpret elements of this metric completion as functions? Why sin(x) and cos(x)? Why, among all the functions in C(S1), are we focusing on en(x) = cos(x) + i sin(x)? The reason ultimately has to do with representation theory of the group G = S1 on C(S1). The spaces Vn = Cen are invariant under the action of G by (gf)(x) = f(g−1x), and they are the only such. So the decomposition of a function into its Fourier components is an attempt to make precise the statement C(S1) = ⊕Vn, in other words an attempt to decompose C(S1) into an infinite sum of its irreducible subspaces for the action of G. 36 8 Differentiation in several variables We now turn to the calculus of functions on domains in Rn, and more general to the theory of smooth maps from Rn to Rm. Differentiation. Let U ⊂ Rn be open. We say f : U → Rm is differentiable at x ∈ U if there is a linear map L : Rn → Rm such that f(x+ t) = f(x) + L(t) + o(|t|). (This means |f(x+ t) − f(x) − L(t)|/|t| → 0 as t→ 0.) Note that t ∈ Rn is a tangent vector in the domain of f , and that L(t) ∈ Rm is a tangent vector in the range. In this can we write f ′(x) = Df(x) = L ∈ Hom(Rn,Rm). Note that the space Hom(A,B) of maps between normed vector spaces inherits a natural norm, given by ‖T‖ = sup v 6=0 |Tv|/|v|. (Use compactness of the unit ball to see this is finite.) We say f ∈ C1(U) if it is differentiable at every point, and the map Df : U → Hom(Rn,Rm) is continuous. In terms of the standard basis, Df is just the matrix (Df)ij = dfi/dxj. Theorem 8.1 If f is differentiable at x then so is each coordinate function, and Df = (dfi/dxj). In other words, ∆f ≈ ∑ j (dfi/dxj)∆xj . From the definitions we easily get: Theorem 8.2 (Chain Rule) If g is differentiable at x and f is differen- tiable at g(x), then D(f ◦ g)(x) = (Dg(x)) ◦Df(g(x)). Theorem 8.3 We have f ∈ C1(U,Rn) iff dfi/dxj exists and is continuous for all i, j. 37 Followup 3: What happens if you have N beetles on a unit circle, equally spaced? You get a logarithmic spiral with a different, smaller angle, namely (2π/N). Followup 4:What if you have N beetles on an ellipse? Explain the relation to curvature. Differential equations. We remark that the solution of a linear differential equation of order n in one variable, with given initial conditions, e.g. f (n)(t) = n−1∑ 0 aif (i)(t), can be reduced to the solution of a particular equation of the form F ′(t) = AF (t) where F : R → Rn, and F (0) is specified. Indeed, we just set F (t) = (Fi(t)) = (f(t), f ′(t), . . . , f (n−1)(t)), and let A be the companion matrix of the associated polynomial. (Put differently, F ′ = (F2, F3, . . . , ∑ aiFi+1).) Theorem 8.5 The equation F ′(t) = AF (t) has a unique solution for a given initial value F (0). Proof. Existence follows using the exponential: set F (t) = exp(tA)F (0). For uniqueness, suppose we have two solutions; then their difference satisfies |F ′(t)| ≤ C|F (t)| and F (0) = 0. But then (log |F |)′ ≤ C and hence |F (t)| ≤ exp(Ct)|F (0)| = 0. Factorization and diagonalization. If A = diag(λi) is diagonal, the solution is very simple — since eA = diag(eλi). Of course this need not be the case — if A = ( 0 1 −1 0 ) then etA = ( cos(t) sin(t) − sin(t) cos(t) ) . As usual, this linear algebra is simpler of the complex numbers. In par- ticular, if P (f) = ( ∑ anD n)(f) factors as ∏ (D− λi), where D = d/dt, then the space of solutions to P (f) = 0 is spanned by eλit. If (D − lambda)n occurs, then we need to multiply eλt by 1, x, . . . , xn−1 to get the full space of solutions to P (f) = 0. This is related to Jordan blocks. 40 The harmonic oscillator. It is useful to note that one of the most fun- damental differential equations, f ′′(t) = −f(t), becomes F ′(t) = AF (t) with A a 90◦ rotation. The solution curves to this equation are circles. Thus in ‘phase space’ a pendulum is moving around a circle — it is only in ‘position space’ that it is momentarily stationary at the end of each swing. Examining the picture in phase space immediately suggests that the quan- tity E = f(t)2 + f ′(t)2 is important, since it remains constant as the pen- dulum moves. In fact this is nothing more than the (sum of the kinetic and potential) energy! Higher derivatives. The most important fact about higher derivate is that the Hessian matrix (Hf)ij = d(df/xi)/dxj is symmetric, if all its entries are continuous. Proof. Consider the case f(x, y). Then df/dx = lim gn(x, y) = limn(f(x+ 1/n, y) − f(x, y)). Thus d2f dy dx = d dy lim gn On the other hand, d2f dx dy = lim dgn dy . So we just need to interchange limits and differentiation. This can be done so long as dgn/dy converges uniformly (on compact sets) as n → ∞. To see this, use the mean value theorem to conclude that dgn dy (x, y) = d2f dxdy (c, y) with |x− c| ≤ 1/n. Since d2f/dxdy is continuous, it is uniformly continuous on compact sets, say with modulus of continuity h(r); and therefore ∣∣∣∣ dgn dy − d2f dxdy ∣∣∣∣ ≤ h(1/n) → 0(c, y) uniformly. 41 The inverse function theorem: Diffeomorphisms. We now prove a nice geometric theorem that allows one to pass from ‘infinitesimal invertibility’ to ‘nearby invertibility’. A map f : U → V between open sets in Rn is a diffeomorphism if it is a homeomorphism and both f and f−1 are C1. Theorem 8.6 Let f be a smooth (C1) map to Rn defined near p ∈ Rn. Suppose Df(p) is an isomorphism, i.e. detDf(p) 6= 0. Then f is a diffeo- morphism near p. For the proof we need two other useful results: 1. If φ : X → X is a strict contraction on a complete metric space, then φ has a fixed point. 2. If U is convex, f : U → Rn is differentiable, and ‖Df‖ ≤ M , then |f(a) − f(b)| ≤M |a− b|. (Cf. Theorem 8.4.) The second follows by applying the mean-valued theorem to the function x 7→ 〈f(x) − f(a), f(b) − f(a)〉 on the interval [a, b]. Proof of the inverse function theorem. Composing with linear maps, we may assume p = 0 = f(0) and Df(0) = I. We may also assume, by continuity, that ‖Df(x) − I‖ < 1/10 when |x| < 1. Now consider for any y0 ∈ Rn the function φ(x) = x+ (y0 − f(x)). This function tries to make f(φ(x)) closer to y0 than f(x) was, by a Newton- like procedure. Note that φ(x) = x iff f(x) = y0, and ‖Dφ‖ = ‖I −Df‖ < 1/10 when |x| < 1. Thus if |y0| < 1/2, φ sends the unit ball into itself, contracting distances by a factor of at least 10. Consequently it has a unique fixed point. Thus f maps the unit ball onto V = B(0, 1/2), and if U is the preimage of V in the unit ball, then f : U → V is a bijection. Since V is open, U is open. Let g : V → U be the inverse. We now notice that |g(y)| is comparable to y. Indeed, the sequence (x0, x1, x2, . . .) defined by x0 = 0, xi+1 = φ(xi) converges rapidly to g(y0). In fact |x0 − x1| = |y0|, so the limit is close to y0; it satisfies |g(y0)| ≥ |y0|/2. The reverse inequality, |g(y0)| ≤ 2|y0|, follows from the fact ‖Df‖ ≤ 2. To complete the proof, we will showDg(0) = I. (ThenDg(y) = (Df)−1(g(y)) is continuous.) To see this, we just note that for y near 0 we have y = f(g(y)) = g(y) + o(|g(y)|) = g(y) + o(|y|), 42 by a finite number of (n− 1)-dimensional hypersurfaces. More accurately, a function in Rn (like in R) is Riemann integrable iff its discontinuities form a set of measure zero. Path integrals. Let us now consider a new kind of integral: ∫ C x dy where C is the unit semicircle in R2 defined by x2 + y2 = 1, y ≥ 0. This integral has a geometric meaning: cut C into little arcs, and the sum up xi∆yi over all of them. If we parameterize C by f(θ) = (cos θ, sin θ), then we obtain ∫ C x dy = ∫ π 0 cos θ d sin θ = ∫ π 0 cos2 θ dθ = π/2. But if we parameterize C by f(x) = (x, √ 1 − x2), then we get ∫ C x dy = ∫ 1 −1 xd √ 1 − x2 = ∫ 1 −1 −x2/ √ 1 − x2 = −π/2. This illustrates (a) that the integral is independent of the choice of param- eterization but (b) that its sign depends on something else — the choice of orientation of C! 3. Differential forms. To make these calculations more precise, we now introduce the formalism of differential forms. First we consider the finite- dimensional algebra A∗(Rn) over R generated by the identity element 1, and by dx1, . . . , dxn, subject to the relations dxi dxj = −dxj dxi. This is a graded algebra with dimAk(Rn) = ( n k ) . (More formally, we have Ak(V ) = ∧kV ∗ = the space of alternating k-forms on V .) We have A0(Rn) = R. It is easy to check that for any α, β ∈ A1(Rn), we have αβ = −βα. (A general sign formula is also easy to guess and prove.) For any open set U ⊂ Rn, the space Ωk(U) is the space of smooth k-forms: ω = ∑ |α|=k fα(x) dxα. 45 Here each fα(x) is a C∞ function. In particular, Ω0(U) is the space of smooth functions f(x), and Ωn(U) is the space of smooth volume elements f(x) dx1 . . . dxn. We can also regard Ωk(U) as the space of C∞ functions f : U → Ak(Rn). The vector space Ω∗(U) thus a graded algebra, by multiplying these functions pointwise. For example, since scalars commute with all elements of A∗(Rn), we have (fα)(gβ) = (fg)(αβ) for any function f, g and forms α, β. Exterior d. The first central piece of structure here is the exterior derivative d : Ωk → Ωk+1, characterized by df = ∑ df/dxi dxi, d(dxi) = 0, and d(αβ) = α dβ + (dα) dβ. Equivalently, d(f dxα) = ∑ (df/dxi) dxi dxα. Theorem 9.3 We have d(dω) = 0. Proof. This follows from symmetry of the Hessian d2f/(dxidxj) and the antisymmetry dxidxj = −dxjdxi. Duals of vectors. What do dxi and df really mean? Intrinsically, dxi ∈ V ∗ = (Rn)∗, so df(x) is dual to the tangent vectors at x. It is simply the linear functional given by df(x)(v) = lim t→0 (f(x+ tv) − f(x))/t. This is a coordinate-free definition. In particular, dxi(ej) = δij . Similarly, the elements of Ak(U) consist of functions α : U → ∧k(V ∗). Determinants. Recalling that ∧nRn is 1-dimensional, we obtain the fol- lowing important formula. Proposition 9.4 For any matrix A, we have V = ∏ i ∑ j Aij dxj = det(A) dx1 · · · dxn. Proof. Since only products of terms dxj with distinct indices survive, V = ∑ Sn ∏ i Ai,σ(i) sign(σ) dx1 · · · dxn. 46 Pullbacks. The second central player is the notion of pullback: if φ : U → V is a smooth map, then we get a natural map φ∗ : Ω∗(V ) → Ω∗(U). This map is characterized by three properties: (1) φ∗(f) = f ◦ φ on functions; (2) φ∗(αβ) = φ∗(α)φ∗(β); and (3) φ∗(df) = dφ∗(f). Proposition 9.5 We have (φ ◦ ψ)∗ = φ∗ ◦ ψ∗. Using the fact that φ∗(dxi) = ∑ φi/dxj dxj , from the formula above we get the followoing crucial result: Theorem 9.6 If dimU = dimV = n, then φ∗(dx1 · · · dxn) = det(Dφ)dx1 · · · dxn. Integration. The third player in the study of differential forms is the notion of integration. It is defined for compactly supported n-forms on Rn by: ∫ f(x) dx1 · · · dxn = ∫ f(x) |dx|. The change of variables formula then implies: Theorem 9.7 If φ : U → V is an orientation-preserving diffeomorphism, then ∫ U φ∗(ω) = ∫ V ω. We can then define integration of a k-form ω over a submanifold Mk by parameterization and pullback. More precisely, if ω is a k-form on Rn defined near an oriented submanifold Mk, and φ : U →Mk is a orientation- preserving parameterization of Mk by a region in Rk, then we define ∫ Mk ω = ∫ U φ∗ω. We have done this for the case of a semicircule M1 = C above. The preceding result shows: Corollary 9.8 The integral of a k-form over an oriented k-dimensional space is independent of how that space is parameterized. 47 DeRham cohomology. We can now give a hint at the connection between differential forms and topology. Given a 1-form ω, a necessary condition for it to be df for some f is that dω = 0. This is essentially saying that the mixed partials must agree. Is this sufficient? Theorem 9.10 If ω is a 1-form on a convex region U ⊂ Rn, and dω = 0, then there exists an f : U → R such that ω = df . Proof. Define f(Q) = ∫ [P,Q] w, and use path-independence to check that df = ω. This simplifies many integrals: e.g. if ω = df , then ∫ γ ω over a compli- cated path from P to Q is still just f(Q) − f(P ). For more general U such an assertion is false. For example, we have seen that on U = R2 − {(0, 0)} there is a closed form ω such that ∫ Cr ω = 2π. Given any region U we define Hk(U) = (k-forms with dω = 0)/(dα : α is a k − 1 form). What we have just shown is that H1(U) = 0 if U is convex, and H1(U) 6= 0 for an annulus. More generally one can show that Hk(Sk) 6= 0, and morally these deRham cohomology groups detect holes in the space U . On the other hand, we might expect that Hk(U) = 0 for convex regions, since they have no holes. This is indeed the case. Theorem 9.11 (Poincaré lemma) If ω is an k-form on a convex region U , then ω = dη for some η iff dω = 0. Div, grad and curl. All the considerations so far have been ‘natural’ in that they do not use a metric on Rn. That is, the integrals transform nicely under arbitrary diffeomorphisms, which can distort length, area and volume. We now explain the relation to the classical theorems regarding div, grad and curl. These depend on the Euclidean metric |x| on Rn. A vector field is just a map v : Rn → Rn. Formally we write ∇ = (d/dx1, . . . , d/dxn). This operator ∇ can be used to define div, grad and curl. 50 The gradient is defined by ∇f = (df/dxi). It characteristic property is that it encodes the directional derivatives of f : 〈∇f, v〉(x) = lim(1/t)(f(x+ tv) − f(x)). In particular, the gradient points in the direction that f is most rapidly increasing, and its rate of increase in that direction is ‖∇f‖. Next, we define the divergence of a vector field by ∇ · v = ∑ dvi/dxi. It measures the flux of v through a small cube. Finally, we define the cross product of two vectors in R3 by a× b = det   e1 e2 e3 a1 a2 a3 b1 b2 b3   = (a2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1), and the curl of a vector field by ∇× v: ∇× v = ( dv3 dx2 − dv2 dx3 , dv1 dx3 − dv3 dx1 , dv2 dx1 − dv1 dx2 ) , The curl also makes sense in R2, but in this case it is a function: ∇× v = det ( d/dx1 d/dx2 v1 v2 ) = dv2 dx1 − dv1 dx2 · Forms and vector fields. To relate differential forms to div grad and curl, using the metric on Rn, we identify vector fields and 1-forms by v = (v1, . . . , vn) ⇐⇒ ω = ∑ vi dxi. It is then clear that v = ∇f ⇐⇒ ω = df. This is because df(v) = 〈∇f, v〉. In other words, ∇f : V → V and df : V → V ∗ correspond under the identification between V and V ∗ coming from the inner product. 51 Vector field v = (vi(x)) ω = ∑ vi(x) dxi Gradient ∇f dω Divergence ∇ · v ∗d ∗ ω Curl ∇× v ∗dω Line integral ∫ M1 v · ŝ ds ∫ M1 ω Surface integral ∫ Mn−1 v · n̂ dA ∫ Mn−1 ∗ω Volume integral ∫ Mn f dV ∫ Mn ∗f Table 3. Dictionary The Hodge star. To address divergence and curl, we need a new operation on differential forms. (See Table 3 for a summary of the correspondence we will obtain.) Namely, we define an operator ∗ : Ak(Rn) → An−k(Rn) on standard basis elements by the requirement that dxα = ±dxβ, where β are the indices from 1 up to n that are not already in α, and dxα (∗dxα) = dx1 · · ·dxn. For example, on R2 we have ∗dx1 = dx2, ∗dx2 = −dx1; and on R3 we have ∗dx1 = dx2 dx3, ∗dx2 = −dx1 dx3, ∗dx3 = dx1 dx2, and ∗dx1 dx2 = dx3, ∗dx1 dx3 = −dx2, ∗dx2 dx3 = dx1. We also have ∗1 = dx1 · · · dxn, and ∗dx1 · · · dxn = 1. To understand what * means we note: Theorem 9.12 If |∇f | = 1 along its level set S in Rn defined by f = 0, then ω = ∗df |S gives the (n − 1)-dimensional volume form on S. In particular, voln−1(S) = ∫ S ∗df . 52 Theorem 9.17 We have ∫ ∂S v · s ds = ∫ S n · (∇× v) |dA|. All these results have a good conceptual explanation: div and curl mea- sure divergence and circulation around tiny loops or boxes, which assemble to give ∂U . The Laplacian. The flow generated by a vector field is volume-preserving on Rn iff ∇ · v = 0. There are a multitude of such vector fields, but many fewer if we require they have no circulation. That is, if add the condition that ∇×v = 0, then v = ∇f (at least locally), and we get Laplace’s equation: ∇ · ∇f = ∆f = ∑ d2fi dx2 i = 0. This equation is of great importance in both mathematics and physics. Its solutions are harmonic functions. They formally minimized ∫ |∇f |2. In electromagnetism, f is the potential of the electric field E = ∇f . It satisfies ∇E = ∆f = ρ where ρ is the charge density. This ∆f = 0 is the equation for the electric potential in a vacuum. To find this potential when the boundary of a region U is held at fixed potentials, one must solve Laplace’s’ equation with given boundary condi- tions. Example: f(r) = 1/rn−2 is harmonic on Rn − {0} and represents the potential of a point charge at the origin. The charge can be calculated in terms of the flux through any sphere. In terms of the Hodge star, we can write ∆f = ∗d ∗ df. 10 Elementary complex analysis Relations of complex analysis to other fields include: algebraic geometry, complex manifolds, several complex variables, Lie groups and homogeneous 55 spaces (C,H, Ĉ), geometry (Platonic solids; hyperbolic geometry in dimen- sions two and three), Teichmüller theory, elliptic curves and algebraic number theory, ζ(s) and prime numbers, dynamics (iterated rational maps). Complex algebra. Let C = R ⊕ Ri where i2 = −1. This set forms a field extension of R. For z = x+ iy ∈ C we define z = x− iy. This is the Galois conjugate of z; it uses the ‘other’ square-root of -1. Then x = Re(z) = (z+z)/2, y = Im(z) = (z−z)/(2i), and x2+y2 = |z|2 = zz. We use the metric d(z, w) = |z − w| to make C into a metric space. From this last expression we see how to invert complex numbers: 1/z = z/|z|2 = (x− iy)/(x2 + y2). Complex multiplication as a matrix. Note that i acts on R2 by J = ( 0 −1 1 0 ) . Similarly T (z) = (x + iy)z is the linear map xI + yJ = ( x −y y x ). We thus obtain an embedding C →M2(R), sending |z| = 1 to SO2(R). Complex differentiation. Calculus with complex functions that are ex- pressed in terms of z and z is fairly simple. First note that dz = x+ iy and dz = x− iy are ordinary complex value differential forms. If we introduce the operators d dz = 1 2 ( d dx − i d dy ) and d dz = 1 2 ( d dx + i d dy ) , then we have, for any smooth function f on a region in C, df = df dz dz + df dz d̄. Complex analytic functions. A region U ⊂ C is an open, connected set. A function f : U → C is analytic if f ′(z) = lim t→0 f(z + t) − f(z) t 56 exists for all z ∈ U . Note that t is allowed to approach 0 is any way whatso- ever. Equivalently we have f(z + t) = f(z) + f ′(z)t+ o(|t|). The following are equivalent: 1. f is analytic: i.e. f ′(z) exists for all z. 2. f is differentiable on U and at each point, Df = ( a b −b a ) for some a, b ∈ R. 3. Df is conformal (angle-preserving or zero), i.e. Df ∈ R · SO(2,R). 4. (Df)J = J(Df), where J = ( 0 −1 1 0 ). 5. f = u + iv and du/dx = dv/dy, du/dy = −dv/dx. (Cauchy-Riemann equations). 6. df/dz = 0. If f(z) is analytic, then f ′(z) = df/dz. Polynomials. To give some examples of analytic functions, let z = x + iy and z = x − iy. There is a natural isomorphism C[x, y] ∼= C[z, z] where z, z are formally treated as independent variables. Under this isomorphism, d/dz and d/dz behave as expected. Thus a polynomial P (x, y) is analytic iff, when expressed in terms of z and z, it only involves z. Theorem 10.1 (Fundamental theorem of algebra) Any polynomial P (z) of degree > 0 has a root in C, and hence can be factored as P (z) = C(z − a1) · · · (z − ad). We will shortly prove this using analysis. Here is a nice application, using logarithmic differentiation. Theorem 10.2 The critical points of a polynomial P (z) are contained in the convex hull of its zeros. Proof. Suppose for example Re ai ≥ 0 for every zero ai of P . Note that Rew < 0 ⇐⇒ Re 1/w < 0. Thus ReP ′/P = Re ∑ 1/(z − ai) < 0 whenever Re z < 0. This shows Re c ≥ 0 for any critical point of P , i.e. c lies in the same halfplane as the zeros ai. Applying the same reasoning to P (eiθz) gives the result above. 57 The series ∑∞ 0 zn = 1/(1−z) is only convergent for |z| < 1, but it admits an analytic extension to C − {1} and then to a rational function. Remarkably, the converse holds: any analytic function is locally a power series, and in particular the existence of one derivative implies the existence of infinitely many. Integrals along paths and boundaries of regions. Suppose f : U → C is continuous. If γ : [0, 1] → C is a path or closed loop, we have ∫ γ f(z) dz = ∫ 1 0 f(γ(t))γ′(t) dt. Note: you can only integrate a 1-form, not a function! More geometrically, if we choose points zi close together along the loop γ, then we have ∫ γ f(z) dz = lim ∑ f(zi)(zi+1 − zi). This definition makes it clear and elementary that ∫ γ dz = 0 for any closed loop γ. Note also that ∣∣∣∣ ∫ ∂U f(z) dz ∣∣∣∣ ≤ (max ∂U |f(z)|) · length(∂U). Cauchy’s Theorem. The most fundamental and remarkable tool in com- plex analysis is Cauchy’s theorem, which allows one to evaluate integrals along loops or more generally boundaries of plane regions. Theorem 10.7 Let U ⊂ C be a bounded region with piecewise-smooth bound- ary, and let f(z) be analytic on a neighborhood of U . Then ∫ ∂U f(z) dz = 0. Provisional proof. Let us assume that f ′(z) is continuous. Then df = (df/dz) dz+(df/dz) dz; hence d(f dz) = 0, and the result follows from Stokes’ theorem. 60 Simple cases: polynomials, power series. Note that if γ is a path from a to b then ∫ γ zn dz = (bn+1 − an+1)/(n+ 1). This proves that ∫ ∂U P (z) dz = 0 for any polynomial P . Taking limits, we find that the same is true for a power series f(z) = ∑ an(z− p)n, so long as γ is contained within its radius of convergence. Cauchy’s integral formula. Because of Cauchy’s theorem, there is only one integral that needs to be explicitly evaluated in complex analysis: ∫ S1 dz z = ∫ S1(a,r) dz z − a = 2πi. From this we obtain: Theorem 10.8 If f is analytic on U and continuous on U , then for all p ∈ U we have: f(p) = 1 2πi ∫ ∂U f(z) dz z − p · Corollary 10.9 More generally, we have f (n)(p) n! = 1 2πi ∫ ∂U f(z) dz (z − p)n+1 · Corollary 10.10 All derivatives of f are analytic. Corollary 10.11 All the derivatives of f exist, i.e. f is a C∞ function on U . Consequences. Theorem 10.12 If f(z) in analytic in B = B(p, R) and continuous on S1(p, R), then ∣∣∣∣ fn(p) n! ∣∣∣∣ ≤ R−n max ∂B |f(z)|. Corollary 10.13 The power series ∑ an(z − p)n, with an = f (n)(p)/n!, has radius of convergence at least R. Corollary 10.14 A bounded entire function is constant. Corollary 10.15 Every nonconstant polynomial p(z) has a zero. Proof. Otherwise 1/p(z) would be a bounded, nonconstant entire function. 61 Theorem 10.16 (Weierstrass-Casorati) If f : C → C is an entire func- tion, then f(C) is dense in C. Proof. If B(p, r) does not meet f(C), then |1/(f(z)− p)| ≤ 1/r on all of C. Corollary 10.17 An entire function with Re f(z) bounded above is constant. Analytic functions from integrals. More generally, for any continuous function g on ∂U , the function F (p) = ∫ ∂U g(z) dz (z − p)m is analytic in U ; indeed, it satisfies F ′(p) = m ∫ ∂U g(z) dz (z − p)m+1 · However it is only when g(z) is analytic that F provides a continuous exten- sion of g. Goursat’s proof. We can now complete Cauchy’s theorem by proving: if f(z) is analytic then f ′(z) is continuous. In fact we will show that if f(z) is analytic in a ball B(c, r) then F (p) =∫ p c f(z) dz is well-defined and satisfies F ′(z) = f(z). Thus F is analytic with a continuous derivative; and hence all its derivatives, including F ′′(z) = f ′(z), are continuous. The key to the definition of F is to show that ∫ ∂R f(z) dz = 0 for any rectangle R. Suppose not. We may assume R has largest side of length 1 and ∫ ∂R f(z) dz = 1. Subdividing, we can obtain a sequence of similar rectangles R = R0 ⊃ R1 ⊃ R2 . . . such that the largest side length of Rn is 2−n and | ∫ ∂Rn f(z) dz| ≥ 4−n. Suppose ⋂ Rn = {p}. Then on Rn we have f(z) = f(p) + f ′(p)(z − p) + en(z), where |en| ≤ ǫn|z − p| and ǫn → 0. Now the first two terms have a zero integral over Rn, while the last term is, on Rn, bounded by 2−nǫn. Moreover the length of ∂Rn is at most 4 · 2−n. Thus ∣∣∣∣ ∫ Rn f(z) dz ∣∣∣∣ ≤ 4ǫn4−n < 4−n when n is large enough — a contradiction. 62 Corollary 10.23 If f is constant along an arc, then f is constant. Theorem 10.24 Suppose fn(z) are analytic functions on U and fn → f uniformly on compact sets. Then f(z) is also analytic. Proof. For any p ∈ U we have f(p) = lim fn(p) = lim 1 2πi ∫ ∂B(p,r) fn(z) z − p dz = 1 2πi ∫ ∂U f(z) z − p dz, by uniform convergence on B(p, r). As remarked earlier, this formula give a holomorphic function on B(p, r) no matter what the continuous function f(z) on ∂B(p, r) is. Theorem 10.25 Any bounded sequence of analytic functions fn ∈ C(U) has a subsequence converging uniformly on compact sets to an analytic function g. Proof. If K ⊂ U and d(K, ∂U) = r, and |f | ≤ M , then for any p ∈ K we find: |f ′(p)| = 1 2π ∣∣∣∣ ∫ S1(p,r) f(z) (z − p)2 dz ∣∣∣∣ ≤M/r. Thus fn|K is equicontinuous and we can apply Arzela-Ascoli. Note: a bounded function need not have a bounded derivative! Consider f(z) = ∑ zn/n2 on the unit disk. Theorem 10.26 If a sequence of analytic functions fn converges to f (lo- cally) uniformly, then for each k, f (k) n (z) → fk(z) (locally) uniformly. What happens with the usual suspects? All of these theorems have counterexamples for functions of a real variable. For example, fn(x) = 1/(1+ (nx)2) is bounded on the whole real axis, but it has no uniformly convergence subsequence on [−1, 1]. What happens if we consider fn(z), z ∈ C? In this case fn(z) has a pole at z = i/n. Thus it is not bounded in C, and not even bounded on a uniform neighborhood of [−1, 1]. This explosion of |f(z)| is necessary to get a counterexample on [−1, 1]. Antiderivatives. A region U ⊂ C is simply-connected if every closed loop in U bounds a disk in U . Equivalently, Ĉ−U is connected (U has no holes). 65 Theorem 10.27 If f(z) is analytic on a simply-connect region U , then there exists an analytic function F : U → C such that F ′(z) = f(z). Corollary 10.28 If γ is any loop in a simply-connected region on which f(z) is analytic, then ∫ γ f(z) dz = 0. Inverse functions. Observe that if Df is conformal and invertible, then (Df)−1 is also conformal. Thus the inverse function theorem already proved for maps on Rn shows: Theorem 10.29 If f is analytic and f(a) = b, then there is an analytic inverse function g defined near b such that g(b) = a and g′(b) = 1/f ′(a). Log and roots. As first examples, we note that log(z) can be defined near z = exp(0) = 1 as an analytic function such that exp(log z) = z and log 1 = 0. Similarly z1/n can be defined near z = 1, as the inverse of zn. More generally zα = exp(α log z) can be defined near z = 1. To examine log more closely, let us try to define the function F (p) = ∫ p 1 dz z on C∗. The problem is that the integral over a loop enclosing z = 0 is 2πi. Nevertheless, on any region U containing z = 1 where F can be defined, we have F ′(z) = 1/z and F (1) = 0. Thus on U we have ( eF (z) z )′ = ( eF (z)(zF ′(z) − 1) z2 )′ = 0 and thus eF (z) = z, i.e. F (z) = log z. A common convention is to take U = C− (∞, 0]. Then F maps U to the strip | Im z| ≤ π. Explicitly, we have log(z) = log |z| + i arg(z) where the argument is chosen in (−π, π). By integration of (1 + z)−1 = ∑ (−1)nzn we obtain the power series log(1 + z) = ∫ dz/(1 + z) = z − z2/2 + z3/3 − · · · , 66 valid for |z| < 1. Once log z has been constructed we can then define zα on U as well. By differentiation we then obtain the power series (1 + z)α = 1 + αz + α(α− 1) 2! z2 + α(α− 1)(α− 2) 3! z3 + · · · , also valid for |z| < 1. In both cases we have a branch-type singularity at z = 0 (not a pole! and not an isolated singularity). 11 Analytic and harmonic functions We begin by observing Cauchy’s theorem implies: Theorem 11.1 (The mean-value formula) If f is analytic on B(p, r), then f(p) is the average of f(z) over S1(p, r). Proof. By Cauchy’s integral formula, we have: f(p) = 1 2πi ∫ 2π 0 f(p+ reiθ) reiθ d(reiθ) = 1 2π ∫ 2π 0 f(p+ reiθ) dθ. Corollary 11.2 (The Maximum Principle) A nonconstant analytic func- tion does not achieve its maximum. For example, if f is analytic on U and continuous on U , and U is compact, then |f(z)| ≤ max ∂U |f(z)|. Proof. If f(z) achieves its maximum at p ∈ U , then f(p) is the average of f(z) over a small circle S1(p, r). Moreover, |f(z)| ≤ |f(p)| on this circle. The only way the average can agree is if f(z) = f(p) on S1(p, r). (Indeed, the average of g(z) = f(z)/f(p) is 1 and |g| ≤ 1 so g(z) = 1 on S1(p, r).) But then f is constant on an arc, so it is constant in U . 67 Corollary 11.9 Any C2 harmonic function is actually infinitely differen- tiable. Corollary 11.10 A harmonic function satisfies the mean-value theorem: u(p) is the average of u(z) over S1(z, p). Corollary 11.11 A harmonic function satisfies the maximum principle. Corollary 11.12 If u is harmonic and f is analytic, then u ◦ f is also harmonic. Theorem 11.13 A uniform limit of harmonic functions is harmonic. Proof. Use Corollary 11.5 and the fact that a uniform limit of analytic functions is analytic. Theorem 11.14 There is a unique linear map P : C(S1) → C(∆) such that u = P (u)|S1 and u is harmonic on ∆. Proof. Uniqueness is immediate from the maximum principle. To see ex- istence, observe that we must have P (zn) = zn and P (zn) = zn. Thus P is well-defined on the span S of polynomials in z and z, and satisfies there ‖P (u)‖∞ = ‖u‖∞. Thus P extends continuously to all of C(S1). Since the uniform limit of harmonic functions is harmonic, P (u) is harmonic for all u ∈ C(S1). Poisson kernel. The map φ can be given explicit by the Poisson kernel. For example, u(0) is just the average of u over S1. We can also say u(p) is the expected value of u(z) under a random walk starting at p that exits the disk at z. Relation to Fourier series. The above argument suggests that, to define the harmonic extension of u, we should just write u(z) = ∑∞ −∞ anz n on S1, and then replace z−n by zn to get its extension to the disk. This actually works, and gives another approach to the Poisson kernel. Laplacian as a quadratic form, and physics. Suppose u, v ∈ C∞ c (C) – so u and v are smooth, real-valued functions vanishing outside a compact set. Then, by integration by parts, we have ∫ ∆ 〈∇u,∇v〉 = − ∫ ∆ 〈u,∆v〉 = − ∫ ∆ 〈v,∆u〉. 70 To see this using differential forms, note that: 0 = ∫ ∆ d(u ∗ dv) = ∫ ∆ (du)(∗dv) + ∫ ∆ u(d ∗ dv). In particular, we have ∫ ∆ |∇u|2 = − ∫ ∆ u∆u. Compare this to the fact that 〈Tx, Tx〉 = 〈x, T ∗Tx〉 on any inner product space. Thus −∆ defines a positive-definite quadratic form on the space of smooth functions. The extension of u from S1 to ∆ is a ‘minimal surface’ in the sense that it minimizes ∫ ∆ |∇u|2 over all possible extensions. Similarly, minimizing the energy in an electric field then leads to the condition ∆u = 0 for the electrical potential. -2 -1 0 1 2 0.0 0.5 1.0 1.5 2.0 Figure 4. Streamlines around a cylinder. Probabilistic interpretation. Brownian motion is a way of constructing random paths in the plane (or Rn). It leads to the following simply inter- pretation of the extension operator P . Namely, given p ∈ ∆, one considers a random path pt with p0 = p. With probability one, there is a first T > 0 such that |pT | = 1; and then one sets u(p) = E(u(pT )). In other words, u(p) is the expected value of u(pT ) at the moment the Brownian path exits the disk. Using the Markov property of Brownian motion, it is easy to see that u(p) satisfies the mean-value principle, which is equivalent to it being harmonic. 71 It is also easy to argue that |p0 − pT | tends to be small when p0 is close to S1, and hence u(p) is a continuous extension of u|S1. Example: fluid flow around a cylinder. We begin by noticing that f(z) = z + 1/z gives a conformal map from the region U ⊂ H where |z| > 1 to H itself, sending the circular arc to [−2, 2]. Thus the level sets of Im f = y(1 − 1/(x2 + y2)) describe fluid flow around a cylinder. Note that we are modeling incompressible fluid flow with no rotation, i.e. we are assuming the curl of the flow is zero. This insures the flow is given by the gradient of a function. Open mapping theorem. Next we explore the local form of an analytic function in more detail, to get a more informative understanding of the max- imum principle. We remark that by the implicit function theorem we have: Theorem 11.15 If f ′(z) 6= 0 then f is a local homeomorphism at z, and its inverse is also analytic. An integral formula for f−1 will be developed later, as a consequence of Rouché’s theorem. For the moment we note that the power series for f−1(z) is easily computed recursively from the power series for f(z). For example, if f(z) = z + cos(z) − 1 = z − z2/2 + z4/24 + · · · then f ′(0) = 1 and f−1(z) = ∑ anz n = z + a2z 2 + a3z 3 + · · · where z = f (∑ anz n ) = z + a2z 2 + a3z 3 − (z + a2z 2)2/2 + z4 +O(z4) = z + (a2 − 1/2)z2 + (a3 − a2)z 3 + · · · , which gives a2 = a3 = 1/2. Theorem 11.16 If f(z) has an isolated zero at z = a, then there is an analytic function g(z) defined near 0 such that g′(0) 6= 0 and f(z) = g(z)n near z = a. Proof. We can assume a = 0. Write f(z) = Aznh(z) with h(0) = 1, and set g(z) = zA1/nh(z)1/n. This makes sense when z is small, since z1/n = exp((log z)/n) is analytic near h(0) = 1. 72 Corollary 12.4 The values of an analytic function are dense in any neigh- borhood of an essential singularity. Proof. If B(p, r) is omitted from f(∆∗), then g(z) = 1/(f(z)−p) is bounded near 0 and hence analytic. Thus f(z) = p+1/g(z) has at worst a pole (which arises if g(0) = 0). Picard’s theorems. Picard’s little and big theorems show more: an entire function can have only one omitted value (and this is sharp for ez), and an analytic function omits at most one value in any neighborhood of an essential singularity. Corollary 12.5 Any analytic function on ∆∗ has the form f(z) = ∑∞ −∞ anz n. The number of negative terms is infinite iff f(z) has an essential singularity at z = 0. Example. On ∆∗ the function e1/z = ∑∞ 0 z−n/n! has an essential singularity at z = 0. So does sin(1/z), etc. Polynomials and rational functions. Theorem 12.6 If f(z) is an entire function and |f(z)| ≤M |z|n, then f(z) is a polynomial of degree at most n. Proof. By Cauchy’s bound, f (n)(z) is bounded and hence constant. Theorem 12.7 If f(z) is entire and continuous at ∞, then f(z) is a poly- nomial. Proof. In this case f(1/z) has at worst a pole at the origin, and so it satisfies a bound as above. Let us say a function f : Ĉ → Ĉ is holomorphic if whenever w = f(z), if w, z ∈ C then f(z) is analytic; if w = ∞, z ∈ C then 1/f(z) is analytic; if w ∈ C and z = ∞ then f(1/z) is analytic; and if w = z = p then 1/f(1/z) is analytic. (This is an example of a map between Riemann surfaces.) Corollary 12.8 Any holomorphic function f : Ĉ → Ĉ is a rational function. Proof. By compactness and isolation, 1/f(z) has only a finite number of zeros in Ĉ, and hence only finitely many in C. Thus we can find a polynomial such that Q(z) = f(z)P (z) is analytic and continuous at infinity. Then Q(z) is also a polynomial. 75 13 Residues: theory and applications Meromorphic functions. A function f : U → Ĉ is meromorphic if it is analytic apart from a discrete set of poles. For example, 1/ sin(z) is mero- morphic. In this section we analyze integrals of meromorphic functions over closed loops. The residue. Let f(z) be a holomorphic function with an isolated singu- larity at p. The residue of f at p, Res(f, p), can be defined in two equivalent ways: 1. Expand f in a Laurent series ∑ an(z − p)n; then a−1, the coefficient of 1/(z − p), is the residue of f at p. 2. Define Res(f, p) = ∫ γ f(z) dz for any small loop around p. The second definition reveals that it is really the 1-form f(z) dz, not the function f(z), which has a residue. For example, we do not have Res(f(g(z)), p) = Res(f(z), g(p))! Instead — if g′(p) 6= 0 — we have Res(f(g(z)), p) = g′(p) Res(f, g(p)). Examples: simple poles. If f(z) has a simple pole at p, then Res(f, p) = lim z→p f(z)(z − p). Put differently: if f(z) has a simple zero at p, then Res(1/f(z), p) = 1/f ′(p). For example, Res(1/(z2 + 1), i) = 1/(2i) = −i/2. If f(z) has a simple pole at p and g(z) is analytic at p, then Res(fg, p) = g(p) Res(f, p). If f(z) vanishes to order exactly k at p, and g(z) to order (k + 1), then f/g has a simple pole and we have Res(f/g, p) = (k + 1) f (k)(p) g(k+1)(p) · This is immediate by writing, e.g. when p = 0, f(z) = zk(ak + O(z)) and g(z) = zk+1(bk+1 +O(z)). For example, Res(z2/(sin z − z), 0) = −6. 76 Logarithmic derivatives. If f(z) has a zero of order k at p then f ′/f has a simple pole at p and we have Res(f ′/f, p) = k. Similarly if f has a pole of order k, then Res(f ′/f, p) = −k. Higher order poles. It is trickier to find the residue when the pole is not simple. For example, Res(z3 cos(1/z), 0) = 1/24. Integrals. The importance of residues comes from: Theorem 13.1 (The residue theorem) Let U be a compact, smoothly bounded region, suppose P ⊂ U is a finite set, and suppose f : U − P → C is contin- uous and analytic on U . Then we have 1 2πi ∫ ∂U f(z) dz = ∑ P Res(f, p). Proof. Immediate by removing from U a small disk around each point P , and then integrating along the boundary of the resulting region. Note: if f(z) is analytic, then Res(f(z)/(z− p), p) = f(p), so the residue theorem contains Cauchy’s formula for f(p). Corollary 13.2 (The argument principle) Suppose f |∂U has no zeros. Then the number of zeros of f(z) inside U is given by N(f, 0) = 1 2π ∫ ∂U f ′(z) dz f(z) · This integral is the same as (1/2πi) ∫ ∂U d log f . It just measure the number of times that f wraps the boundary around zero. The topological nature of the argument principle: if a continuous f : ∆ → C has nonzero winding number on the circle, then f has a zero in the disk. Letting N(f, a) denote the number of solutions to f(z) = a, we have another proof of: Corollary 13.3 (Open mapping theorem) The function N(f, a) is con- stant on each component of C− f(∂U). (It simply gives the number of times that f(∂U) winds around a.) 77 Definite integrals 1: rational functions on R. Whenever a rational function R(x) = P (x)/Q(x) has ∫ R |R(x)| dx is finite, we can compute this integral via residues: we have ∫ ∞ −∞ R(x) dx = 2πi ∑ Im p>0 Res(R, p). (Of course we can also compute this integral by factoring Q(x) and using partial fractions and trig subsitutions.) Example. Where does π come from? It emerges naturally from rational functions by integration — i.e. it is a period. Namely, we have ∫ ∞ −∞ dx 1 + x2 = 2πiRes(1/(1 + z2), i) = 2πi(−i/2) = π. Of course this can also be done using the fact that ∫ dx/(1+x2) = tan−1(x). More magically, for f(z) = 1/(1 + z4) we find: ∫ ∞ −∞ dx 1 + x4 = 2πi(Res(f, (1 + i)/ √ 2) + Res(f, (1 + i)/ √ 2) = π√ 2 · Both are obtain by closing a large interval [−R,R] with a circular arc in the upper halfplane, and then taking the limit as R → ∞. We can even compute the general case, f(z) = 1/(1 + zn), with n even. For this let ζk = exp(2πi/k), so f(ζ2n) = 0. Let P be the union of the paths [0,∞)ζn and [0,∞), oriented so P moves positively on the real axis. We can then integrate over the boundary of this pie-slice to obtain: (1 − ζn) ∫ ∞ 0 dx 1 + xn = ∫ P f(z) dz = 2πiRes(f, ζ2n) = 2πi/(nζn−1 2n ), which gives ∫ ∞ 0 dx 1 + xn = 2πi n(−ζ−1 2n + ζ+1 2n ) = π/n sin π/n · Here we have used the fact that ζn 2n = −1. Note that the integral tends to 1 as n→ ∞, since 1/(1 + xn) converges to the indicator function of [0, 1]. Definite integrals 2: rational functions of sin(θ) and cos(θ). Here is an even more straightforward application of the residue theorem: for any rational function R(x, y), we can evaluate ∫ 2π 0 R(sin θ, cos θ) dθ. 80 The method is simple: set z = eiθ and convert this to an integral of an analytic function over the unit circle. To do this we simple observe that cos θ = (z + 1/z)/2, sin θ = (z − 1/z)/(2i), and dz = iz dθ. Thus we have: ∫ 2π 0 R(sin θ, cos θ) dθ = ∫ S1 R ( 1 2i ( z − 1 z ) , 1 2 ( z + 1 z )) dz iz · For example, for 0 < a < 1 we have: ∫ 2π 0 dθ 1 + a2 − 2a cos θ = ∫ S1 i dz (z − a)(az − 1) = 2πi(i/(a2 − 1)) = 2π 1 − a2 · Definite integrals 3: fractional powers of x. ∫∞ 0 xaR(x)dx, 0 < a < 1, R a rational function. For example, consider I(a) = ∫ ∞ 0 xa 1 + x2 dx. Let f(z) = za/(1 + z2). We integrate out along [0,∞) then around a large circle and then back along [0,∞). The last part gets shifted by analytic continuation of xa and we find (1 − 1a)I(a) = 2πi(Res(f, i) + Res(f,−i)) and Res(f, i) = ia/(2i), Res(f,−i) = (−i)a/(−2i). Thus, if we let ia = ω = exp(πia/2), we have I(a) = π(ia − (−i)a) (1 − 1a) = π ω − ω3 1 − ω4 = π ω + ω−1 = π 2 cos(πa/2) · For example, when a = 1/3 we get I(a) = π/(2 cos(π/6)) = π/ √ 3. Residues and infinite sums. The periodic function f(z) = π cot(πz) has the following convenient properties: (i) It has residues 1 at all the integers; and (ii) it remains bounded as Im z → ∞. From these facts we can deduce some remarkable properties: by integrating over a large rectangle S(R), we find for k ≥ 2 even, 0 = lim R→∞ 1 2πi ∫ S(R) f(z) dz zk = Res(f(z)/zk, 0) + 2 ∞∑ 1 1/nk. 81 Thus we can evaluate the sum ∑ 1/n2 using the Laurent series cot(z) = cos(z) sin(z) = 1 − z2/2! + z4/4! − · · · z(1 − z2/3! + z4/5! − · · · ) = z−1(1 − z2/2! + z4/4! − · · · )(1 + z2/6 + 7z4/360 + · · · ) = z−1 − z/3 − z3/45 − · · · (using the fact that (1/(3!)2 − 1/5! = 7/360). This shows Res(f(z)/z2, 0) = −π2/3 and hence ∑ 1/n2 = π2/6. Similarly, 2ζ(2k) = −Res(f(z)/z2k, 0). For example, this justifies ζ(0) = 1 + 1 + 1 + · · · = −1/2. Little is known about ζ(2k + 1). Apéry showed that Z(3) is irrational, but it is believed to be transcendental. We note that ζ(s) = ∑ 1/ns is analytic for Re s > 1 and extends an- alytically to C − {1} (with a simple pole at s = 1). In particular ζ(0) is well-defined. Because of the factorization ζ(z) = ∏ (1 − 1/ps)−1, the be- havior of the zeta function is closely related to the distribution of prime numbers. The famous Riemann hypothesis states that any zero of ζ(s) with 0 < Re s < 1 satisfies Re s = 1/2. It implies a sharp form of the prime number theorem, π(x) = x/ log x+O(x1/2+ǫ). The zeta function also has trivial zeros at s = −2,−4,−6, . . .. Hardy’s paper on ∫ sin(x)/x dx. We claim I = ∫ ∞ 0 sin x dx x = π 2 · Note that this integral is improper, i.e. it does not converge absolutely. Also, the function f(z) = sin(z)/z has no poles — so how can we apply the residue calculus? The trick is to observe that −2iI = lim r→0 ∫ r<|x|<1/r eix dx x · We now use the fact that |eix+iy| ≤ e−y to close the path in the upper halfplane, and conclude that 2iI = lim r→0 ∫ S1(r)+ eiz dz z · Since Res(eiz/z, 0) = 1, we find 2iI = (2πi)(1/2) and hence I = π/2. 82 3. Or, regarding Ĉ as PC2, define the length of w at v to be ‖w‖ = 2|v∧w|/|v|2. Then if we map C into C2 by z 7→ (1, z), we get v = (1, z), w = (0, dz), and ‖w‖ = 2|dz|/(1 + |z|2). In the last version we have used the Hermitian structure on C2. Note that orthogonal vectors in C2 determine antipodal points in Ĉ. Topology. Some topology of projective spaces: RP2 is the union of a disk and a Möbius band; the Hopf map S3 → S2 is part of the natural projection C2 − {0} → Ĉ. Gauss-Bonnet for spherical triangles: area equals angle defect. Prove by looking at the three lunes (of area 4θ) for the three angles of a triangle. General form: 2πχ(X) = ∫ X K + ∫ ∂X k. Isometries. The isometry group of (Ĉ, ρ) is PU(2) ⊂ PSL2(C). Every finite subgroup of Aut(Ĉ) is conjugate into PU(2). Thus the finite subgroups are Z/n, D2n, A4, S4 and A5. 3. The unit disk. We first remark that ∆ ∼= H, e.g. by the Möbius transformation I(z) = i(1 − z)/(1 + z). Thus Aut(∆) ∼= Aut(H). Theorem 14.5 Every automorphism of ∆ or H extends to an automorphism of Ĉ. Proof. Let G(H) = Aut(H) ∩ Aut(Ĉ) and similarly for G(∆). Note that I ∈ Aut(Ĉ) so G(∆) ∼= G(H) under I. Now G(H) obvious contains the transformations of the form g(z) = az + b with a > 0, b ∈ R, which act transitively. So G(∆) also acts transitively. Thus if f ∈ Aut(∆) then f(0) = 0 after composition with an element of G(∆). But then f(z) = eiθz by the Schwarz Lemma, so f ∈ G(∆). Corollary 14.6 The automorphism group of ∆ is PU(1, 1), the group of Möbius transformations of the form g(z) = az + b bz + a with |a|2 − |b|2 = 1. Corollary 14.7 The automorphism group of H is given by PSL2(R). 85 These automorphisms preserve the hyperbolic metrics on ∆ and H, given by ρ∆ = 2|dz| 1 − |z|2 and ρH = |dz| Im z · Hyperbolic geometry. Geodesics are circles perpendicular to the circle at infinity. Euclid’s fifth postulate (given a line and a point not on the line, there is a unique parallel through the point. Here two lines are parallel if they are disjoint.) Gauss-Bonnet in hyperbolic geometry. (a) Area of an ideal triangle is∫ 1 −1 ∫∞√ 1−x2(1/y 2)dydx = π. (b) Area A(θ) of a triangle with two ideal vertices and one external angle θ is additive (A(α+ β) = A(α) +A(β)) as a diagram shows. Thus A(α) = α. (c) Finally one can extend the edges of a general triangle T in a spiral fashion to obtain an ideal triangle containing T and 3 other triangles, each with 2 ideal vertices. Classification of automorphisms of H2, according to translation distance. The Schwarz lemma revisited. Theorem 14.8 Any holomorphic map f : H → H is a weak contraction for the hyperbolic metric. If |Df | = 1 at one point, then f is an isometry. Dynamical application of Schwarz Lemma. Theorem 14.9 Let f : Ĉ → Ĉ be a rational map. Then the immediate basin of any attracting cycle contains a critical point. Corollary 14.10 The map f has at most 2d− 2 attracting cycles. Conformal mapping. Examples: 1. Möbius transformations. Lunes and tangent-lunes = pie-slices and strips. 2. All pie-slices are isomorphic to H using zα. 3. All strips are isomorphic to H using exp. 4. A half-strip is isomorphic to a lune using exp, and hence to a half-plane. 5. Putting it all together: sin(z) maps the region above [−π/2, π/2] to H. 86 The Riemann mapping theorem. Theorem 14.11 For any simply-connected region U ⊂ C, U 6= C, and any basepoint u ∈ U , there is a unique conformal homeomorphism f : (U, u) → (∆, 0) such that f ′(u) > 0. Proof. let F be the family of univalent maps (U, u) → (∆, 0). Using a square-root and an inversion, show F is nonempty. Also F is closed under limits. By the Schwarz Lemma, |f ′(u)| has a finite maximum over all f ∈ F . Let f be a maximizing function. If f is not surjective to the disk, then we can apply a suitable composition of a square-root and two automorphisms of the disk to get a g ∈ F with |g′(u)| > |f ′(u)|, again using the Schwarz Lemma. Uniformization of annuli. Theorem 14.12 Any doubly-connected region in the sphere is conformal iso- morphic to C∗, ∆∗ or A(R) = {z : 1 < |z| < R}. The map from H to A(R) is z 7→ zα, where α = log(R)/(πi). The deck transformation is given by z 7→ λz, where λ = 4π2/ log(R). Reflection. The Schwarz reflection principle: if U = U∗, and f is analytic on U ∩ H, continuous and real on the boundary, then f(z) extends f to all of U . This is easy from Morera’s theorem. A better version only requires that Im(f) → 0 at the real axis, and can be formulated in terms of harmonic functions (cf. Ahlfors): If v is harmonic on U∩H and vanishes on the real axis, then v(z) = −v(z) extends v to a harmonic function on U . For the proof, use the Poisson integral to replace v with a harmonic function on any disk centered on the real axis; the result coincides with v on the boundary of the disk and on the diameter (where it vanishes by symmetry), so by the maximum principle it is v. Reflection gives another proof that all automorphisms of the disk extend to the sphere. Univalent functions. The class S of univalent maps f : ∆ → C such that f(0) = 0 and f ′(0) = 1. Compactness of S. The Bieberbach Conjecture/de Brange Theorem: f(z) = ∑ anz n with |an| ≤ n. The area theorem: if f(z) = z + ∑ bn/zn is univalent on {z : |z| > 1}, then ∑ n|bn|2 < 1. The proof is by integrating fdf over the unit circle and 87