Final Exam Solutions - Analysis - Spring 2008 | MATH 131A, Exams of Mathematical Methods for Numerical Analysis and Optimization

Download Final Exam Solutions - Analysis - Spring 2008 | MATH 131A and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! Math 131a Final Exam Lecture 2 Spring 2008 Name: Instructions: • There are 7 problems. Make sure you are not missing any pages. • Give complete, convincing, and clear answers (or points will be deducted). • No calculators, books, or notes are allowed. • Answer the questions in the spaces provided on the question sheets. If you run out of room for an answer, continue on the back of the page. Question Points Score 1 10 2 10 3 10 4 10 5 10 6 10 7 10 Total: 70 1. (10 points) Let X, Y be nonempty sets and let f be a bounded, nonnegative, real-valued function defined on X × Y. For each x ∈ X define G(x) = sup{f(x, y) : y ∈ Y }. Prove that sup{G(x) : x ∈ X} = sup{f(x, y) : (x, y) ∈ X × Y }. Solution: First, we prove that sup{G(x) : x ∈ X} ≤ sup{f(x, y) : (x, y) ∈ X × Y }. By definition of least upper bound, it suffices to show that sup{f(x, y) : (x, y) ∈ X × Y } is an upper bound for {G(x) : x ∈ X}. Suppose x0 ∈ X. Since (x0, y) ∈ X × Y for every y ∈ Y, we have that sup{f(x, y) : (x, y) ∈ X × Y } is an upper bound for {f(x0, y) : y ∈ Y } and so G(x0) = sup{f(x0, y) : y ∈ Y } ≤ sup{f(x, y) : (x, y) ∈ X × Y }. Now, we prove that sup{G(x) : x ∈ X} ≥ sup{f(x, y) : (x, y) ∈ X × Y }. Again, it suffice to show that sup{G(x) : x ∈ X} is an upper bound for {f(x, y) : (x, y) ∈ X×Y }. Suppose (x0, y0) ∈ X×Y. Then f(x0, y0) ≤ sup{f(x0, y) : y ∈ Y } = G(x0) ≤ sup{G(x) : x ∈ X}. 4. (10 points) Suppose E is a subset of R and f, g : E → R are continuous at p ∈ E. Define h : E → R by h(x) = max{f(x), g(x)}. Prove that h is continuous at p. Solution: Let  > 0. We need to find δ > 0 such that |h(x) − h(p)| <  for every x ∈ E satisfing |x− p| < δ. There are three possible cases: f(p) > g(p), f(p) < g(p), f(p) = g(p). We will first consider the case when f(p) > g(p). Let 1 = min(, (f(p) − g(p))/2). Since f is continuous at p, there exists a δ1 > 0 such that |f(x) − f(p)| < 1 for every x ∈ E satisfying |x − p| < δ1. Since g is continuous at p, there exists a δ2 > 0 such that |g(x) − g(p)| < 1 for every x ∈ E satisfying |x − p| < δ2. Let δ = min(δ1, δ2) and suppose x ∈ E with |x − p| < δ. Since |x − p| < δ ≤ δ1, we have f(x) > f(p) −  > f(p)−(f(p)−g(p))/2) = (f(p)+g(p))/2. Since |x−p| < δ ≤ δ2, we have g(x) < g(p)+ < g(p) + (f(p) − g(p))/2) = (f(p) + g(p))/2. Thus, f(x) > g(x) and so |h(x) − h(p)| = |min(f(x), g(x)) − min(f(p), g(p))| = |g(x) − g(p)| < 1 < . For the second to last inequality, we again use the fact that |x− p| < δ2. The case when f(p) < g(p) follow by an argument analogous to that for the case f(p) > g(p). Finally suppose f(p) = g(p). Find δ > 0, as above, so that if x ∈ E and |x − p| < δ we have |f(x) − f(p)| <  and |g(x) − g(p)| < . Let x ∈ E with |x − p| < δ. There are two possible cases: f(x) ≥ g(x) and f(x) < g(x). In the former case, we have |h(x) − h(p)| = |min(f(x), g(x)) − min(f(p), g(p))| = |g(x) − g(p)| < . In the latter case, we have |h(x)− h(p)| = |min(f(x), g(x))−min(f(p), g(p))| = |f(x)− f(p)| < . 5. (10 points) Suppose that E is a bounded subset of R and that f : E → R is uniformly continuous on E. Prove that f is bounded on E. You are allowed to use theorems from the book (such as the Heine-Borel theorem). Solution: Since f is uniformly continuous on E, we may find a δ > 0 such that |f(x)− f(x′)| < 1 whenever x, x′ ∈ E and |x− x′| < δ. Clearly {Nδ(x)}x∈E is an open cover of E. We claim that it also covers the closure of E, E. Recall that E is the union of E with its limit points. Suppose y is a limit point of E. Then there is an x ∈ E with x ∈ Nδ(y), so y ∈ Nδ(x). We know that E is bounded, say by M . By an argument similar to that in the preceding paragraph, we have that E is bounded by M + δ. Since E is a closed bounded set of real numbers, it is compact (by the Heine-Borel theorem). Thus, E and hence E is covered by a finite subcollection {Nδ(x1), . . . , Nδ(xn)} of {Nδ(x)}x∈E. Let B = 1 + max(|f(x1)|, . . . , |f(xn)|). We claim that f is bounded by B. Indeed, suppose x ∈ E. Then x ∈ Nδ(xi) for some i. Since |x − xi| < δ, we have |f(x)| = |f(x)− f(xi) + f(xi)| ≤ |f(x)− f(xi)|+ |f(xi)| < 1 + |f(xi)| ≤ B. 6. (10 points) Suppose f : (−1, 1) → R is bounded. Define g : (−1, 1) → R by g(x) = x3f(x). Prove that g is differentiable at 0. What is g′(0)? Solution: We claim that g is differentiable at 0 and that g′(0) = 0. In otherwords, that limx→0 g(x)−g(0) x−0 = 0. Let  > 0. We need to find δ > 0 so that |g(x)−g(0) x−0 −0| <  whenever 0 < |x−0| < δ. Since f is bounded on (−1, 1), we may find M > 0 such that |f(x)| < M for every x ∈ (−1, 1). Let δ = min(1, /M) and suppose that 0 < |x− 0| < δ. Then |g(x)−g(0) x−0 − 0| = | x3f(x) x | = |x||x||f(x)| < 1  M M = .