Download Midterm Exam Questions Solutions - Introduction to Graduate Analysis | Math 205 and more Exams Mathematics in PDF only on Docsity! Math 205 Spring Term 2006 Midterm Examination - Solutions 1. Let M be a complete metric space and f : M → M a contraction. Denoting by x0 the fixed point of f , prove that d(x, x0) ≤ 1 1− r d(x, f(x)) for any x ∈ M where r ∈ (0, 1) is the Lipschitz constant of f . Solution: Since f(x0) = x0 we see that d(f(x), x0) = d(f(x), f(x0)) ≤ r d(x, x0) and, therefore, by the triangular inequality d(x, x0) ≤ d(x, f(x)) + d(f(x), x0) ≤ d(f(x), x) + rd(x, x0) which readily implies the claim. 2. Let f, g ∈ C1(Rn, R) be positive functions. Show that fg ∈ C1(Rn, R) and compute D(fg). Show that, if fg attains a minimum at x, then ∇f(x) and ∇g(x) are linearly dependent. Solution: By assumption we have that both f and g possess continuous partial derivatives in all direction. Then ∂j(fg)(x) = f(x)∂jg(x) + g(x)∂jf(x) , j = 1, . . . , n and thus all partial derivatives of fg exist and are continuous. We conclude that fg is differentiable and, by the above formula for its partial derivatives, we infer that ∇(fg)(x) = f(x)∇g(x) + g(x)∇f(x) . At a point of minimum we would have that 0 = ∇(fg)(x) = f(x)∇g(x) + g(x)∇f(x) which would make the gradients linearly dependent since f and g never vanish. 1 3. Let f ∈ C1(Rn, R) and assume that x ∈ L := f−1(5) := {y ∈ Rn | f(y) = 5} . If γ ∈ C1 ( (0, 1), L ) is a curve through x, show that∇f(x) is orthogonal to the curve γ at x. Solution: Let γ be a curve through x with the above properties and say that γ(0.5) = x. Then f ( γ(t) ) = 5 ∀ t ∈ (0, 1) and therefore, by taking one derivative, ∇f ( γ(t) ) · γ̇(t) = 0 ∀ t ∈ (0, 1) . In particular we have that ∇f(x) · γ̇(0.5) = 0 which gives the desired result since γ̇(0.5) is clearly tangent to the curve at x. 4. Let f ∈ C2(Rn, R) with D2f(x) > 0 for some x ∈ Rn. Show that, in a neighborhood of x, the graph Gf := {(x, f(x)) |x ∈ Rn} of f lies above its tangent plane at x. Solution: Since D2f(x) is positive definite and D2f is continuous, we can find δ > 0 such that D2f(y) is still positive if |y − x|2 ≤ δ. Then, by Taylor expansion with remainder, we obtain that f(y) = f(x) + Df(x)(y − x)︸ ︷︷ ︸ equation for tangent plane + 1 2 (y − x)T D2f(z)(y − x) for any y ∈ B(x, δ) and some z on the segment between x and y. Thus the claim follows since 1 2 (y − x)T D2f(z)(y − x) > 0 regardless of y, z ∈ B(x, δ). Alternatively, consider the function g : Rn → R , y 7→ f(y)− f(x)−Df(x)(y − x) . 2
