Bonus Homework. Math 766 Spring ) For E 1,E 2 R n, define E 1 + E 2 = {x + y : x E 1,y E 2 }.

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1 Bonus Homework Math 766 Spring ) For E,E R n, define E + E = {x + y : x E,y E }. (a) Prove that if E and E are compact, then E + E is compact. Proof: Since E + E R n, it is sufficient to prove that E + E is closed and bounded. E + E is bounded: Since E and E are compact, they are bounded. So there exists M > such that x < M for all x E and y < M for all y E. Then for z E + E, there exist x E and y E such that z = x + y. So z x + y < M. Hence E + E is bounded. E + E is closed: Let {z n } E + E such that z n z for some z R n. There exist x n E and y n E such that z n = x n + y n for each n. Since E is compact and {x n } E, there exists a subsequence {x nk } {x n } and x E such that x nk x as k. ow since E is compact and {y nk } E, there exists a subsequence {y nkl } {y k } and y E such that y nkl y as l. Then take z nkl {z n } and since z n z z = lim l z nkl = lim l x nkl + y nkl = x + y. But since x E and y E, it follows that z = x + y E + E. Therefore E + E is closed. Hence E + E is closed and bounded. (b) There exists a closed set E R such that E + is not closed Proof: Define E = {a n } n where a n = n n For all n, a n > n n+ and a n as n. Then it follows that [ ] E c = (a, ) (a n+,a n ) which is an open set. So E is closed. ow for each n Z, we have n E + since ( ) n = n n + n E +. n

2 But n as n and E +. It can be seen that / E + since for all k,n k n < n n + k < k + n, which implies that E + R\Z. Then E + is not closed. ) If f n converges pointwise to a continuous function f on [,] and every f n is continuous and non-negative on [,], then f n converges uniformly to f on [,]. Proof: Let ε > and define for K = K (ε) = For each x [,], since f n for all n f (x) = { x : f (x) f (x) ε f n (x) f n (x) for every. Since f n = f pointwise on [,], for each x [,] there exists = (x,ε) such that > implies f (x) f n (x) = f (x) f n (x) < ε. That is for each x [,], there exists = (x,ε) such that x / K. Then K = /. Also K + K for all since for any x K + we have f (x) } + f n (x) f (x) f n (x) ε. Since f n and f are continuous, so is f f n(x), and so { K = x : f (x) f (x) ε is a closed set. Also K [,], so K is compact. Then K are a nested sequence of compact sets with empty intersection. Therefore by Cantor s intersection theorem, there exists = (ε) such that K = /. Then if n >, then for all x [,] f (x) f n (x) = f (x) f n (x) < ε. }. Hence f n(x) converges uniformly to f on [,].

3 3 3) Let ψ C[,] and define for f,g C[,]. If ψ(x) > for all x [,], then ρ ψ is a metric on C[,] If ψ(x) = for x / and ψ(x) = x / for / < x, then ρ ψ is not a metric on C[,]. Proof: For any f,g C[,], ρ ψ ( f,g) is well defines since ψ(x) f (x) g(x) is a continuous and hence integrable function on [,]. Also = ψ(x) g(x) f (x) dx = ρ ψ (g, f ). ow assume that f g. Then there exists x [,] such that f (x ) g(x ). Since f (x) g(x) is continuous at x and f (x ) g(x ) >, there exists δ > such that for all x [,] such that x x < δ f (x) g(x) f (x ) g(x ) < f (x ) g(x ) which implies that for x [,] such that x x < δ f (x) g(x) f (x ) g(x ) Also since ψ is continuous with ψ(x) >, there exists x [x δ,x + δ] [,] such that ε = inf ψ(x). x [x δ,x +δ] Without loss of generality assume that x. Then x +δ x εδ f (x ) g(x ). A symmetric argument holds if x = using that x. On the other hand if f = g, then f (x) = g(x) for all x [,] and =. So if and only if f = g. Finally for f,g,h C(,) ψ(x) f (x) h(x) dx + ψ(x) h(x) g(x) dx = ρ ψ ( f,h) + ρ ψ (h,g).

4 4 So ρ ψ is a metric on C(,). If ψ(x) = for x / and ψ(x) = x / for / < x, then to see that ρ ψ is not a metric define { x f (x) = x [, ] x (,] g(x) = which are both continuous on [,]. But the we have that f g and (x ) f (x) g(x) dx =. So ρ ψ cannot be a metric. 4) Let (X,ρ) be a metric space and E X be closed. Then f defined f (x) = inf{ρ(x,y) : y E} is continuous and f (x) = if and only if x E. Proof: Let x X and ε >. For any z B ε/ (x) and y E, by the triangle inequality we have Then ρ(x,y) ρ(x,z) + ρ(z,y) < ε + ρ(z,y) ρ(y,z) ρ(y,x) + ρ(x,z) < ρ(y,x) + ε. f (x) = inf ρ(x,y) ρ(x,z) + inf ρ(z,y) ε + f (z) < ε + f (z) y E y E f (z) = inf ρ(y,z) inf ρ(y,x) + ρ(x,z) f (x) + ε < f (x) + ε. y E y E Hence z B ε/ implies that f (x) f (z) < ε and f is continuous on X. If f (x) =, then for all n there exists y n E such that ρ(x,y n ) < inf y E (x,y) + n = f (x) + n = n. Then ρ(x,y n ) as n which means that y n x in X as n. But E is closed, so x E. On the other hand, if x E, then inf ρ(x,y) ρ(x,x) =. y E Then f (x) = if and only if x E. 5) Let f : U V be a continuously differentiable function between two nonempty open sets U,V R n. Suppose that the Jacobian determinant of f is never zero on U, that f (K) is compact for any compact set K V, and that V is connected. Then f (U) = V Proof: First I claim that f (U) is an open set in R n and hence an open set in the subspace topology of V. Fix y = f (x) f (U) where x U. Since f (x), by the inverse function theorem, there exist open neighborhoods U x U of x and V y V of y such that f is - from U x onto V y and f

5 5 is continuously differentiable on V y. That is V y = f (U x ) is an open neighborhood of y contained in f (U). Hence f (U) is open. ow I claim that f (U) is a closed set in the subspace topology of V. Let y n f (U) where y n y as n for some y V. Define K = {y n } n {y} V. Since y n y, it follows that K is compact. There exist x n f (K) U such that f (x n ) = y n. By assumption f (K) U is also compact, so there exists a subsequence {x nk } {x n } converging to some x f (K) U as k. Since f is continuous, f (x ) = lim k f (x nk ) = lim n y n = y. Therefore y f (U), which proves that f (U) is closed in V. Then f (U) and V \ f (U) are open sets in V. Moreover f (U) (V \ f (U)) = / and f (U) (V \ f (U)) = V. Since V is connected, either f (U) = / or V \ f (U) = /. Since U is nonempty, f (U) /. Therefore V \ f (U) = / and it follows that V f (U). The oposite inclusion, f (U) V, follows trivially since f : U V. So we have that f (U) = V.