A linear map T : X –> Y is cts iff it is bounded
If T cts, there is 1/delta > 0, if |x| < 1/delta, |T(x)-T(0)|=|T(x)| < 1. Thus if |x| < 1, |T(x)| < delta.
If T bounded, there is delta > 0, |x|<1 ==> |T(x)| < delta. So if eps > 0, |x-y| < eps/delta ==>|T(x-y)| < eps.
The dual space of an NVS V is Banach.
- Let T_n Cauchy in V*.
- For each v in V, |T_n(v)-T_m(v)| <= ||T_n-T_m|| ||v|| is Cauchy in R, thus converges. Define T(v) = lim T_n(v), noting that T is linear.
- We prove T is bounded. Let ||v|| <= 1. Pick N so that all n, m > N satisfy ||T_n-T_m|| < 1. So |T_n(v)| = |T_n(v)-T_m(v)+T_m(v)| <= ||T_n-T_m|| ||v|| +||T_m|| ||v|| <= 1+||T_m||. Since T_n Cauchy, it is bounded.
- Now we show ||T_n-T|| –> 0. Pick ||v|| <= 1. Let N such that all n,m>N satisfy ||T_n-T_m|| < eps. Thus |T_n(v)-T_m(v)| < eps. Thus |T_n(v)-T(v)| <= eps. Thus ||T_n-T|| <= eps.
If T in B(X, Y), ||T|| <= ||T||. In particular, T is bounded.
||T|| = sup_{g in Y nonzero, ||g|| <= 1} ||T(g)|| = sup_{g in Y nonzero} sup_{x in V nonzero, ||x|| <= 1} ||g(T(x))|| <= sup_g sup_x ||T(x)|| = ||T||
Hahn Banach V1 Statement
Let V be a real NVS , and W a subspace. Let p : V –> R positive homogenous (p(av)=ap(v) if a > 0) and subadditive (p(u+v) <= p(u)+p(v)), and f : W –> R linear with f(w) <= p(w). Then f extends to a linear map on V upper bounded by p everywhere.
Hahn Banach V2
Let V an NVS, W <= V and f in W. Then f extends to f’ in V with the same norm.
Show that for a nonzero v in an NVS, that there is some functional f of norm 1 mapping v to its norm. Show that u,v distinct iff they are separated by some functional. Show also that the double dual map is an isometry.
If V an NVS, v nonzero, some f in V* has ||f|| = 1 and f(v)=||v||. Thus V* is nontrivial, and ker(V*) = {0}. Thus if u,v distinct, they are separated by some dual vector. Note also that this shows the norm of phi(v) is at least 1 for any nonzero v, thus it is 1.
If T in B(V,W), then ||T*||=||T||.
Proof.
Define nowhere dense, first category, meagre, second category, residual, comeagre. State and prove Baire’s category theorem.
- Let X a topological space. A subset A of X is nowhere dense is its closure has empty interior. A is first category or meagre if it is a ctble union of nowhere dense sets. A is second category or non-meagre otherwise. A is residual if X\A is meagre.
- Baire’s Category theorem says that if X is a complete metric space, X is not meagre. Equivalently, the intersection of dense open sets is non-empty. In fact it is dense.
- Proof. Suppose not. Let X be a ctble union of nowhere dense sets. Then the empty set is an intersection of open dense sets U_n.
- Construct a sequence x_n, eps_n such that the closed ball B(x_n,eps_n) lies in U_n and the closed balls are nested. It is clearly possible.
- Then the intersection of the closed balls is non-empty. But it lies in all of U_n. Contradiction
The NVS of all real eventually zero sequences under the L1 norm is not Banach.
Let E_n the subspace of x such that x_m=0 for all m at least n. Then E_n is closed and E_n has empty interior. But V is the union of E_n, so V is first category. Thus V is not complete.
There exists a cts nowhere differentiable function f : [0,1] –> R
- We let V = C([0,1]) under the uniform norm and show that the set of functions differentiable at some point A is contained in a meagre set. Then since V is complete, result follows. Define A_n as the set of f such that there is some x, such that whenever |y-x| < 1/n, |f(x)-f(y)|/|x-y| <= n. Clearly if f differentiable at x, then picking n > |f’(x)| ensures that |f(x)-f(y)|/|x-y| < n for |x-y| sufficiently close.
- A_n is closed. Indeed let f_k a sequence in A_n converging to f. Then there is x_k such that for all y in [0, 1] intersect (x_k-1/n,x_k+1/n) that
|f_k(x_k)-f_k(y)| <= n|x_k-y|. Now pick a subsequence of x_k going to x. wlog x_k –> x. Then pick y with |y-x| < 1/n. For k sufficiently large, y lies in (x_k-1/n, x_k+1/n). Then taking the limit of above and noting f_k(x_k) –> f(x), we get |f(x)-f(y)| <= n|x-y|. - A_n has empty interior. Indeed, suppose B(f,eps) lies in A_n. Then f is uniformly cts, so we can pick a piecewise affine f_0 in B(f,eps/2). i.e., there is delta>0 so that |x-y| < delta implies |f(x)-f(y)| < eps/4. Then B(f_0, eps/2) lie in B(f,eps). But just do a bunch of zig zags.
Find a meagre set that is not measure 0. Find a measure zero set that is not meagre.
Let q_n a listing of the rationals and let D_n be the union of k>=n of (q_k-2^{-n-k}, q_k+2^{-n-k}). So the measure of D_n goes to 0, thus D defined as the intersection of D_n is measure 0. D_n complement is closed and has empty interior, so the union of D_n^c is meagre. Thus D is comeagre. Thus D is not meagre.
State and prove the uniform boundedness principle.
- If V is a Banach space and W an NVS, and f_{alpha} is a collection in B(V,W). If the functions are pointwise bounded, then they are uniformly bounded.
- Let E_n be the the set of v in V such that ||f_{alpha}(v)|| <= n for every alpha.
- E_n is closed, since for a fixed f_{alpha}, the map v to ||f_alpha(v)|| is cts and we are taking an intersection of the preimages of [-n, n].
- Since V is complete, some E_n has non-empty interior. So there is u in V and eps > 0 such that for each v in V, sup_{alpha} ||f_alpha(v)|| <= n.
- Pick v in B(0,1) arbitrarily. Then u + eps/2 v is in B(u,eps). Thus sup_{alpha} ||f(u) +eps/2f(v)|| <= n. But ||f(u)+eps/2f(v)|| >= eps/2 ||f(v)||-||f(u)||. So we get a bound for sup||f(v)|| <= 2/eps(n+sup_{alpha}||f(u)||) < infinity.
State and prove the open mapping theorem.
- If V Banach, and T in B(V,W) surjective has second-category image, then T is open. In particular, if W is Banach and T is surjective, T is open.
- T is open iff T(B(0,1)) contains B(0,eps) for some positive eps. Indeed, for the reverse direction, consider p in open U with T(p)=q. Then T(U) contains T(B(p,delta)) = q +deltaT(B(0,1)) which contains q+delta B(0,eps) = B(q,eps).
- im T is the union of closure(T(B(0,n)) ) so one of these has non-empty interior. i.e., for some n, there is v,eps>0, B(v,eps) is closure(T(B(0,1))) = C. Thus B(-v,eps) lies in C. Claim that B(0, eps) lies in C. Indeed, if w in B(0, eps), then there is T(u_n) converging to v+w, T(v_n) converging to -v+w. So T(1/2(u_n+v+n)) –> w, where we note 1/2(u_n+v_n) lies in B(0,1). So we’ve shown that B(0,eps) is contained in the closure of T(B(0,1)). By scaling the norm of W, we wlog that B(0,1) contained in cl(T(B(0,1)).
- If B(0,1) in cl(T(B(0,1))), then B(0,1/2) in T(B(0,1)). Proof: Pick w in B(0,1/2). We pick v_n in B(0,1/2^n) such that |T(v_1+…+v_n)-w| < 1/2^{n+1}. This can be done since we note that B(0,delta) is in cl(T(B(0,delta))). Thus v_1+..v_n is Cauchy hence converges to v. And ||v|| < 1. And T(v)=w.
State and prove the inverse mapping theorem.
- Let V,W Banach and T in B(V,W). Then if T is injective and surjective, T has a cts inverse.
- Immediate from open mapping theorem. T is an open map, so T^{-1} is cts.
State and prove the closed graph theorem
- Let V,W be Banach spaces and T:V–>W be linear. Then if the graph Gamma={(x,T(x)) x in V} is closed in the product topology (which can be induced by the norm ||(v,w)||=max(||v||,||w||)), then T is bounded/cts.
- Consider phi: Gamma –> V. Note that Gamma is a closed subspace of VxW, which is complete, hence Gamma is complete. Next note phi is injective and surjective. Finally phi is bounded, since ||phi(v,T(v))||=||v|| <= max{||v||,||Tv||} = ||(v,Tv)||. So by the inverse mapping theorem, phi^{-1} is bounded, and so T is bounded.
State and prove Urysohn’s lemma.
- If X is a normal topological space and C0 C1 are two disjoint closed sets, there is a cts f in C(X) such that f is 0 on C1, 1 on C2, and f(x) lies in [0, 1] for all x in X.
- Normality implies that if C contained in U, then there is U’, C’ with C in U’ in C’ in U.
- Therefore we construct U1 = X \ C1. So C0 in U1. Therefore C0 in U_{1/2} in C_{1/2} in U1. Then define f(x) = inf q, x in U_q.
State and Prove Arzela-Ascoli Theorem
If K is compact Hausdorff and F a subset of C(K), then F is pre-compact iff F is bounded and equicontinuous.
Proof:
- We take as given that F is pre-compact iff F is totally bounded.
- Let F be totally bounded. Then F is certainly bounded. Since F is totally bounded, for any eps > 0, there exist f_1, …, f_n in F with every f in F strictly within eps of some f_i. Pick x in K. Each f_i is cts hence there exist U_i so that the deviation of f_i from x is at most eps in U_i. Now consider U, the intersection of the U_i which is a nhood of x. So if y in U, for f in F, let f be within eps of f_i. Then |f(x)-f(y)| <= |f(x)-f_i(x)|+|f_i(x)-f_i(y)|+|f_i(y)-f(y)| < 3eps. So F is equicts.
- Now let F bounded and equicts. So we get U_x such that all f deviate from f(x) by at most eps in U_x. U_x is an open cover of K hence U_{x_i} is a finite cover. Then if we restrict F to {x_i}, we get that F is a bounded subset of R^n under the sup norm, so F is pre-compact, so totally bounded. So we can pick f_j in F as the eps-cover over the x_i. Then if f in F, let f within eps of f_j when restricted to {x_i}. |f(x)-f_j(x)| <= |f(x)-f(x_i)| + |f(x_i)-f_j(x_i)| + |f_j(x_i)-f_j(x)| <= eps+eps+eps.
State and Prove the real Stone-Weierstrass Theorem
Let A an algebra in C(K) (K compact Hausdorff) that separates points. Then either cl(A)=C(K) or for some x in K, cl(A) is the set of functions in C(K) that vanish at some x.
- Note that cl(A) is an algebra. If f in cl(A), then |f| in cl(A). Hence cl(A) closed under max and min. Proof: Pick eps > 0. Then expand sqrt(x+eps) in a Taylor series around 1/2. This gives a sequence of polynomials converging uniformly to sqrt(x+eps) on [0,1] so we get some polynomial S(x) such that |S(x)-sqrt(x+eps)| < eps on [0,1]. Let f in A and consider S(f^2)-S(0), has no constant term, therefore is in A. ||f|-(S(f^2)-S(0))| <= | |f|-sqrt(f^2+eps)|+|sqrt(f^2+eps)-S(f^2)|+|S(0)| which is small.
- Fix g in C(K). If for all x, y distinct, there is f_x,y differing from g at x,y by at most eps, and we’re closed under taking max and min, then we can uniformly eps approximate g by a function in A.
Proof: We prove first that fixing x, we can find f_x with f_x < g+eps everywhere, and f_x, g differing at x by at most eps. This is actually very obvious. For each y, just let U_y a nhood of y such that |f_x,y(z)-g(z)| < eps in U_y (this is clearly possible by picking f_x,y differing from g at y by eps/2, then using continuity). Then take a finite subcover and take the min of those functions. Now, given f_x, we let U_x a nhood of x be such that |f_x(z)-g(z)|< eps. Then take a finite subcover, then take the max of those f_i. Its bounded above by g+eps since each f_x_i is. It’s at least g-eps by contruction. - To complete the proof, consider case 1, that every x in K has some f in A st f(x)!=0. Then, for any x,y distinct we can construct h_{x,y} separating x,y. Then h_x nonzero at x, h_y nonzero at y. So some linear combination of h_x, h_y and h_{x,y} is nonzero at x,y and separates x,y. So let h be that linear combination. Then (h(x),h(y)), (h(x)^2, h(y)^2) are lin. indep. So some lin combo gives equality to g at x,y. This completes this part of the proof. Otherwise, add the identity to A. Then done.
State the parallelogram law.
||u+v||^2+||u-v||^2=2||u||^2+2||v||^2.
How to remember? It’s the parallelogram. And for a square, 2+2 = 2*(1+1).
State and prove the orthogonal decomposition theorem.
Let E a Euclidean space and F a complete subspace. Then E is the direct sum of E and its orthogonal complement.
Proof: Given x in E, let d=inf_{y in F} d(y,x). Let y_n in F be such that d(x,y_n) decreases to d. We claim that y_n is Cauchy. Indeed,
||(y_n-x)-(y_m-x)||^2+||y_n+y_m-2x||^2 = 2||y_n-x||^2+2||y_m-x||^2, and noting that ||y_n+y_m-2x|| >= 2d, we get that ||y_n-y_m||^2 <= 2||y_n-x||^2+2||y_m-x||^2-4d^2 --> 0. Thus y_n-->y as F is complete, where y in F. Finally, x-y is orthogonal to E. Indeed, if not we could change y to decrease the distance to x.State and prove the Riesz representation theorem.
Let H a Hilbert space. Then, the map phi: H –> H* given by v |–> is anti-linear, isometric and bijective.
- For isometric, |phi(v)(w)| <= ||v|| ||w|| by Cauchy Schwarz, so ||phi|| <= ||v||. And |phi(v)(v)| =||v||^2 so we have equality.
- For surjectivity, pick f in H* nonzero. Observe that im(f) is 1-dimensional, so H/ker(f) is one-dimensional. Thus ker(f)^{perp} is 1-dimensional. We can prove this directly as follows. Let u, v in ker(f)^{perp} both nonzero. Then f(u), f(v) != 0. So af(u)+bf(v)=0. So 0=<u>=. So ||au+bv||^2=0.</u>
- Let ker(f)^{perp} = where v has norm 1. We claim of course that f = , for some nonzero a. Then f(v)=a||v||^2 =a, so we claim f=.
Let x in H. Note that (x-v) lies in ^{perp} which is the double perp of ker(f). But ker(f) is closed, so this is just ker(f).
f(x)=f(v+ (x - v))=f(v)0=.</u>
Prove that the Fourier series of a periodic function converges under the L2 norm.
More precisely, let f : S^1 –> C be cts and define an inner product by = 1/2pi integral fg.
Then let f_k = Let S_n be the sum from i = -n to n of f_k e^{ikx}. Note that e^{ikx} are orthonormal, hence S_n is just orthogonal projection onto U_n spanned by e^{ikx}, -n <= k <= n. Then note that the algebra of e^{ikx} are dense in C(S^1) by Stone-Weirstrass. Finally, note that || <= || for an orthogonal projection.
Define a Hilbert basis and prove an equivalent statement.
If H a Hilbert space, a subset B is a Hilbert basis if it is a maximal orthonormal set. Equivalently, B is an orthonormal set with a dense span. Indeed, cl(span(B)) is closed, so cl(span(B))+cl(span(B))^{perp}=H. B is a Hilbert basis iff cl(span(B))^{perp}={0} iff cl(span(B))=H.
State and prove Bessel’s inequality. State and prove a result about component-wise inner products and deduce Parsevel’s equality. Hence prove all separable Hilbert spaces are l^2.
Bessel’s inequality: If e_n is an orthonormal set on any Euclidean space (can be finite or infinite), then for all x, x_i = satisfies
sum_{n} |x_i|^2 <= ||x||^2.
Proof: For a finite sum, sum x_i e_i is just orthogonal projection onto span , so it is trivial. for infinite n, we just take the limit as n–>infty.
If H a Hilbert space, e_n a Hilbert basis, then =sum_n x_i bar(y_i), x_i = . Moreover this sum converges absolutely.
Proof:
- It suffices to consider {e_n} an infinite basis, as the finite case is trivial.2
- x = sum{n>=1} x_n e_n. First, this series is Cauchy, since sum |x_n|^2 <= ||x||^2 by Bessel. Hence it converges to some y in H. Consider x-y. Then x-y is orthogonal to every e_n. Hence x-y is in ^{perp} = (cl())^{perp} = {0}. Hence x=y.
- Finally, by continuity of the inner product, = lim = lim sum_{n <= N} x_n y_n^{bar}
- To show absolute convergence, note that sum_{n <= N} |x_n||y_n| <= ||sum_{n <= N} x_n||^2||sum_{n <= N} y_n||^2 <= ||x||^2 ||y||^2 by Cauchy Schwarz and Bessel’s inequality.
Parsavel’s equality states that ||x||^2 = sum |x_n|^2, which is immediate from the above.
We have the map H to l^2 given by x |–> . It is clearly linear. Parsavel’s equality shows it is injective and moreover an isometry. It remains to show it is surjective, which is trivial. Indeed, if a_n in l^2, just define x = sum a_n e_n. This sum is cauchy, hence converges. And =lim =a_n. Thus x maps to a_n.
