Linear Analysis

Question 1

A linear map T : X –> Y is cts iff it is bounded

Answer 1

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.

Question 2

The dual space of an NVS V is Banach.

Answer 2

1. Let T_n Cauchy in V*.
2. 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.
3. 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.
4. 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.

Question 3

If T in B(X, Y), ||T|| <= ||T||. In particular, T is bounded.

Answer 3

||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||

Question 4

Hahn Banach V1 Statement

Answer 4

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.

Question 5

Hahn Banach V2

Answer 5

Let V an NVS, W <= V and f in W. Then f extends to f’ in V with the same norm.

Question 6

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.

Answer 6

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.

Question 7

If T in B(V,W), then ||T*||=||T||.

Answer 7

Proof.

Question 8

Define nowhere dense, first category, meagre, second category, residual, comeagre. State and prove Baire’s category theorem.

Answer 8

1. 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.
2. 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.
3. 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.
4. 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.
5. Then the intersection of the closed balls is non-empty. But it lies in all of U_n. Contradiction

Question 9

The NVS of all real eventually zero sequences under the L1 norm is not Banach.

Answer 9

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.

Question 10

There exists a cts nowhere differentiable function f : [0,1] –> R

Answer 10

1. 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.
2. 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|.
3. 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.

Question 11

Find a meagre set that is not measure 0. Find a measure zero set that is not meagre.

Answer 11

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.

Question 12

State and prove the uniform boundedness principle.

Answer 12

1. 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.
2. Let E_n be the the set of v in V such that ||f_{alpha}(v)|| <= n for every alpha.
3. 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].
4. 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.
5. 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.

Question 13

State and prove the open mapping theorem.

Answer 13

1. 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.
2. 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).
3. 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)).
4. 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.

Question 14

State and prove the inverse mapping theorem.

Answer 14

1. Let V,W Banach and T in B(V,W). Then if T is injective and surjective, T has a cts inverse.
2. Immediate from open mapping theorem. T is an open map, so T^{-1} is cts.

Question 15

State and prove the closed graph theorem

Answer 15

1. 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.
2. 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.

Question 16

State and prove Urysohn’s lemma.

Answer 16

1. 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.
2. Normality implies that if C contained in U, then there is U’, C’ with C in U’ in C’ in U.
3. 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.

Question 17

State and Prove Arzela-Ascoli Theorem

Answer 17

If K is compact Hausdorff and F a subset of C(K), then F is pre-compact iff F is bounded and equicontinuous.

Proof:

1. We take as given that F is pre-compact iff F is totally bounded.
2. 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.
3. 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.

Question 18

State and Prove the real Stone-Weierstrass Theorem

Answer 18

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.

1. 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.
2. 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.
3. 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.

Question 19

State the parallelogram law.

Answer 19

||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).

Question 20

State and prove the orthogonal decomposition theorem.

Answer 20

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.

Question 21

State and prove the Riesz representation theorem.

Answer 21

Let H a Hilbert space. Then, the map phi: H –> H* given by v |–> is anti-linear, isometric and bijective.

1. For isometric, |phi(v)(w)| <= ||v|| ||w|| by Cauchy Schwarz, so ||phi|| <= ||v||. And |phi(v)(v)| =||v||^2 so we have equality.
2. 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>
3. 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>

Question 22

Prove that the Fourier series of a periodic function converges under the L2 norm.

Answer 22

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.

Question 23

Define a Hilbert basis and prove an equivalent statement.

Answer 23

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.

Question 24

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.

Answer 24

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:

1. It suffices to consider {e_n} an infinite basis, as the finite case is trivial.2
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.
3. Finally, by continuity of the inner product, = lim = lim sum_{n <= N} x_n y_n^{bar}
4. 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.

Question 25

Define the spectrum, resolvent set of an operator. Define the point spectrum and approximate point spectrum.

Answer 25

Let X be a Banach space and T in B(X) a bounded operators. The spectrum of T is the set of complex lambda such that T-lambda is not invertible, where here invertible means no continuous inverse. In fact, if there were a set theoretic inverse, by the inverse mapping theorem it would have a continuous inverse. The resolvent set is the complement of the spectrum. The point spectrum is the set of eigenvalues of T. The approximate point spectrum is the set of lambda such that there are vectors x_n in X with (T-lambda)(x_n) converging to zero. Note this contains the point spectrum.

Question 26

State a result about the spectrum of an operator.

Answer 26

Let X a Banach space and T a bounded operator. The spectrum of X is a non-empty, closed, subset of B(0,||T||).

1. If ||T|| < 1, then 1-T is invertible.
   Proof: Do the obvious thing, which is define U=1+T+T^2+T^3+…. Well first this series is absolutely convergent. Since B(X) is Banach, U exists. Now (1-T)U (x) = (1-T) lim(x+T(x)+T^2(x)+…T^n(x)) = lim(1-T^{n+1}(x)) = x. Check the other side as well.
2. If S_1 in B(X) is invertible, then all S_2 in B(S_1, ||S_1^{-1}||) is also invertible.

Proof: Write S_2 = S_1(I-S_2^{-1}(S_1-S_2)) and use the previous lemma!

3.

Question 27

Show that l_p is Banach

Answer 27

1. Let x_n Cauchy. Extract a limit x.
2. Fixing eps > 0, we have N s.t. for all n,m>N, ||x_n-x_m|| < eps. Then fixing M, n>N we have
   (sum_{j=1 to M} ||x_j-x_n,j||^p)^(1/p) <= (sum ||x_j-x_m,j||^p)^(1/p)+ ||x_n-x_m||.

Indep of M. So done.

To show x is indeed in l^p, just note that if it weren’t, then surely ||x|| <= ||x-x_n||_p+||x_n||, where this ineq holds for sum of the first M terms, hence for all terms.

Question 28

Show that if T : H–>H is self adjoint in a Hilbert space H, ||T|| = sup ||, for ||x|| <= 1.

Answer 28

We have || <= ||Tx|| ||x||, so sup || <= ||T||.

For the converse, note that ||Tx|| = /||Tx|| = = , where ||y||=1.

So if M = sup ||, we need to show that
|| <= M. It suffices to prove Re <= M.

But Re = 1/4 (-) <= 1/4* M ( ||x+y||^2+||x-y||^2) = 1/4M2*(||x||^2+||y||^2) <= M.

We know the parallelogram law, which says

||Tx||

Question 29

The double dual map phi : V –> V** has ||phi|| <= 1. Intuition is it is easy to show that things in the dual are smaller.

Answer 29

Easy.

Question 30

Show that the adjoint of a bounded map T in a Hilbert space exists and is unique.

What can we say about ker(T)?

Question 31

Some easy results about compact self-adjoint operators

1. Every nonzero element of the spectrum is an eigenvalue
2. Show that the restriction of T to the orthogonal complement of an not-0) eigenspace is a compact & self-adjoint operator.
3. Show that the only accumulation point of the point spectrum is 0.
4. Show that there are countably many eigenvalues.

Answer 31

1. Recall ker(T)=(Im T*)^{perp}=(Im T)^{perp} whenever T is self-adjoint. Note that T-lambda is self-adjoint. Hence if lambda is in the spectrum but not an eigenvalue, Im(T-lambda)^{perp}=0.

We show that Im(T-lambda) is closed. It suffices to prove T-lambda is bounded below. Say x_n norm 1, (T-lambda)(x_n)–>0. Then T compact, so some subsequence Tx_n–>y. So x_n–>y/lambda. So lambda is in the point spectrum, contradiction. Hence H=Im(T-lambda)+Im(T-lambda)^{perp}=Im(T-lambda).

2. For compactness, note that if =0 for all y in E_{lambda}, then =0 for all y in E_{lambda}, as lambda !=0. Hence the image of the unit ball in E_{lambda}^{perp} is E_{lambda}^{perp} interseced with the image of the usual unit ball.
3. If lambda_{n} are infinitely many distinct eigenvalues accumulating to delta, we get distinct orthonormal v_{n}. Then by compactness, |Tv_{phi(n)}-Tv_{phi(m)}| = sqrt(2)(delta-eta) –> 0, a contradiction.
4. It is immediate that there are countably many eigenvalues, since if the only eigenvalue which is not isolated is 0.

Question 32

Give an example of a Hilbert space and a proper subspace which has trivial orthogonal complement.

Answer 32

l^2 with the subspace of eventually zero sequences.

Question 33

State the spectral theorem for compact self-adjoint operators.

Explain why it is an easy deduction from the main lemma: If T compact self-adjoint, then either ||T|| or -||T|| is an eigenvalue.

Answer 33

If T compact self-adjoint in H, and T nonzero, then T has countably many eigenvalues and H decomposes as the orthogonal sum of the eigenspaces of T. Moreover, the only accumulation point of the eigenvalues is 0.

Question 34

Given K a compact non-empty subset of C, find a Hilbert space H and an operator T with spectrum K.

Answer 34

file:///C:/Users/steve/Documents/General/Maven/Notes%20on%20Spectral%20Theorem.pdf

Question 35

Easy results for normal operators

1. Define a normal operator
2. ||Tv||=||T*v||
3. 1 kerT=kerT=(ImT)^{perp}=(ImT)^perp
4. The residual spectrum is empty
5. sigma_p(T*)=sigma_p(T)^conj
6. sigma_ap(T*)=sigma_ap(T)^conj

Answer 35

1. TT=TT
2. ===
3. 1 kerT=kerT* by 2. x in kerT* iff =0 for all y iff =0 for all y iff x in T*^perp.
4. Let T-lambda injective. We show T-lambda has dense image. We know T-lambda is normal, hence 0=ker(T-lambda)=(Im(T-lambda))^perp. Hence Im(T-lambda) is dense.
5. lambda in sigma_p(T) iff Tx=lambda*x, some nonzero x iff for all y, ==, so Ty=lambda^conj y.

Question 36

Show that if T is self-adjoint, a is real, then
||(T+ia)x||^2 = ||Tx||^2+|a|^2 ||x||^2.

Let a nonzero. Show that T+ia is injective. Show that T+ia is surjective.

Explain why we’ve shown that if T is self-adjoint, then the spectrum of T is real.

Answer 36

Note (T+ia)* = T-ia, hence

||(T+ia)x||^2 = =||Tx||^2+a^2||x||^2.

Question 37

Show that ||T|| = ||T||. Deduce that if T_n –> T, then T_n –> T*.

Question 38

Explain why T finite rank ==> T* finite rank. Deduce that T compact ==> T* compact.

Question 39

Does T**=T, where T* denotes adjoint in a Hilbert space?

Answer 39

Yes, of course. ==. By uniqueness of adjoint, T**=T.

Question 40

In an NVS, suppose A is a subset. Show that span(A) is dense iff the annihilator of A is 0.

Answer 40

Let cl(span(A))=X. If f annihilates A, then f annihilates cl(span(A))=X, hence f=0.

Now let the annihilator of A be 0, and suppose span(A) is not dense.
Hence there exists an open ball B(x,eps) outside of cl(span(A)).
Consider the function d(y) = inf_{a in A, ||y-a||}.
1. d(y_1+y_2) <= d(y_1)+d(y_2)
2. d(lambda y)=lambda d(y) for lambda > 0 as A a subspace.
Hence by Hahn-Banach, we get a bounded linear functional f with f(x)=eps/2, and f <= d. Hence f is 0 on A but f nonzero.