A sort of diary for Math 507, spring 2004

But I told Professor Vogelius (scheduled to teach a successor course in the fall) that I would do some version of the Spectral Theorem for bounded operators. So I discussed what a "spectral theorem" was. In finite dimensions, this seemed easy. If dimension is no longer finite, then ... there are many versions. I recommended Riesz-Nagy's book (now available in a very reasonably priced paperback edition) as an original source. I would do a version of the continuous functional calculus as in Reed & Simon, and then a version of the holomorphic functional calculus as in Conway's book. And, finally (after some applications), I would do a measurable functional calculus verson.

A continuous linear bijection between Banach spaces must be a homeomorphism. Or: if two Banach space norms on a vector space are comparable, then they are equivalent.

Digression
All about finite dimensional (normed) vector spaces.

1. Any two norms on a finite-dimensional vector space are equivalent. (Students simplified my clumsy proof! I thank them.)
2. Any finite-dimensional subspace of a normed space is closed.
3. Any linear mapping of a finite dimensional vector space to a normed space is continuous. (We showed that finite-codimensional subspaces need not be closed.)
4. (F. Riesz) Any locally compact normed space is finite dimensional.

Stated the Closed Graph Theorem. The Reed-Simon ansatz:
Suppose V and W are Banach spaces, and T:V->W is a linear transformation. Then:
T is continuous
IF
   i) When v_j->v in V
THEN
   ii)T(v_j) converges to w.
   AND
   iii) T(v)=w.
The Closed Graph Theorem asserts that T will be continuous if you can proved the following implication:
IF
   i) When v_j->v in V
   AND
   ii)T(v_j) converges to w.
THEN
   iii) T(v)=w.
And of course this allows us to assume more.

I applied this result to prove the Hellinger-Toeplitz Theorem.

I went over the theory of Fourier series on the circle, or the interval [0,2Pi] where the ends were identified. My references were Introduction to Abstract Harmonic Analysis by Katznelson and some material by Varadhan. I did not refer to Zygmund's "bible" on Fourier series.

When reading various references, some attention should be paid to whether the integral has a 1/2Pi or not, etc. I outlined the background for the Fejér and Dirichlet kernelts (summation methods), and used Uniform Boundedness to "produce" (?) an example of a continous function whose Dirichlet-summed Fourier series did not converge at 0. The Fejér sums, of course, behaved very well.

We (Mr. Speck and I) stated the Open Mapping and Closed Graph Theorems. I'll prove the Open Mapping Theorem, and Mr. Speck will deduce the Closed Graph Theorem from it. (I claimed that the two results were in duality much as the Inverse/Implicit Function Theorems were. Sigh. I also attempted to prove a false result. Oh well.)

We stated Montel's Theorem, proved Vitali's Theorem (which spreads convergence) and proved Osgood's Theorem (if a sequence of holomorphic functions converges pointwise in an open set, then in a dense open set the sequence converges u.c.c. to its limit, a holomorphic function). (Proof from Ash, Complex Analysis).

We proved the uniform boundedness/Banach-Steinhaus Theorem. The proof was very analogous to Osgood's Theorem. The instructor unfortunately stated a false lemma, but was saved from disaster by alert listeners. The proof sort of followed Reed & Simon's Functional Analysis. Completeness is needed (R^infty, T_j(v)=<v,je_j>.).

We verified that separate continuity implies joint continuity for bilinear maps (where one factor is complete). This was contrasted with the non-linear case.

Briefly defined the density character of a topological space (the "smallest" cardinality of a dense subset) and showed that the density character of X was always less than or equal to that of X^*.

I discussed the canonical (isometric) embedding of X into X^** and mentioned James's example of a Banach space X isomorphic to its double dual but not reflexive (reflexive means that the X-->X^** is a bijection). Here (at least from Rutgers browsers) is a link to James' original paper. This paper is not easy reading for me -- it is rather brief. More information can be found in Lindenstraus & Tzafriri, "Classical Banach Spaces", and in Megginson's recent functional analysis book.

It is time to leave H-B and begin to struggle with the phenomenal consequences of completeness. So I stated a version of the Baire Category Theorem. And another version of the Baire Category Theorem. We attempted (and I think succeeded!) in showing that these results were identical.

So I wanted to discuss a consequence of the theorem in a familiar area, and therefore stated a result called the Vitali-Saks-Osgood Theorem which I will attempt to prove next time.

Then I went back to look at Sturm-Liouville problems one more time. I remarked that the dimension of each eigenspace was 1. I further said that the kernel had a nice eigenspace expansion. We looked at a kernel to solve the initial value problem, and I compared a graph of a it with the kernel graph below. "Clearly" one is self-adjoint and the other is not. The algebra resulting from the fact that this kernel is not self-adjoint is irritating. I should ask people to compute the commutator as a homework assignment.

I looked briefly at the functional calculus for these operators. We could take the square root of a compact self-adjoint operator by just taking the square root of the eigenvalues on each eigenspace. We could define cosine of a compact self-adjoint operator -- we will do more of this when we do spectral theory of general self-adjoint operators, where the point spectrum (eigenvalues) is much smaller than the whole spectrum.

The spectral theory of normal compact operators is a small amount of additional work, which I will not do. The idea is: start with a compact operator, T, on H, which is normal: [T,T^*]=0. Here {A,B]=AB-BA. The analogy to keep in mind is that self-adjoint operators are like real numbers, and normal ones are like complex numbers. So T=A+iB where A and B are self-adjoint. A is (T+T^*)/2 and B is (T-T^*)(2i). A and B commute, and on, say, A's eigenspaces, B is compact, so B has invariant eigenspaces. Therefore T "diagonalizes". Etc. In fact, there is a mapping from functions holomorphic on T's spectrum to operators. This mapping is an algebra homomorphism with kernel 0, and maps z to T. The resulting functional calculus has many uses.

We waved goodbye to the beautiful S-L theory. Now to the core of basic functional analysis:

• the Hahn-Banach Theorem (proof: "obvious" baby step + "Zornicate")
• the consequences of completeness
  • Open Mapping/Closed Graph
  • Uniform Boundedness

Examples
Hilbert space, L^p, C[0,1], and their duals and the theorems (some with intricate or long proofs) designed to give isometric isomorphic identification of the duals. Finite dimensional vector spaces.

Intuition from algebra
V a vector space over a field. Look at the algebraic dual of V. If dim V is finite, that's just another copy of V. If dim V is infinite, then the algebraic dual has lots of things in it. For example, suppose that {x_a} (a in A) is a Hamel basis for V. Then there's a dual basis, {x_a^*} (a in A) at least in V^{* alg}. But as Ms. Mau pointed out, if B is an infinite subset of A, then SUM_{a in B}x_a^* is a distinct element of V^{* alg}. So if dim V is infinite, then the duals keep getting bigger and bigger in terms of cardinality of the Hamel basis, the only invariant.

Do the duals of Banach spaces keep getting bigger? How should we measure "bigger"? It will turn out that the Hamel basis are uncountable always. What about separable? Equivalent to the whole space being the closure of the rational linear span of a countable number of linear independent vectors. We saw that the complex Borel measures and L^infty were not separable, but the other examples were.

I remarked that the sequence of duals of Banach spaces had 3 types of behavior: either they were all the "same" (Hilbert space), or they alternated (L^p for 1<p<infty), or they were all different. Examples for the last behvior turn out to be L¹ or C([0,1]) but this is not clear nor is the general description just given!

More generally
We considered L^1/2([0,1]). I gave it a translation-invariant metric in which the space is complete. What's its dual space?
Then I gave the holomorphic functions on an open set of C, a certain translation-invariant metric which is equivalent to uniform convergence on compact sets. Call this H(U). What's its dual space?
I described sequential convergence on the collection of test functions. A sequence converges if it is concentrated in a compact set, and if the sequence and its derivatives all converge uniformly (though not necessarily equiuniformally!) to the function and its derivatives. This is enough to define continuous linear map to the reals. (The actually topology on the test functions is more complicated.) What's the dual space?

The first example has dual space {0}. The second has a Frechet space structure, and it has neighborhoods of 0 which are convex. Its dual space is big. The dual space in the third case is the distributions (Schwartz) is very big. The test functions are also locally convex. So that is why there is a sublinear functional in the proof of the H-B theorem. That's next time!

The most routine example is L(y)=y'' on [0,Pi] with y(0)=0 and y(Pi)=0. Then the Green function can be explicitly computed. Here is part of a Maple session to see what the Green function looks like:

f:=(x,t)->piecewise(t>x,x*(t-Pi),t*(x-Pi));
B:=plot3d(f(x,t),x=0..Pi,t=0..Pi,color=red,grid=[100,100]):
A:=spacecurve([t,1.5-t,f(t,1.5-t)],t=0..1.5,thickness=4,color=black,numpoints=200):
display3d({A,B});

What's going on?

1 0
1 1

I remarked that the bulk of the story would be in step 1, as follows:
Step 1
If T is a compact self-adjoint operator on a Hilbert space H, then either ||T|| or -||T|| is an eigenvalue of T.
We proved this by using the result #5 from last time combined with #4. We took a sequence {h_n} which satisfied #4, and then got a subsequence so that T(the subsequence) converged and converged, and then looked at ||Th_n-h_n||².

Now we iterated step 1, and got a sequence of finite dimensional, mutually orthogonal eigenspaces. The eigenvalues decreased. We showed that they must decrease to 0 with a similar compactness argument.

A compact self-adjoint operator on an infinite dimensional H must have 0 in its spectrum, or else Id (which is T composed with T^-1) will be compact. And we further know: 0 can be the only accumulation point of the spectrum, which is real. Every non-zero number is an eigenvalue corresponding to mutually perpendicular collection of finite dimensional eigenspaces. And if P_n is the orthogonal projection corresponding to the eigenspace L_n, then T=SUM_n=1^infty L_nP_n.

On a non-separable Hilbert space, the kernel of a compact self-adjoint T is not 0.

In each eigenspace we could use Gram-Schmidt, so if T is 1-1, then the associated collection of o.n. eigenvectors is complete. This is a nice way to prove completeness of an o.n. set.

I tried to analyze, almost word-by-word, a paragraph from Reed & Simon's Functional Analysis book on the Fredholm alternative. So this is from p.201 of that text: "The basic principle which makes compact operators important is the Fredholm alternative: If A is compact, then either Af=f has a solution of (I-A)^-1 exists. This is not a property shared by all bounded linear operators. For example, if A is the operator (Af)(x)=xf(x) on L²[0,1], then Af=f has no solutions by (I-A)^-1 does not exist (as a bounded linear operator). In terms of "solving equations" the Fredholm alternative is especially nice: It tells us that if for any f there is at most one g with g=f+Ag,k then there is always exactly one. That is, compactness and uniqueness together imply existence ..." (For a multiplier operator on an L², the spectrum will coincide with the "essential support" of a function.) We discussed these sentences.

If {L_n} is any real sequence decreasing to 0, then there must be a compact operator in l² with 0 and those numbers as spectrum. Here the spectral radius is r_T=||T|| exactly!

Back again to K(x,y), now continuous in [0,1]]². It gives rise to a compact map T from L²[0,1] to itself. But also all of the outputs of T are continuous on [0,1]. So eigenfunctions are continuous. And there is a further restriction on such operators: SUM_n=1^infinty |L_n|²<=the L² norm of K(x,y) on [0,1]]². So the sequence of eigenvalues must be square-summable (a Hilbert-Schmidt operator). In fact, it turns out that the eigenfunction expansion for Tf must converge absolutely and uniformly to Tf, not a totally obvious fact (need a bit of Banach space compact operator theory).

0 0
1 0

I finished proving that an L² kernel in an integral operator defines a compact linear map, completing another verification of the Ansatz previously mentioned. Here we approximated the map by one made up of a finite Fourier series (giving a finite rank map) which approximated the integral operator. We could have done something similar for the continuous case, using Stone-Weierstrass to get, say, polynomials approximating the kernel. These would give finite rank operators uniformly close to the integral operators being considered.

For the long-desired Spectral Theorem for Compact Self-Adjoint Operators on a Hilbert Space, I listed the following results:

1. Eigenvalues are real.
2. Eigenspaces corresponding to different eigenvalues are orthogonal.
3. Eigenspaces corresponding to a non-zero eigenvalue are finite dimensional.
4. If T is compact and self-adjoint, sup|<T(h),h>| when ||h||=1 is the same as ||T||.
5. If T is compact and self-adjoint, and L (corresponds to lambda) is non-zero, and if inf||T(h)-Lh|| for ||h||=1 is 0, then L is an eigenvalue of T.

I tried to give a proof of this last result, and happily accepted help from Ms. Bao. Take a minimizing sequence, and, since T is compact, pass to a subsequence, which I will call {h_j} with ||h_j||=1 so that ||T(h_j)-Lh_j||-->0 and T(h_j)-->f. Then I claim that f is not 0. If it were, then the minimizing requirement ||T(h_j)-Lh_j||-->0 would fail, since then h_j=(1/L)[Lh_j-T(h_j)+T(h_j)]-->(1/L)f, and therefore ||h_j||-->||(1/L)f|| and the sequence is 1 always. But since h_j-->(1/L)f, we see that T(h_j)-->(1/L)T(f). But then f=(1/L)T(f) and we are done.

You can see some references to the Palais-Smale condition, which I mentioned in class, in this book review.

I stated a version of the Spectral Theorem. More to come!

I finally defined compact operator: the transform of a bounded sequence has a convergent subsequence. So the image of the unit ball is relatively compact. I went over conditions related to compactness in a metric space, included being totally bounded.

The identity mapping is compact iff dim H <infty (so eigenspaces of compact maps are finite dimensional). A compact mapping is continuous. The sum of compact mappings is compact. The compact maps are a two-sided ideal in L(H). The compact mappings are norm-closed in L(H). A map in L(H) is compact iff it is the limit of maps with finite dimensional range. The proofs used 3 epsilon arguments and the totally bounded part of compactness. Several statements made which would become homework problems.

We analyzed the spectrum of T(e_j)=(1/j)e_j, a compact map. We introduced the Hilbert cube (sum of a_je_j with |a_j|<=1/j), a compact set with empty interior.

I stated the Ansatz (used as Professor Zeilberger does, meaning a principal or motivating idea) that "Integral operators are compact." (Of course, the Bergman kernel immediately shows this is false.)

I began two true versions of this ansatz:
Suppose we have a space of functions, A, on [0,1], and K(x,y) is a function on [0,1]², then define for f in A, Tf(x)=₀¹K(x,y)f(y) dy. Then T is compact. This is true when A is the collection of continuous functions with the sup norm, and K is continuous on [0,1]². I verified that the hypotheses of the Arzela-Ascoli Theorem are satisfied. This is also true when A is the collection of L² functions with the L² norm, and K is L² on [0,1]². I need to prove this. And the Spectral Radius Formula. And the Spectral Theorem for compact self-adjoint operators.

3 0
0 2

If A is in L(H), then ker(A) is the orthogonal complement of im(A^*). It is not generally true that the orthogonal complement of the ker(A) is the image: we got an example (just e_j-->(1/j)e_j) where the image is dense in H.

Definition of the spectrum, the resolvent, and the point spectrum. How self-adjoint operator T in Cⁿ has eigenvalue at least one of +/-||T||. If self-adjoint, eigenvalues are real, and eigenvectors of different eigenvalues are orthogonal. Example in Cⁿ of how a Hermitean matrix has an o.n. basis of eigenvectors. The use of the Rayleigh-Ritz method to approximate eignevalues and eigenvectors by multiplications of the matrix on a "random" vector, together with genericity assumptions. (Can also be done for other transformations, including a large class of integral operators.)

Then a quote from p. 196 of Conway's book on how holomorphic functions clearly extend (!?). There was some discussion of these statements, including how to define integration of vector-valued continuous functions on curves in C.

BIG Theorem
The spectrum of T is always a compact non-empty subset of C.

Proof used (I-T)^-1=I+T+T²+T³+... when ||T||<1 several times, coupled with an amazing citation of a vector-valued version of Liouville's Theorem.

A spectral theorem in this context seeks to reconstruct T from information about the spectrum of T, maybe the resolvent of T, maybe auxiliary assumptions about T. Evidence that some hypotheses are needed is the map in C² given by

0 0 
1 0

L(H) is the set of continuous linear maps from a Hilbert space to itself, the operators on H. What do we know? L(H) has the operator norm, in which it is a Banach space. Also, if composition is used as multiplication, and S and T are operators, so is ST and ||ST||<:=||S|| ||T||: L(H) is a normed algebra. Although all Hilbert spaces are "the same"(depending on the cardinality of an orthonormal basis) [I mean all "places" in H are sort of the same] the operators are very very very different and complicated. We will see this.

Simple examples of operators: multiply by an L^infty function; integrate against a kernel (with uniform L¹ bounds in each variable); d/dx in an incomplete space; rightshift (an injective not surjective isometry) (same as multiplication by z in H²) .

We will try to prove a spectral theorem for compact self-adjoint operators. I've got to discuss everything in green.

What's here follows If u(h,k) is a bounded sesquilinear mapping from HxH to the ground field, then u(h,k)=<Ah,k>=<h,Bk> for some unique operators A and B, with bounds by M on both norms. This can be used to define A^*. A is self-adjoint if A=A^*. A is normal if AA^*=A^*A. Adjoints for the examples above were given, in addition to noting that the adjoint of a finite dimensional operator given by a matrix is the transpose/conjugate.

Here are some simple facts: (A+B)^*=A^*+B^* and (aA)^*=(a-bar)A^* and (AB)^*=B^*A^* and A^**=A.

Less obvious: ||A||=||A^*||=sqrt{||AA^*||} (start with ||A(h)||² and compute. Perhaps even less obvious: If H is a complex Hilbert space, then A is self-adjoint if and only if <Ah,h> is real for all h in H. (Proof: "expand" both <A(h+k),h+k> and <A(h+ik),h+ik>. Note that the last result is false for real Hilbert spaces: the matrix

0  1
-1 0

• {w_k}_{k in K} is a Hilbert space basis.
• If h is in H and <h,w_k>=0 for all k, then h=0.
• If h is in H, then h=(unordered!)SUM_{k in K}<h,w_k>w
• If h₁ and h₂ are in H, then <h₁,h₂>=(unordered!)SUM_{k in K}<h₁,w_k> <h₂,w_k>
• If h is in H, then ||h||²=(unordered!)SUM_{k in K}|<h,w_k>|²