# Bounded operator and Compactness problem

Let $H$ be a Hilbert space with orthonormal basis $(e_{n})_{n\in\mathbb{N}}$. Furthermore, let $T\colon H\rightarrow C[a,b]$ be a bounded operator.

a) Let $x\in [a,b]$. Show that there is a unique $g_{x}\in H$ with $\langle f,g_x\rangle=(Tf)(x)$ and all $f \in H$.

Updated version 1. So far I have:

Let $x\in[a,b]$, define the linear continuous map $L_{x}:H\rightarrow \mathbb{K}$, with $\mathbb{K}=\mathbb{C}$ or $\mathbb{K}=\mathbb{R}$ by $$L_{x}(f):=T(f)(x).$$

Since $H$ is an Hilbert space and $L_{x}:H\rightarrow \mathbb{K}$ is a bounded linear functional on $H$ we can apply the Riesz-Frechet theorem. According to the Riesz-Frechet theorem there exists a unique $g_{x}\in H$ such that for $x\in[a,b]$ and all $f\in H$ $$L_{x}(f)=T(f)(x)=(Tf)(x)=\langle f,g_{x}\rangle.$$

Question 1: have I proven it correctly? Or am I missing some important details?

b) Show that $$\sup_{x\in [a,b]} \sum_{j=1}^{\infty}|(Te_{j})(x)|^{2}<\infty.$$

A hint is that $||g_{x}||\leq ||T||_{H \rightarrow C[a,b]}$.

Updated version 1.

Question 2: how do I prove this hint? I want to prove it before I make use of it.

For the rest of the problem I have this so far:

Using part a), we have that: $$\sup_{x\in [a,b]} \sum_{j=1}^{\infty}|(Te_{j})(x)|^{2}=\sup_{x\in [a,b]} \sum_{j=1}^{\infty}|\langle e_{j},g_{x}\rangle|^{2}.$$ Since the $e_{j}$ form an orthonormal basis in $H$, Bessel's inequality yields $$\sup_{x\in [a,b]}\sum_{j=1}^{\infty}|\langle e_{j},g_{x}\rangle|^{2}\leq \sup_{x\in [a,b]}||g_{x}||_{2}^{2}.$$

Now by definition of the operator norm we have: $$\sup_{x\in [a,b]}\sum_{j=1}^{\infty}|\langle e_{j},g_{x}\rangle|^{2}\leq \sup_{x\in [a,b]}||g_{x}||_{2}^{2}\leq ||T||\cdot ||g_{x}||.$$

Using the hint and the fact that operator $T$ is bounded we get: $$\sup_{x\in [a,b]}\sum_{j=1}^{\infty}|\langle e_{j},g_{x}\rangle|^{2}\leq \sup_{x\in [a,b]}||g_{x}||_{2}^{2}\leq ||T||\cdot ||g_{x}||\leq ||T||\cdot ||T||_{{H \rightarrow C[a,b]}} <\infty.$$

Question 3: Is the proof now complete or am I missing a detail/making a mistake? In class for example for Bessel's inequality we usually had this form $\sum_{j=1}^{\infty}|\langle f,e_{j}\rangle|^{2}\leq ||f||_{2}^{2}<\infty$.

c) Show that $\sum_{j=1}^{\infty} ||Te_{j}||_{L^{2}}^{2}< \infty$.

Updated version 1. What I have so far:

We define the function $$x\mapsto\sum_{j=1}^{\infty}|Te_{j}(x)|^{2}.$$ We have from part b): $$\sup_{x\in [a,b]} \sum_{j=1}^{\infty}|(Te_{j})(x)|^{2}<\infty.$$

Integrating the function over $(a,b)$ yields

$$\int_{a}^{b}\sum_{j=1}^{\infty}|Te_{j}(x)|^{2}dx<\infty.$$

Now making use of the Fubini-Tonelli theorem we can interchange limit and integral and get:

$$\int_{a}^{b}\sum_{j=1}^{\infty}|Te_{j}(x)|^{2}dx=\sum_{j=1}^{\infty}\int_{a}^{b}|Te_{j}(x)|^{2}dx=\sum_{j=1}^{\infty}||Te_{j}||_{L^{2}}^{2}<\infty.$$

Question 4: Is the proof now complete or am I missing a detail/making a mistake? In fact the interchange is not clear to me. We did treat Fubini in class but only shorty to change integrals not integral and sum. Usually when changing integral and summation we used monotone convergence.

d) Show that $T\colon H\rightarrow L^{2}(a,b)$ is compact.

We have to get the result by making use of estimates.

Updated version 1. I could prove it without estimating and had:

We note that $H$ and $L^{2}(a,b)$ are Hilbert spaces and the operator $T\colon H\rightarrow L^{2}(a,b)$ is a bounded linear operator as given before. Furthermore in part c) we obtained an inequality. Thus we have that $T\colon H\rightarrow L^{2}(a,b)$ is an abstract Hilbert-Schmidt operator. Now we can apply a certain theorem that states that every abstract Hilbert-Schmidt operator is compact.

Now for the proof making use of estimates:

To show that $T$ is compact, we have to approximate $T$ in the operator norm by finite rank operators. For $N\in\mathbb{N}$ define the linear operator $T_{N}\colon H\rightarrow L^{2}(a,b)$.

Question 5: I don't know how to show that $T_{N}$ is of finite-rank? Normally you would have something along the lines that $T_{N}$ has range within $span\{f_{1},\ldots,f_{N}\}$ and hence is of finite-rank. But for this problem I don't see it. I think you can say that $ran(T)\subseteq C[a,b]$, but $C[a,b]$ is infinite diminesional so this confuses me.

Question 6: How do I show that $||T-T_{N}||\rightarrow 0$, so I can conclude that $T$ is a compact operator.

-
Welcome to math.stackexchange. I upvoted your question because it gives an interesting problem and you give your thought about it in a detailed way. – Davide Giraudo Mar 25 '13 at 16:43

(a) You got the idea, but used it in the reversed direction. Define the linear continuous map $L_x\colon H\to \Bbb R$ by $L_x(f):=T(f)(x)$, and then use Riesz-Frechet theorem.
(b) Your hint will give the conclusion after a use of Bessel's equality. To see you hint works, notice that $\lVert g_x\rVert^2\leqslant \lVert T\rVert\cdot \lVert g_x\rVert$ by definition of the operator norm.
(c) Integrate over $(a,b)$ the function $x\mapsto \sum_{j=1}^{+\infty}|T(e_j)(x)|^2$, then use Funbini-Tonelli's theorem to switch the sum and the integral.
(d) Approximate $T$ by a sequence of finite-ranked operators.