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Trying to solve this question: Let $\mathcal{H},(\cdot,\cdot)$ be a real Hilbert space, and $\{e_n\}_{n=1}^\infty$ an orthonormal basis on $\mathcal{H}$. Let $F:[0,1]\rightarrow \mathcal{H}$ be continuous. Show that there exists a unique positive self-adjoint operator $T \in B(\mathcal{H})$ such that:

$$(Tx,y) = \int_0^1(F(t),x)(F(t),y)dt \quad \text{ for all } x,y \in \mathcal{H}.$$

Also show that $T$ is compact.

As hint we got that we may use $\lim_{N\rightarrow \infty}\int_0^1\sum_{n=N+1}^\infty|(F(t),e_n)|^2\,dt = 0$.

So far all I've come up with is that $\int_0^1 (F(t),F(t))dt = \lim_{N \rightarrow \infty}\int_0^1\sum_{n=1}^N|(F(t),e_n)|^2dt$ but I have no idea where I'm going with this.

Is there a smarter way to doing this than actually finding an operator which fits all the conditions?

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3 Answers 3

Hint (1): Show that $\int_0^1 (F(t),x)(F(t),y)dt$ is a bounded bilinear operator. The rest would follow from Riesz Representation Theorem.

Edit: I add an extra hint.

Hint (2): A bilinear map $\sigma: H\times H\to\mathbb R$ is bounded if there exists $M$ such that $|\sigma(x,y)|\leq M\|x\| \|y\|$. If $\sigma$ is a bounded bilinear map then there exists a bounded operator $u:H\to H$ so that $\sigma(x,y)=\langle u(x), y\rangle$ (here $\langle, \rangle$ denotes the inner product of $H$). Moreover, if $\sigma$ is positive definite then $u$ is positive and self-adjoint. This follows (easily) from the Riesz Representation Theorem you mentioned in the comments.

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All right, I'll see if I can work this out. –  BallzofFury Jul 18 '12 at 12:41
    
How its stated in my book doesn't seem applicable here to me. For $\mathcal{H}$ a Hilbert space and $f \in B(\mathcal{H},\mathcal{R})$ there exists a unique $y \in \mathcal{H}$ such that $f_y(x) = (x,y)$ for all $x \in \mathcal{H}$. If we do this in each argument, doesn't it just prove it for a single point in each argument, and not for all combinations of $x$ and $y$? –  BallzofFury Jul 18 '12 at 13:08
    
@BallzofFury That's not the version I had in mind. I'll add a further hint. –  azarel Jul 18 '12 at 17:45
    
Right, worked that out. Right now I have everything except for the boundedness of $T$. I tried working it out as $|Tx|^2=\langle Tx,Tx\rangle=\langle TTx,x\rangle \leq |T||Tx||x|$ but I can't get any further. –  BallzofFury Jul 18 '12 at 21:26
    
@BallzofFury To prove that the integral is bounded just apply Caucy-Schwarz inequality and the compactness of $[0,1]$. –  azarel Jul 18 '12 at 21:56
up vote 1 down vote accepted

For the interested: To prove the existence of such an operator, let $x \in \mathcal{H}$ and define the following function $f_x(y):= \int_0^1 (F(t),x)(F(t),y)\,dt$. We will show that this function is bounded, namely: \begin{align*} |f_x(y)| &= |\int_0^1 (F(t),x)(F(t),y)\,dt| \\ &\leq \int_0^1 |(F(t),x)(F(t),y)|\,dt \\ & = \int_0^1 |(F(t),x)||(F(t),y)|\,dt \\ &\leq |x|\int_0^1 |F(t)|^2\,dt |y| \quad \text{(Cauchy-Schwarz)} \\ &\leq|x| \max_{t \in [0,1]}|F(t)|^2 |y| \end{align*} which gives that $f$ is a bounded operator since $f$ is continuous in the compact interval $[0,1]$, so attains a maximum. Due to the fact that $f$ is bounded, by the Riesz-Fr\'echet representation theorem there exists a unique $z \in \mathcal{H}$ such that $f(x) = (z,y)$. We then define $T$ such that $T(x) = z$. This shows that there exists a unique $T$ such that for all $x,y \in \mathcal{H}$ we have $(Tx,y) = \int_0^1 (F(t),y)(F(t),x)\,dt$.

Now we want to show that $T \in B(\mathcal{H})$, so that it is linear and bounded. For linear, let $x_1,x_2 \in \mathcal{H}$ and $\lambda,\mu \in \mathcal{R}$. Then: \begin{align*} (T(\lambda x_1 + \mu x_2),y) &= \int_0^1 (F(t),\lambda x_1 + \mu x_2)(F(t),y)\,dt \\ &= \int_0^1 (\lambda(F(t),x_1) + \mu(F(t),x_2))(F(t),y)\,dt \\ &= \lambda\int_0^1 (F(t),x_1)(F(t),y)\,dt + \mu\int_0^1 (F(t),x_2)(F(t),y)\,dt \\ &= \lambda(Tx_1,y) + \mu(Tx_2,y) = (\lambda Tx_1,y) + (\mu Tx_2,y) \end{align*} which proves linearity. Now for the boundedness of $T$, take $x \in \mathcal{H}$ then we have by Cauchy-Schwarz that: \begin{align*} |Tx|^2 &= (Tx,Tx) \\ &= \int_0^1 (F(t),x)(F(t),Tx)\,dt \\ &= \int_0^1 (F(t),x)\int_0^1 (F(t),x)(F(t),F(t))\,dt\,dt \\ &= \int_0^1 (F(t),x)^2 |F(t)|^2 \,dt \\ &\leq \left(\int_0^1 |F(t)|^4\,dt\right) |x|^2 \leq \max_{t \in [0,1]}|F(t)|^4 |x|^2 \end{align*} which in turn implies that $|Tx| \leq \max_{t\in[0,1]}|F(t)|^2 |x|$. Therefore $T \in B(\mathcal{H})$.

We will now show that $T$ is a positive operator. Let $x \in \mathcal{H}$, then: \begin{align*} (Tx,x) &= \int_0^1 (F(t),x)(F(t),x)\,dt \\ &= \int_0^1 (F(t),x)^2\,dt \geq 0 \end{align*} since $(F(t),x)^2 \geq 0$ for all $x \in \mathcal{H}$ and all $t \in [0,1]$. For the self-adjoint condition, this is easily seen since we are working on a real Hilbert space. Then: \begin{align*} (Tx,y) &= \int_0^1 (F(t),x)(F(t),y)\,dt \\ &= \int_0^1 (F(t),y)(F(t),x)\,dt \\ &= (Ty,x) = (x,Ty). \end{align*}

As the last required property, we will show that $T$ is a compact operator. Look at the following: \begin{align*} \sum_{n=1}^\infty |Te_n|^2 &= \sum_{n=1}^\infty (Te_n,Te_n) \\ &= \sum_{n=1}^\infty \int_0^1 (F(t),e_n)\int_0^1 (F(t),e_n)(F(t),F(t))\,dt\,dt \\ &= \int_0^1 |F(t)|^2 \sum_{n=1}^\infty (F(t),e_n)(e_n,F(t)) \,dt \\ &= \int_0^1 |F(t)|^2 |F(t)|^2\, dt \quad \text{(by Parseval relation Q3.24)} \\ &\leq \max_{t \in [0,1]}|F(t)|^4 \end{align*} which is bounded by the compactness of $[0,1]$, which means that $T$ is a Hilbert-Schmidt operator and therefore compact.

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This is just a hint.

If the operator exists, then uniqueness, positivity and self-adjointedness (?) all follow from the formula.

For existence, consider the numbers $\alpha_{n,m} = \int_0^1(F(t),e_n)(F(t),e_m)dt$. Show that $\sum |\alpha_{n,m}|^2 < K$, for some $K$. Then define $T$ so that $\alpha_{n,m} = \langle T e_n, e_m \rangle$. Show $T$ is bounded.

To show compactness, show that $T$ is the norm limit of a sequence of finite rank operators (each of which is compact by finite rank). The norm limit of a sequence of compact operators is compact. The form of $T$ should suggest an obvious finite rank operator.

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