Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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.

share|cite|improve this answer
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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.