Prove existence/uniqueness of a certain operator on Hilbert Space 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?
 A: 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.  
A: 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.
A: 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.
