Integration Representation of Bounded Bilinear Functional on $L_2([0,1])$ Suppose $B(f,g)$ is a bilinear functional on $L_2([0,1])$, satisfying $|B(f,g)|\leq M \|f\|_2\|g\|_2$ for some $M>0$. My question is whether there always exists a function $\xi_B(x,y)$ (Maybe essentially bounded?) such that 
$$B(f,g) = \int_{[0,1]^{\otimes2}}\xi_B(x,y)f(x)g(y)dxdy.$$
One possible idea to prove it is to use Riesz Representation theorem first and have 
$B(f,g) = <\tau_B f,g>$, where $\tau_B$ is a bounded linear operator from $L_2([0,1])$ to $L_2([0,1])$ and then there might be some results for an integration representation of $\tau_B$. But I am not sure where can I find results about it.
P.S.: In my research, $B(f,g)$ is also symmetric, i.e. $B(f,g)=B(g,f)$, or, as a result, $\tau_B$ is self-adjoint. It will also be good if you can help me proving integration representation with this extra condition. I don't think it will be very useful though.
 A: Take $B(f,g)=\langle f,g\rangle$. Then, for all $f,g$,
$$
\int_{[0,1]} f(x)g(x)\,dx=\langle f,g\rangle=\int_{[0,1]^{\otimes2}}\xi (x,y)f(x)g(y)dxdy.
$$
It follows that, for all $g$,
$$
\int_0^1\xi(x,y)\,g(y)\,dy=g(x).
$$
Take in particular $g(x)=1_{[0,c]}$ for some $c\in (0,1)$. Then, for all $x\in [0,c]$, 
$$
\int_0^c\xi(x,y)\,dy=\int_0^1\xi(x,y)\,1_{[0,c]}(y)\,dy=1_{[0,c]}(x)=1. 
$$
Now, for any $c,d\in(0,1)$, 
$$
\int_c^d\xi(x,y)\,dy=\int_0^d\xi(x,y)\,dy-\int_0^c\xi(x,y)\,dy=1-1=0.
$$
Then Lebesgue differentiation tells us that, for all $x$, $\xi(x,y)=0$ a.e. $(y)$. 
In summary, $\xi_B$ does not exist in general.
A: (I will interpret $L_2([0,1])$ as a real Hilbert space since you are assuming $B$ is bilinear, not sesqui-liear.)
As @Martin Argerami's argument shows, the statement is in general false. However, under the assumption that $\tau_B$ is a Hilbert-Schmidt operator, we can find $\xi_B \in L_2([0,1]\times[0,1])$ such that
$$
\tau_B f(x) = \int_{[0,1]} \xi_B(x,y)f(y)dy,
$$ and
$$
B(f,g) = \int_{[0,1]\times[0,1]} \xi_B(x,y)g(x)f(y)dxdy.
$$ A Hilbert-Schmidt operator is defined as a bounded linear operator $T$ satisfying
$$
\lVert T\rVert^2_{\text{HS}}=\sum_{i\in I} \lVert T\phi_i\rVert^2 = \sum_{i,j\in I} |\langle T\phi_i, \phi_j\rangle|^2<\infty
$$ for some orthonormal basis $\{\phi_i\}_{i\in I}$ of $H$. We can show that the above definition does not depend on the choice of $\{\phi_i\}_{i\in I}$ because if $\{\varphi_k\}_{k\in K}$ is another orthonormal basis, it holds that
$$\begin{eqnarray}
\sum_{i\in I} \lVert T\phi_i\rVert^2 &=& \sum_{i\in I,k\in K} |\langle T\phi_i, \varphi_k\rangle|^2 \\&=& \sum_{i\in I,k\in K} |\langle \phi_i, T^*\varphi_k\rangle|^2 \\&=& \sum_{l,k\in K} |\langle \varphi_l, T^*\varphi_k\rangle|^2\\&=&\sum_{l,k\in K} |\langle T\varphi_l, \varphi_k\rangle|^2 \\&=& \sum_{l \in K} \lVert T\varphi_l\rVert^2,
\end{eqnarray}$$ by Parseval's identity. Suppose $T$ is a Hilbert-Schmidt operator such that
$$
T: \phi_j \mapsto \sum_{i\in I} a_{ij}\phi_i,
$$ for all $j\in I$. Then its Hilbert-Schmidt norm is
$$
\lVert T\rVert^2_{\text{HS}}=\sum_{j\in I} \lVert T\phi_j\rVert^2 = \sum_{i,j\in I} |\langle T\phi_j, \phi_i\rangle|^2= \sum_{i,j\in I} |a_{ij}|^2 <\infty.
$$ Now, let us define $$\xi(x,y) = \sum_{i,j\in I} a_{ij}\phi_i(x)\overline{\phi_j(y)}.$$ Since $\{\phi_i(x)\overline{\phi_j(y)}\}_{i,j\in I}$ forms an orthonormal subset (in fact basis) of $L_2([0,1]\times[0,1])$, it follows that $\xi \in L_2([0,1]\times[0,1])$. We can observe that 
$$
\int_{[0,1]} \xi(x,y)\phi_j(y)dy = \sum_{i\in I} a_{ij}\phi_i(x)=T\phi_j(x),
$$as desired. Finally, for $B$ to admit Hilbert-Schmidt operator $\tau_B$, it is sufficient and necessary that
$$
\lVert \tau_B\rVert^2_{\text{HS}}=\sum_{i,j\in I} |\langle \tau_B\phi_i, \phi_j\rangle|^2 = \sum_{i,j\in I} |B(\phi_i,\phi_j)|^2 <\infty,
$$ which is stronger than the original assumption $$|B(f,g)|\leq M\lVert f\rVert\lVert g\rVert.$$
A: For every $g$ the mapping $f\mapsto B(f,g)$ is a bounded linear functional, so by Riesz Representation Theorem there exists a unique $\phi_g\in L^2$ such that $B(f,g)=\langle f,\phi_g\rangle$ for all $f$. Notice that $\|\phi_g\|_2\leq M\|g\|_2$, so $g\mapsto\phi_g$ is bounded and anti-linear and hence continuous.
Let $\{e_n\}$ be an orthonormal basis of $L^2$. Then for all $f,g\in L^2$ we have
\begin{align}
B(f,g)
&=\langle f,\phi_g\rangle
\\&=\left\langle f,\sum\phi_{\langle g,e_n\rangle e_n}\right\rangle
\\&=\sum\langle g,e_n\rangle\langle f,\phi_{e_n}\rangle
\\&=\sum\int_0^1g(y)\overline{e_n(y)}\mbox{ d}y\int_0^1f(x)\overline{\phi_{e_n}(x)}\mbox{ d}x
\\&=\sum\int_{[0,1]^2}\overline{\phi_{e_n}(x)e_n(y)}f(x)g(y)\mbox{ d}x\mbox{ d}y
\end{align}
Hence if we define $E_n(x,y):=\overline{\phi_{e_n}(x)e_n(y)}$, then if $\sum E_n$ converges to some $\xi_B$, then this has the desired properties.
However, I did not find out whether this series converges. Not even for a simple example such as $B(f,g):=\langle f,\overline{g}\rangle=\int fg$ with $e_n=e^{2\pi in\cdot}$. Here $\phi_{e_n}=e^{-2\pi in\cdot}$. You can verify that, indeed, $B(f,g)$ equals the sum of the integrals. However, I do not know whether $\sum E_n$ converges.
