Cfrgtkky

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.

Take $B(f,g)=langle f,grangle$. Then, for all $f,g$,
$$
int_{[0,1]} f(x)g(x),dx=langle f,grangle=int_{[0,1]^{otimes2}}xi (x,y)f(x)g(y)dxdy.
$$
It follows that, for all $g$,
$$
int_0^1xi(x,y),g(y),dy=g(x).
$$
Take in particular $g(x)=1_{[0,c]}$ for some $cin (0,1)$. Then, for all $xin [0,c]$,
$$
int_0^cxi(x,y),dy=int_0^1xi(x,y),1_{[0,c]}(y),dy=1_{[0,c]}(x)=1.
$$
Now, for any $c,din(0,1)$,
$$
int_c^dxi(x,y),dy=int_0^dxi(x,y),dy-int_0^cxi(x,y),dy=1-1=0.
$$
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.

(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
$$
lVert TrVert^2_{text{HS}}=sum_{iin I} lVert Tphi_irVert^2 = sum_{i,jin I} |langle Tphi_i, phi_jrangle|^2<infty
$$ for some orthonormal basis ${phi_i}_{iin I}$ of $H$. We can show that the above definition does not depend on the choice of ${phi_i}_{iin I}$ because if ${varphi_k}_{kin K}$ is another orthonormal basis, it holds that
$$begin{eqnarray}
sum_{iin I} lVert Tphi_irVert^2 &=& sum_{iin I,kin K} |langle Tphi_i, varphi_krangle|^2 \&=& sum_{iin I,kin K} |langle phi_i, T^*varphi_krangle|^2 \&=& sum_{l,kin K} |langle varphi_l, T^*varphi_krangle|^2\&=&sum_{l,kin K} |langle Tvarphi_l, varphi_krangle|^2 \&=& sum_{l in K} lVert Tvarphi_lrVert^2,
end{eqnarray}$$ by Parseval's identity. Suppose $T$ is a Hilbert-Schmidt operator such that
$$
T: phi_j mapsto sum_{iin I} a_{ij}phi_i,
$$ for all $jin I$. Then its Hilbert-Schmidt norm is
$$
lVert TrVert^2_{text{HS}}=sum_{jin I} lVert Tphi_jrVert^2 = sum_{i,jin I} |langle Tphi_j, phi_irangle|^2= sum_{i,jin I} |a_{ij}|^2 <infty.
$$ Now, let us define $$xi(x,y) = sum_{i,jin I} a_{ij}phi_i(x)overline{phi_j(y)}.$$ Since ${phi_i(x)overline{phi_j(y)}}_{i,jin I}$ 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
$$
int_{[0,1]} xi(x,y)phi_j(y)dy = sum_{iin I} a_{ij}phi_i(x)=Tphi_j(x),
$$as desired. Finally, for $B$ to admit Hilbert-Schmidt operator $tau_B$, it is sufficient and necessary that
$$
lVert tau_BrVert^2_{text{HS}}=sum_{i,jin I} |langle tau_Bphi_i, phi_jrangle|^2 = sum_{i,jin I} |B(phi_i,phi_j)|^2 <infty,
$$ which is stronger than the original assumption $$|B(f,g)|leq MlVert frVertlVert grVert.$$

For every $g$ the mapping $fmapsto B(f,g)$ is a bounded linear functional, so by Riesz Representation Theorem there exists a unique $phi_gin L^2$ such that $B(f,g)=langle f,phi_grangle$ for all $f$. Notice that $|phi_g|_2leq M|g|_2$, so $gmapstophi_g$ is bounded and anti-linear and hence continuous.

Let ${e_n}$ be an orthonormal basis of $L^2$. Then for all $f,gin L^2$ we have
begin{align}
B(f,g)
&=langle f,phi_grangle
\&=leftlangle f,sumphi_{langle g,e_nrangle e_n}rightrangle
\&=sumlangle g,e_nranglelangle f,phi_{e_n}rangle
\&=sumint_0^1g(y)overline{e_n(y)}mbox{ d}yint_0^1f(x)overline{phi_{e_n}(x)}mbox{ d}x
\&=sumint_{[0,1]^2}overline{phi_{e_n}(x)e_n(y)}f(x)g(y)mbox{ d}xmbox{ 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 $e_n=e^{2pi incdot}$. Here $phi_{e_n}=e^{-2pi incdot}$. You can verify that, indeed, $B(f,g)$ equals the sum of the integrals. However, I do not know whether $sum E_n$ converges.