I want to show that if we have $T \in L(L^2([a,b])))$ an Hilbert-schmidt operator then we have that there exists $k\in L^2([a,b] \times [a,b])$ such that $T(\phi)(t)= \int_a^{b}k(t,s)\phi(s)ds.$
I was thinking maybe from the operator $T$ we can find a functional where we can apply the Riez Representation theorem maybe, but i got nowhere , any hints are aprecciated, just something to get me started, thanks.
Consider the mapping $\Phi: L^2([a,b] \times [a,b]) \rightarrow \mathcal{L}_{HS}(L^2([a,b]))$ given by $((\Phi(k))(f))(t) = \int_a^b{k(t,s)f(s)\,ds}$.
It is known that $\|\Phi(k)\|_{HS}=\|k\|_L^2$. Thus $\Phi_k$ is injective with a complete (thus closed) image.
One easily sees that a finite rank operator is in the image of $\Phi$ (if we have an orthonormal basis of the image $u_1,\ldots,u_n$, then $\langle Tf,\,u_i\rangle=\langle f,\,g_i\rangle$ for some $g_i \in L^2$ and $k(t,s)=\sum_{i=1}^n{\overline{g_i(s)}u_i(t)}$ works).
Finite rank operators are dense in $\mathcal{L}_{HS}(L^2([a,b]))$ for the Hilbert-Schmidt norm, so $\Phi$ is surjective (as its image is closed).