Let $A\subset\mathbb{R}$ and $k\in L^2(A^2;\mathbb{C})$. I am trying to understand how we can conclude that the integral operator $K$ defined to act on $L^2(A;\mathbb{C})$ by $Ku(x) := \int_Ak(x,y)u(y) \mathrm dy$ is a Hilbert-Schmidt operator, namely that $\sum_{i=1}^\infty \|Ke_i\|_{L^2(A)}^2 < \infty$, if $\{e_j\}$ is a Hilbert basis for $L^2(A;\mathbb{C})$. It is clear to me that the product functions $(e_ie_j)(x,y) := e_i(x)e_j(y)$ form a Hilbert basis for $L^2(A^2;\mathbb{C})$ and that the kernel function $k(.,.)$ is an $L^2(A;\mathbb{C})$ function for almost every fixed $x,y$, meaning that $k(.,y), k(x,.)\in L^2(A;\mathbb{C})$ for almost every $x,y\in A$.
My current problem is that 1.) I don't know what I should compute $\|Ke_i\|_{L^2(A)}$ to be and/or 2.) the only upper bound I can find for the terms $\|Ke_i\|_{L^2(A)}$ is $k$'s $L^2$ norm, $\int_{A^2}|k(x,y)|^2dxdy$. To be more precise,
we have
$$\|Ke_i\|_{L^2(A)}^2 = \int_A\left|\int_Ak(x,y)e_i(y) \mathrm dy\right|^2 \mathrm dx$$
and while we can surely represent
$$k = \sum_{j,l=1}^\infty \langle k,e_je_l\rangle e_je_l$$
I don't know how to use this, since the extra $e_j,e_l$ terms are not conjugated, so we just get
$$\|Ke_i\|_{L^2(A)}^2 = \int_A\left|\sum_{j,l=1}^\infty\langle k,e_je_l\rangle \int_Ak(x,y)e_i(y)e_j(x)e_l(y) \mathrm dy\right|^2 \mathrm dx$$
and so I can't see any way to use the orthonormality of the $e_i$s. What should I do?
$$k=∑K(\bar e_n)⊗e_n$$ and the $K(\bar e_n)⊗e_n$s are pairwise orthogonal, hence $$\|k\|_2^2=\sum\|K(\bar e_n)\otimes e_n\|_{L^2(A^2)}^2=\sum\|K(\bar e_n)\|_{L^2(A)}^2=:\|K\|_{HS}^2.$$