Help understanding proof that Hilbert-Schmidt operators are compact I am trying to understand a proof that Hilbert-Schmidt operators are compact. From the book that I am reading it is stated that for a Hilbert Schmidt operator $K$ with kernel
$$k=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\langle k,e_n\otimes \overline{e_m}\rangle e_n\otimes \overline{e_m}$$
Here $e_n\otimes \overline{e_m}=e_n(x)\overline{e_m(y)}$ is an orthonormal basis for $L^2(I\times I)$ (this has been proven and I also understand this).
Then one can consider the finite rank operators $K_{NM}$ with kernel
$$k_{NM}=\sum_{n=1}^{N}\sum_{m=1}^{M}\langle k,e_n\otimes \overline{e_m}\rangle e_n\otimes \overline{e_m}$$
The book proceeds to state that since $\|K\|\leq \|k\|$ (has also been proven and I understand this as well) that it is possible to approximate $K$ by the finite rank operators $K_{NM}$ in operator norm. This is what I'm struggling with. I can't seem to be able to show that 
$$\|K-K_{NM}\|\rightarrow 0$$
as $N,M\rightarrow \infty$. I've tried but I can't seem to get from $\|K-K_{NM}\|$ to a situation in which I can use that $\|K\|\leq \|k\|$. Does anyone have any hints?
 A: Sketch: Observe
\begin{align}
Kx = \sum^\infty_{n=1} \sum^\infty_{m=1} \langle e_m\mid ke_n\rangle \langle e_n\mid  x\rangle\ e_m 
\end{align}
then it follows
\begin{align}
(K-K_{NM})x =&\ \left(\sum^\infty_{n=N+1} \sum^\infty_{m=M+1}+\sum^N_{n=1}\sum^\infty_{m=M+1}+\sum^\infty_{n=N+1}\sum^M_{m=1} \right) \langle e_m\mid ke_n\rangle \langle e_n\mid  x\rangle e_m\\\
=:&\ HH+LH+HL.
\end{align}
Take the $L^2$-norm, we get
\begin{align}
\|HH \|_2^2 =&\ \left\|\sum^\infty_{m=M+1} \left(\sum^\infty_{n=N+1}\langle e_m\mid ke_n\rangle \langle e_n\mid x\rangle \right) e_m\right\|_2^2\\
=&\ \sum^\infty_{m=M+1}\left|\sum^\infty_{n=N+1}\langle e_m\mid ke_n\rangle \langle e_n\mid x\rangle \right|^2\\
\leq&\ \sum^\infty_{m=M+1}\left(\sum^\infty_{n=N+1}|\langle e_m\mid ke_n\rangle|^2 \right)\left(\sum^\infty_{n=N+1}|\langle e_n\mid x\rangle|^2\right)\\
\leq&\ \left(\sum^\infty_{m=M+1}\sum^\infty_{n=N+1}|\langle e_m\mid ke_n\rangle|^2\right)\|x\|^2_2
\end{align}
and
\begin{align}
\|HL\|_2^2 =&\ \left\|\sum^M_{m=1} \left(\sum^\infty_{n=N+1}\langle e_m\mid ke_n\rangle \langle e_n\mid x\rangle \right) e_m\right\|_2^2\\
=&\ \sum^M_{m=1}\left|\sum^\infty_{n=N+1}\langle e_m\mid ke_n\rangle \langle e_n\mid x\rangle \right|^2\\
\leq&\ \sum^M_{m=1}\left(\sum^\infty_{n=N+1}|\langle e_m\mid ke_n\rangle|^2 \right)\left(\sum^\infty_{n=N+1}|\langle e_n\mid x\rangle|^2\right)\\
\leq&\ \left(\sum^M_{m=1}\sum^\infty_{n=N+1}|\langle e_m\mid ke_n\rangle|^2\right)\|x\|^2_2.
\end{align}
Hence it follows
\begin{align}
\|K-K_{NM}\|_\text{op} \leq \left(\sum^\infty_{m=M+1}\sum^\infty_{n=N+1}|\langle e_m\mid ke_n\rangle|^2+\sum^M_{m=1}\sum^\infty_{n=N+1}|\langle e_m\mid ke_n\rangle|^2+\sum^\infty_{m=M+1}\sum^N_{n=1}|\langle e_m\mid ke_n\rangle|^2\right)^{1/2}\rightarrow 0
\end{align}
as $N, M \rightarrow \infty$. 
Additional Remark: For intuition,  consider the case when $V$ is a finite-dimensional inner product space. 
Let $T:V\rightarrow V$ be a linear map and $\mathcal{B}=\{e_k\}^n_{k=1}$ be an orthonormal set. Then we see that $T$ has a matrix representation given by
\begin{align}
[T]_\mathcal{B}=A:=
\begin{pmatrix}
\langle e_1\mid Te_1 \rangle &  \langle e_1\mid Te_2 \rangle & \cdots & \langle e_1\mid Te_n \rangle\\
\langle e_2\mid Te_1 \rangle &  \langle e_2\mid Te_2 \rangle & \cdots & \langle e_2\mid Te_n \rangle\\
\vdots& \vdots & \ddots & \vdots\\
\langle e_n\mid Te_1 \rangle &  \langle e_n\mid Te_2 \rangle & \cdots & \langle e_n\mid Te_n \rangle\\
\end{pmatrix}
\end{align}
since
\begin{align}
Tx=&\ T\left(\sum^n_{i=1}\langle e_i\mid x\rangle e_i \right)= \sum^n_{i=1}\langle e_i\mid x\rangle Te_i= \sum^n_{i=1}\langle e_i\mid x\rangle \left(\sum^n_{j=1}\langle e_j \mid Te_i\rangle e_j \right)\\
=&\ \sum^n_{j=1}\left(\sum^n_{i=1}\langle e_j \mid Te_i\rangle\langle e_i\mid x\rangle\right)e_j
\end{align}
