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?

  • $\begingroup$ $K-K_{NM}$ is again a kernel operator. Apply the inequality to this operator instead of the original one. $\endgroup$
    – MaoWao
    Feb 5, 2019 at 19:26
  • $\begingroup$ @MaoWao That's what I've been trying to do, but I can't see how to proceed. I can only seem to use the reverse triangle inequality, i.e., $$||K-K_{NM}||\geq |\ ||K||-||K_{NM}||\ |$$ But this doesn't seem to help me. $\endgroup$
    – James
    Feb 5, 2019 at 19:28

1 Answer 1


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}

  • $\begingroup$ Thank you for the response! I am not sure I get the first line. It's the inner products I can't understand. Wouldn't it be $\langle x,e_n\otimes\overline{e_m}\rangle e_n\otimes\overline{e_m}$? The rest of your proof makes sense to me. $\endgroup$
    – James
    Feb 5, 2019 at 19:40
  • $\begingroup$ Do you know how $e_n\otimes \overline{e_m}$ acts on $x=\sum \langle e_\ell \mid x\rangle e_\ell$? $\endgroup$ Feb 5, 2019 at 19:46
  • $\begingroup$ I am not sure. I only know that $e_n\otimes \overline{e_m}=e_n(x)\overline{e_m(y)}$. One thing I did forget to mention is that $\{e_n\}$ itself is an orthonormal basis for $L^2(I)$. $\endgroup$
    – James
    Feb 5, 2019 at 19:49
  • 1
    $\begingroup$ Let me give you a hint. There's a difference between $e_n\otimes \overline{e_m}$ as an operator vs. the kernel of $e_n\otimes \overline{e_m}$ which is $e_n(x)\overline{e_m(y)}$. $\endgroup$ Feb 5, 2019 at 19:51
  • $\begingroup$ Thanks. I can't quite see it, but I'm assuming what you mean is that I'm missing how we can express $K$ operating on an element $x\in L^2(I)$, and I'm only using the expansion for the kernel $k$? Also, do you have any material or references you can advice me to read to get a better understanding? $\endgroup$
    – James
    Feb 5, 2019 at 20:41

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