An operator $T:\mathcal H\to\mathcal H$ on an infinite-dimensional Hilbert space $\mathcal H$ is said to be compact if it can be written in the form $$ T=\sum_{n=1}^\infty\lambda_n\langle f_n,\cdot\rangle g_n $$ where $f_1,f_2,\ldots$ and $g_1,g_2,\ldots$ are (not necessarily complete) orthonormal sets, and $\lambda_1,\lambda_2,\ldots$ is a sequence of positive numbers with limit zero, called the singular values of the operator.

A compact operator is not necessarily linear. Is that right?

However, a compact operator is necessarily bounded. Is it possible to show that a compact operator is bounded using the above definition? What I would have to show is that $$ \frac{\|Th\|}{\|h\|}\le K $$ with some $K>0$ for all non-zero $h\in\mathcal H$. I could proceed in the following way \begin{align*} \|Th\|&=\biggl\|\sum_{n=1}^\infty\lambda_n\langle f_n,h\rangle g_n\biggr\|\\ &\le\sum_{n=1}^\infty\lambda_n|\langle f_n,h\rangle|\\ &\le\|h\|\sum_{n=1}^\infty\lambda_n. \end{align*} But this requires that $\sum_{n=1}^\infty\lambda_n<\infty$, which is not necessarily true for a compact operator. It is true for a trace class operator, but not all compact operators are trace class operators.

Any help is much appreciated!

  • $\begingroup$ See the definition of compact operator in Wikipedia. It is linear by definition, and because it carries bounded sets to completely bounded sets, it is continuous as well. $\endgroup$ – астон вілла олоф мэллбэрг Dec 2 '16 at 8:59
  • $\begingroup$ @астонвіллаолофмэллбэрг But we don't need the linearity in the definition, do we? Why do we choose to define them as lilnear operators? What is the reason behind this? $\endgroup$ – Cm7F7Bb Dec 2 '16 at 11:07
  • 1
    $\begingroup$ No, we do not need linearity, as you have pointed out. I think it is there just to preserve structure. But once it is linear, it must be continuous as well. $\endgroup$ – астон вілла олоф мэллбэрг Dec 2 '16 at 11:55

We have (Pythagoras !) with $s=\sup\{|\lambda_j|: j \in \mathbb N\}$

$||Th||^2=\sum_{n=1}^\infty|\lambda_n|^2|<f_n,h>|^2 \le s^2||h||^2$

  • $\begingroup$ Thanks for the answer (+1)! $s$ is actually equal to the operator norm of the operator $T$, right? Is it possible to define a compact operator that is not necessarily a linear operator? $\endgroup$ – Cm7F7Bb Dec 2 '16 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.