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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!

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

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

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  • $\begingroup$ 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

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