# Linearity and boundedness of a compact operator on Hilbert space

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!

• 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. Commented Dec 2, 2016 at 8:59
• @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? Commented Dec 2, 2016 at 11:07
• 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. Commented Dec 2, 2016 at 11:55

We have (Pythagoras !) with $s=\sup\{|\lambda_j|: j \in \mathbb N\}$
$||Th||^2=\sum_{n=1}^\infty|\lambda_n|^2|<f_n,h>|^2 \le s^2||h||^2$
• 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? Commented Dec 2, 2016 at 13:37