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!
