In my book it is said that compact operators are automatically bounded since compact operator maps $\lVert u \rVert=1$ which is bound to compact set which is also bounded. (Linear is not mentioned) But I believe this definition of bounded operator, rather than $\lVert Lu\rVert \leq K\lVert u\rVert ~\forall u$, requires linearity.
In this solution, I wonder if $\lVert u_n\rVert \rightarrow 0$, then $\lVert Au_n\rVert$ seems don't have to go to infinity.
However I saw the statement in many places: "Every compact operator on a Hilbert space is a bounded operator."
Please help.
A non-linear operator may easily be "compact" (in the sense that is maps bounded sets to relatively compact sets) without being continuous. Take for example the map $f\colon \mathbb R\to\mathbb R$ defined by $$f(x)=\begin{cases} 1 & x\ge 0 \\ 0& x<0.\end{cases}$$
This is why, in the non-linear case, it is necessary to assume that the operator is continuous, other than compact. See for example the statement of the Schauder fixed point theorem.
In the linear case the continuity assumption is redundant. This is due to a fundamental theorem which asserts that a linear operator is continuous if and only if it maps bounded sets to bounded sets. Since a compact subset of a metric space is necessarily bounded, a compact linear operator always satisfies this condition, and so it is always continuous.