Definition of Compact Mapping
I was reading around the other day and came across the term "compact mapping". After googling, I saw the following two definitions:
Let $X$ be a topological space. Then a mapping $f:X \to X$ is compact if $f^{-1}(\{x\})$ is compact for every $x \in X$.
Let $X$ be a Banach space. Then a mapping (not necessarily linear) $f:X \to X$ is compact if the closure of $f(Y)$ is compact whenever $Y \subset X$ is bounded.
Are these definitions equivalent if $X$ is a Banach space? If not, what is the usual meaning in the context of Banach spaces? For example, Schaefer's Fixed Point Theorem states
If $X$ is a Banach space and $f:X \to X$ is a continuous and compact mapping such that $$\{x \in X: x = \lambda f(x) \mbox{ for some } 0 \leq \lambda \leq 1\}$$ is bounded then $f$ has a fixed point.
Which definition is meant? Sorry if I am missing something obvious here.
$\endgroup$ 32 Answers
$\begingroup$The first is true and the second is the definition of a Compact operator (different from a mapping at all and necessarily linear) on a Banach space by this changes: Let $X$ be a Banach space. Then a linear operator ( necessarily linear) $f:X\to X $ is compact if the closure of $f(Y)$ is relatively compact whenever $Y\subset X$ is bounded.
$\endgroup$ 1 $\begingroup$The question seems to be answered in comments. For the sake of not leaving question unanswered, let me copy here the texts of the comments:
Grasshopper: They're not equivalent. Consider the function $f: \mathbb{R} \to \mathbb{R}$ that is constantly $0$. $f$ satisfies condition 2 but not condition 1.
Philip Brooker: Also, the identity mapping on an infinite dimensional Banach satisfies condition 1, but not condition 2.
Adam Smith: In the context of Banach spaces (for instance, in the statement of the fixed point theorem you quoted), the second definition is (almost) always what is meant.
