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I was reading around the other day and came across the term "compact mapping". After googling, I saw the following two definitions:

  1. Let $X$ be a topological space. Then a mapping $f:X \to X$ is compact if $f^{-1}(\{x\})$ is compact for every $x \in X$.

  2. 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.

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    $\begingroup$ 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. $\endgroup$ – Grasshopper Nov 30 '11 at 6:45
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    $\begingroup$ Also, the identity mapping on an infinite dimensional Banach satisfies condition 1, but not condition 2. $\endgroup$ – Philip Brooker Nov 30 '11 at 7:16
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    $\begingroup$ 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. $\endgroup$ – Adam Smith Nov 30 '11 at 7:18
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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.

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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.

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  • $\begingroup$ Welcome to Math SE. Please try to use LaTeX to format math. $\endgroup$ – Martin Argerami Dec 25 '12 at 7:37

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