# 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:

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.

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

## 2 Answers

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.

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.

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