Show $||f(x)-x||$ has a minimum value for $x$ in a compact space.

Question

Let $V$ be a normed vector space and $U \subseteq V$. Assume $f: U \rightarrow V$ is continuous and $U$ is compact. I am trying to show that $||f(x)-x||$ achieves a minimum value for some $x \in U$, but am a bit stuck.

I tried creating a convergent subsequence for $x_n$ and $f(x_n)$, but the intersection of these two subsequences is not guaranteed to be a subsequence, so I couldn't proceed there.

Any help/hints would be greatly appreciated!

Answer 1

I assume $U \subset V$ are subsets of a normed space. Since $U \ni x \mapsto x \in V$ is continuous, so is $x \mapsto \|f(x) - x\|$ (because $x \mapsto \|x\|$ is continuous as well by the triangular inequality and adding preserves continuity). Since $U$ is also compact, the result follows immediately.

Answer 2

Two basic facts:

1. If $g: U \to Y$ is continuous and $U$ is compact then $g[U]$ is compact in $Y$.

2. A compact subset of $\Bbb R$ has a maximum and a minimum.

From this the result follows once you shown that $x \to \|x - f(x)\|$ is continuous whenever $f$ is continuous, which can be shown quite easily (the map $x \to \|x\|$ is continuous, and so is $(x,y) \to x-y$ etc.)