I know that if $f: [a,b] \to [a,b]$ is continuous then by the intermediate value theorem it must have a fixed point. However, I recently saw this theorem stating that given a continuous function $f: \mathbb{R}^n \to \mathbb{R}^n $, if there is some $M >0$ such that $x \in B(0,M): d_2(x,f(x)) < \epsilon$ for every $\epsilon > 0$. Then $f$ has a fixed point.
Intuitively I would say that the fixed point is $x$. But I wouldn't know how to actually show that $f(x) - x = 0$. Any hints?
On
This criteria works only for compact domain and does not necessarily work unbounded domains. Since $[a,b]$ is compact, we have, Let $\epsilon_n = \epsilon/2^n$. Let $M$ be such that we can find $x_n$ such that $|x_n| \leq M$ and $|x_n-f(x_n)| < \epsilon_n$. Now $x_n$ is an infinite sequence in compact set hence has a subsequence which converges. Let $x^*$ be the limit of the subsequence $x_{n_k}$ mentioned above.
Then $\lim_{k \rightarrow \infty} |x_{n_k} - f(x_{n_k})| = 0$. But since $x_{n_k} \rightarrow x^*$, we have $\lim_{k \rightarrow \infty} |x_{n_k} - f(x_{n_k})| = |x^* - f(x^*)|$.
Hence $|x^* - f(x^*)| = 0$ and hence $x^*$ is the fixed point.
The proof is general and works even for continuous functions over compact set in $\mathbb{R}^n$.
Let's prove the following :
Let $X$ is a compact metric space, and $f : X \rightarrow X$ a continuous function such that for every $\varepsilon > 0$, there exists $x \in X$ such that $d(x,f(x)) < \varepsilon$. Then $f$ has a fixed point.
Proof : Let $F : X \rightarrow \mathbb{R}$ defined by $F(x)=d(x,f(x))$.
$F$ is continuous over the compact set $X$, therefore, $F(X)$ is compact.
But by assumption on $f$, for every $n \in \mathbb{N}^*$, there exists $x_n \in X$ such that $F(x_n) < \dfrac{1}{n}$.
The sequence $F(x_n)$ therefore converges to $0$, and since $F(X)$ is closed, then $0 \in F(X)$.
This means exactly that there exists $x \in X$ such that $F(x)=0$, i.e. such that $\boxed{f(x)=x}$.