# Two fixed point theorems in Banach space

Here is a famous fixed point theorem in finite dimension by Brouwer:

Theorem 1 (Brouwer fixed point theorem) Let ${M \subset {\mathbb R}^n}$ be a convex compact set, for any continuous function ${f : M \rightarrow M}$, there exists a point ${x_0 \in M}$ such that ${f (x_0) = x_0}$.

There are couple of ways to extend this theorem to Banach spaces. First recall that a mapping between two Banach spaces is called compact, if the mapping is continuous (not necessarily linear) and the images of bounded sets are pre-compact.

# Hausdorff Measure and Capacity

Definition 1 Let ${E \subset {\mathbb R}^n, 0 \le d < \infty, 0 < \delta \le \infty}$. Define

$\displaystyle \mathcal{H}^d_\delta(E) : = \inf \bigg\lbrace (1/2)^d \alpha(d) \sum_{i=1}^\infty diam(A_i)^d | E\subset \bigcup_{i=1}^\infty A_i, diam(A_i)\le \delta \bigg\rbrace,$

where ${\alpha(d) := \frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2} + 1)}}$.

$\displaystyle \mathcal{H}^d (E) := \lim_{\delta \rightarrow 0} \mathcal{H}^d_\delta(E) = \sup_{\delta >0} \mathcal{H}^d_\delta(E)$

Then ${\mathcal{H}^d}$ is called d-dimensional Hausdorff measure in ${{\mathbb R}^n}$.

Remark 1 Hausdorff measure is a Borel outer measure, it is not Radon for ${0 \le d < n}$ because ${{\mathbb R}^n}$ is not ${\sigma-}$finite with respect to ${\mathcal{H}^d}$. And it is a measure if restricted on Lebesgue measurable sets (by Caratheodory condition: We say a set ${E}$ satisfies Caratheodory condition, if for any set ${A \subset {\mathbb R}^n}$, ${\mathcal{L}^n(A) = \mathcal{L}^n(A \cap E) + \mathcal{L}^n(A \setminus E)}$). Moreover, note that ${\alpha(d)}$ gives the volume of unit ball in ${d}$ dimension if ${d}$ is an integer, we naturally have (not trivially) ${\mathcal{H}^n = \mathcal{L}^n }$, where ${\mathcal{L}^n}$ is the n-dimensional Lebesgue measure.

# Kolmogorov-Riesz theorem

I will show an ${L^1}$ version of Arzela-Ascoli theorem, which is called Kolmogorov-Riesz theorem, by using a fact in topology as following.

Definition In a metric space, a set is totally bounded, if for any fixed ${\varepsilon>0}$, the set can be covered by finitely many open balls of radius ${\varepsilon}$.

Lemma In a metric space, a set ${E}$ is compact if and only if ${E}$ is complete and totally bounded.

Proof of the lemma can be find here.

Theorem 1 (Kolmogorov Riesz Theorem (finite measure domain)) Let ${\mathcal{F}}$ be a bounded set of ${L^1({\mathbb R}^n)}$ functions, Assume that ${\forall \varepsilon >0}$, there exists a ${\delta >0}$ such that

$\displaystyle \| f(\cdot + y) - f \|_{L^1({\mathbb R}^n)} \le \varepsilon$

whenever ${|y| < \delta}$, for all ${f \in \mathcal{F}}$. Then ${\mathcal{F}_{| \Omega}}$ has a compact closure in ${L^1(\Omega)}$ for any finite measure set ${\Omega \subset {\mathbb R}^n}$.

Definition ${X,Y}$ are Banach spaces, we say ${X}$ is embedded in ${Y}$, denoted by ${X \hookrightarrow Y}$, if ${X \subset Y}$ and ${\| \cdot \|_Y \le C \| \cdot \|_X}$.
Notation: ${\langle \xi \rangle := \sqrt{1+\xi^2}}$.