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.

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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.

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Harnack inequality implies Holder continuity

We will show a basic argument why Harnack inequality implies Holder continuity. Assume {u \in H^1(B_2)} is a weak solution to a uniform elliptic equation

\displaystyle -\partial_i(a^{ij} \partial_j u) = 0,

in {B_2}. Assume we have Harnack inequality (Indeed we do have due to Moser), i.e. for any positive weak solution {u} we will have

\displaystyle \sup_{B_1} u \le C \inf_{B_1} u,

where {C} is a constant independent of {u}.

First one will know {u \in L^\infty(B_1)}, see this note.

If {u} is a weak solution, so are {\sup_{B_2} u - u} and {u - \inf_{B_2} u}. And they are strictly positive in {B_1} due to the strong maximum principle, otherwise {u} will be a constant. By Harnack inequality, we will have

\displaystyle \sup_{B_2} u - \inf_{B_1} u \le C(\sup_{B_2} u - \sup_{B_1} u),

\displaystyle \sup_{B_1} u - \inf_{B_2} u \le C(\inf_{B_1} u - \inf_{B_2} u).

Adding these two inequalities up we will have

\displaystyle \text{osc}_{B_2} u + osc_{B_1} u \le C(osc_{B_2} u - osc_{B_1}),


\displaystyle osc_{B_1} u \le \frac{C-1}{C+1} osc_{B_2} u.

Denote {\theta = \frac{C-1}{C+1} < 1}, and choose {\alpha > 0} such that {\theta = 2^{- \alpha}}, by interation we know

\displaystyle osc_{B_{2^{-k}}} u \le \theta^k osc_{B_1} u \le 2 \theta^k \|u\|_{L^\infty(B_1)}.

Then for any {0 < r < 1}, choose {k \in {\mathbb N}} such that {2^{-k-1} < r \le 2^{-k}}, we will have

\displaystyle osc_{B_r} u \le osc_{B_{2^{-k}}} u \le 2^{-\alpha k + 1} \|u\|_{L^\infty(B_1)} = 2^{1+\alpha} \cdot 2^{- \alpha ( k +1)} \|u\|_{L^\infty(B_1)}< 2^{1+\alpha} \|u\|_{L^\infty(B_1)} r^\alpha.

So Holder continuity follows.