# 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.

Lemma 2 Let ${X}$ be a Banach space, ${M \subset X}$ be a bounded nonempty set, and ${T : M \rightarrow X}$ be a compact operator. Then there exists a sequence of continuous operator ${T_k : M \rightarrow X}$ such that

$\displaystyle \sup_{x \in M} \|T_k(x) - T(x)\| \le \frac{1}{k}.$

Proof: Since ${T(M)}$ is relatively compact, there exist ${N = N(k)}$ balls ${\{B(x_i)\}_{i = 1}^N}$ of radius ${\frac{1}{k}}$ that cover ${T(M)}$, with ${x_i \in T(M)}$ for all ${i}$. Consider ${T_k : M \rightarrow X}$ defined by

$\displaystyle T_k (x) := \frac{\sum_{i=1}^N dist (T(x) , T(M) \setminus B(x_i)) x_i}{\sum_{i=1}^N dist (T(x) , T(M) \setminus B(x_i))}.$

One can easily check the denominator has a positive lower bound, and both numerator and denominator are continuous (since ${T}$ is continuous and the composition of two continuous functions is continuous). Hence ${T_k}$ is continuous. One can also easily check that the denominator is positive only when ${T(x) \in B(x_i)}$ for some ${i}$, i.e. ${\| T(x) - x_i\| \le \frac{1}{k}}$. Therefore

$\displaystyle |T_k(x) - T(x)| \le \frac{\sum_{i=1}^N dist (T(x) , T(M) \setminus B(x_i)) \| x_i - T(x)\|}{\sum_{i=1}^N dist (T(x) , T(M) \setminus B(x_i))} \le \frac{1}{k}.$

$\Box$

Theorem 3 (Schauder fixed point theorem) Let ${X}$ be a Banach space, ${M}$ be a bounded closed convex set. If ${T : M \rightarrow M}$ is a compact operator, then there exists an ${x_0 \in M}$ such that ${T(x_0) = x_0}$.

Proof: Let ${T_k}$ be the sequence of continuous operator from the previous lemma. Let ${X_k}$ be the convex hull of ${\{x_1 , \cdots, x_{N(k)} \}}$, the center of covering balls. Set ${M_k = X_k \cap M}$, then by the definition of ${T_k}$, one can see ${T_k}$ maps ${M_k}$ into itself. Notice that ${M_k}$ is finite-dimensional, convex and compact, we can apply Brouwer fixed point theorem to obtain a fixed point for ${T_k}$, say ${y_k}$.

Since ${y_k \in T(M)}$ and ${T(M)}$ is pre-compact, there exists a subsequence, still denote by ${y_k}$, that converges to ${x_0 \in M}$. Then

$\displaystyle \|x_0 - T(x_0)\| = \lim_{k\rightarrow \infty} \|y_k - T(y_k) \| =\lim_{k\rightarrow \infty} \| T_k(y_k) - T(y_k)\| \le \lim_{k\rightarrow \infty} \frac{1}{k} = 0.$

$\Box$

Remark 1 From the proof we can see the boundedness of ${M}$ and compactness of ${T}$ have only been used to deliver the pre-compactness of ${T(M)}$. So if we know ${T(M)}$ is pre-compact, we can drop those 2 conditions, and of course, ${T}$ still needs to be continuous.

Theorem 4 (Leray-Schauder fixed point theorem) Let ${X}$ be a Banach space, ${T}$ be a compact operator from ${X}$ into itself, and suppose there is an a-priori bound

$\displaystyle \|x\| < M$

for all ${x \in X}$ and ${\sigma \in [0,1]}$ satisfying ${x = \sigma T(x)}$. Then ${T}$ has a fixed point.

Proof: Define a mapping ${T^* : \bar{B}_M \rightarrow \bar{B}_M}$ by

\displaystyle T^* (x) = \left\{ \begin{aligned} &T(x), &\text{if } \|T(x)\| \le M,\\ &\frac{M T(x)}{\|T(x)\|}, &\text{if } \|T(x)\| \ge M. \end{aligned} \right.

Then ${T^*}$ is a continuous mapping. Since ${T}$ is compact, so is ${T^*}$ (one can test against a bounded sequence to see this). By Schauder fixed point theorem, there exists a fixed point ${x}$ for ${T^*}$. It remains to show ${x}$ is also a fixed point for ${T}$. Assume ${\|T(x)\| \ge M}$, then ${x = T^*(x) = \sigma T(x)}$, with ${\sigma = \frac{M}{\|T(x)\|}}$. And we will have ${\|x\| = \|T^*(x)\| = M}$, which contradicts the a-priori estimate. Therefore ${\|T(x)\| < M}$ and ${x = T^*(x) =T(x)}$. $\Box$

Remark 2 None of the above fixed point theorems give uniqueness of fixed point, which is different from Banach (contracting) mapping theorem.

Remark 3 If one can reduce an equation to a fixed point problem of a compact operator (especially for integral equation), an a-priori estimate of the type above will give the existence of solution through Leray-Schauder fixed point theorem, but not the uniqueness!