# Leray-Schauder fixed point theorem

By | January 8, 2018

First, we derive the Schauder fixed point theorem.

Theorem 1 (Schauder fixed point theorem). Let be a compact convex set in a Banach space and let be a continuous mapping of into itself. Then has a fixed point, that is, for some .

Proof. Fix . Since is compact, has a finite subcover which covers . Write .
Let . Define by

The map is well-defined. Since is continuous, is continuous. Also,

Note that for such with , . Thus, for such , we have . Hence . Now is continuous, is compact and convex set. So is homeomorphic to . Hence by the Brouwer fixed point theorem, for some .

Now since and is compact, there exists a convergent subsequence which converges to in . We claim that is a fixed point. Note

as . Hence .

Using this theorem, we obtain another version of Schauder fixed point theorem.

Theorem 2. Let be a bounded and closed convex set in a Banach space and let be a compact mapping of into itself. Then has a fixed point.

Proof. Let be the closed convex hull of Then is convex and since the closed convex hull of a compact set is itself compact, is compact. Note that because and is closed. So . Since is convex and is the convex hull of , .
Thus, by the Schauder fixed point theorem, has a fixed point in . So we are done.

Now we prove the Leray-Schauder fixed point theorem. Here we consider the simplest form.

Theorem 3 (Leray-Schauder). Let be a Banach space and let be a compact mapping of into itself. Also, suppose there exists a constant such that

for all and satisfying . Then has a fixed point.

Proof. We may assume . Define by

Let . Then maps to . We show that
is compact. First, is continuous. For those with , we
have

From the continuity of , is continuous on .

Next, let be a sequence in the ball . There are two cases:

•  there exists a subsequence satisfying  for all ;
•  there exists a subsequence satisfying  for all .

For the first case, since and is compact, there exists a subsequence such that converges strongly in . So converges strongly in .

For the second case, choose a subsequence so that and as by Bolzano-Weierstrass theorem on real numbers and compactness of . Hence

Thus, is compact.

Hence by Theorem 2, has a fixed point . We claim that is also a fixed point of . Suppose with . Then

and so , which contradicts the assumption. Hence and consequently .