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

Theorem 1 (Brouwer fixed point theorem)Let be a convex compact set, for any continuous function , there exists a point such that .

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 2Let be a Banach space, be a bounded nonempty set, and be a compact operator. Then there exists a sequence of continuous operator such that

*Proof:* Since is relatively compact, there exist balls of radius that cover , with for all . Consider defined by

One can easily check the denominator has a positive lower bound, and both numerator and denominator are continuous (since is continuous and the composition of two continuous functions is continuous). Hence is continuous. One can also easily check that the denominator is positive only when for some , i.e. . Therefore

Theorem 3 (Schauder fixed point theorem)Let be a Banach space, be a bounded closed convex set. If is a compact operator, then there exists an such that .

*Proof:* Let be the sequence of continuous operator from the previous lemma. Let be the convex hull of , the center of covering balls. Set , then by the definition of , one can see maps into itself. Notice that is finite-dimensional, convex and compact, we can apply Brouwer fixed point theorem to obtain a fixed point for , say .

Since and is pre-compact, there exists a subsequence, still denote by , that converges to . Then

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

Theorem 4 (Leray-Schauder fixed point theorem)Let be a Banach space, be a compact operator from into itself, and suppose there is an a-priori boundfor all and satisfying . Then has a fixed point.

*Proof:* Define a mapping by

Then is a continuous mapping. Since is compact, so is (one can test against a bounded sequence to see this). By Schauder fixed point theorem, there exists a fixed point for . It remains to show is also a fixed point for . Assume , then , with . And we will have , which contradicts the a-priori estimate. Therefore and .

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

Remark 3If 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!