Let be a metric space with . Recall that a mapping is non-expansive if satisfies

for all .

Theorem. Let be a nonempty, closed, convex subset of a normed linear space with nonexpansive and a subset of a compact set of . Then has a fixed point.

Proof. Let . For defineSince is convex and , we see that and it is clear that is a contradiction. Therefore each has a unique fixed point . That is

.

In addition, since lies in a compact subset of , there exist a subsequence of integers and a with as in .

Thus as in . By continuity, as in which claims that .

The main theorem of this topic is a result proved independently by Browder, Gohde and Kirk in 1965. We state it as follows.

Theorem.Let be a nonempty, closed, bounded, convex set in a (real) Hilbert space . Then each nonexpansive map has at least one fixed point.

**Remark. **Notice that uniqueness need not hold as the example, , shows.

**Remark.** In fact the above Theorem, it is enough to assume that is a uniformly convex Banach space.

In the proof of the above Theorem, we will need the following two technical results.

**Claim 1.** Let be a Hilbert space with , and let be constants with . If there exists an with

then

.

*Proof***.** This comes from the so-called parallelogram law, that is,

.

**Claim 2.** Let be a Hilbert space, a bounded set and a nonexpansive map. Suppose and . Let denote the diameter of and let with and . Then

.

Proof of main theorem.Assume that . (A modified argument from the one given below holds for any , therefore for simplicity we let .) Also assume that (otherwise we are finished). For each , notice thatis a contradiction. Now the first theorem guarantees that there exists a unique with

.

Thus,

.

For each , let

.

Now

is a decreasing sequence of nonempty closed sets. Let

and since the s are decreasing we have

with for each . Consequently, with .

Next let

,

where

.

Now is a decreasing sequence of closed, nonemplty sets. We now show that

.

To see this, let . Then

.

Also since we have

.

Thus

.

This implies

and

.

It is easy to check that is non-expansive. As a result

is a non-expansive map. This guarantees that there exists with . If , then

and has a fixed point. If does not belong to then

with

.

Source: Ravi P. Agarwal, Maria Meehan, Donal O’Regan, *Fixed point theory and applications*, Cambridge University Press, 2001.