# “Comparing” fixed point Theorems.

In class, I saw Banach's (Picard) fixed point theorem:

Given a complete metric space and a contractive mapping, it admits a unique fixed point.

And Brouwer's:

Given a continuous function in a convex compact subset of a Banach space, it admits a fixed point.

Now I tried "comparing" these theorems to see if one is "stronger" than the other.
For instance, contractive is Lipschitz and so it's continuous.
Or, compact implies complete.
Given that both are in a bigger group of theorems (Fixed point), does comparing them make any sense?
If so, between Brouwer's and Banach's, which one requires less contraints?

• The fixed point in Brouwer's theorem need not be unique. – Omar Antolín-Camarena Nov 3 '15 at 21:01
• Thank you @OmarAntolín-Camarena , I edited accordingly. – Peter Calvo Nov 3 '15 at 21:22

Brouwer actually proved the theorem for $\mathbb{R}^n$, and for subsets that are homeomorphic to the closed unit ball $B^n$ (not necessarily convex). What you state as Brouwer's theorem is a generalisation due to Schauder.
• Continuous functions are actually a "small" subset of the possible functions on a subset of a complete metric space (with the usual topology). Then uniform continuity is even rarer, with really common maps like $x^2$ failing to be uniformly continuous. Then there is another subset called "Hölder Continuous" which, informally, is saying "uniformly continuous in this prescribed way (i.e. an exponent $\alpha)$". A subset of those are when the $\alpha = 1$, and that would be Lipschitz. In this sense, we say that being Lipschitz is "strong" although such quantifiers are rarely rigorous. – Marcus Aurelius Jan 6 '18 at 2:07