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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?

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

Neither is stronger than the other: one is defined in metric terms (contractions), the other assumes just continuity. But Schauder assumes a vector space structure (or convex makes no sense). I'd say the second one is more restrictive in the space (we need a Banach space, not just a complete metric space), but says something for every compact convex subset for any continuous function. Banach gives a fixed point for the whole space and just contractive maps. Unicity of the fixed point is then a bonus again. So you could argue either way, and both have their place.

Both assume some form of completeness (metric completeness, or compactness), which is a common theme in fixed point theorems.

But there are situations where one can be applied, and the other cannot, and vice versa. But fixed point theorems are quite useful (especially in proving all sorts of existence results), there are quite a few of them, all applicable in different settings (e.g. ordered sets, as well, e.g., or for multivalued maps (Kakutani) etc.), See Wikipedia, e.g., or Dugundji and Granas' classic book "Fixed Point Theory".

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    $\begingroup$ I would add that Banach's Fixed Point Theorem is constructive and provides an algorithm to approximate the fixed point, with explicit error bounds. $\endgroup$ – Julián Aguirre Nov 3 '15 at 21:22
  • $\begingroup$ I can sense that comparing theorems is not "kosher" but still, I cannot avoid thinking about how I've been told "Lipschitz is a very strong condition" or, "convexity is very rare" in my Topology classes. Do these "quantifiers" translate into something for statements? Thank you @HennoBrandsma $\endgroup$ – Peter Calvo Nov 3 '15 at 21:30
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    $\begingroup$ 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. $\endgroup$ – Marcus Aurelius Jan 6 '18 at 2:07

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