# Browder's Fixed Point Theorem in uniformly convex Banach spaces with non-identical image

This is in fact an exercise from Dirk Werner's book "Funktionalanalysis", but I do think that the result is quite interesting and up to now, I can only partly solve this problem. From the point of my view, this exercise has required some knowledge which is not given in the book and some deep understanding in fixed point theorem, so I also wanna discuss this topic here. The exercise is given as follows:

Let $$X$$ be a uniformly convex Banach space, and $$F:B_X\to X$$ is a nonexpansive mapping, i.e., it is $$1$$-Lipschitz, where $$B_X$$ is the closed unit disc in $$X$$. Suppose that the mapping has no fixed point, then there is some $$x\in S_X$$ and some $$\lambda>1$$ such that $$F(x)=\lambda x$$, where $$S_X$$ is the unit sphere in $$X$$.

In particular, using this theorem one immediately obtains the following sufficient (and interestingly formulated) condition for the existence of a FP:

If $$F(S_X)\subset B_X$$, then $$F$$ has a FP.

All the assumptions indicate that we should construct some nonexpansive mapping, having identical domain and codomain, to apply the Browder's FPT. The problem here is that the image of $$F$$ is not equal to $$B_X$$ (and otherwise the existence of a FP will directly follow from the Browder's FPT), so we can not directly use $$F$$ as the nonexpansive mapping in the Browder's FPT. So far, I have solved this problem for the case that $$X$$ is a Hilbert space, see this post on MSE. However, this technique can not be used for the Banach space case, since we have used the fact that the metric projection onto $$B_X$$ in a Hilbert space is nonexpansive, which is in general not true for a Banach space. However, using the so called generalized metric projection concept introduced by Alber (see here), it seems possible to extend this result to a uniformly convex space. But to be honest, since this is an exercise in a textbook, I would wonder if some additional knowledge which is not introduced in the book should be used to solve this problem. So I want to ask if this result is also true for a uniformly convex space and if so, is there an easier way to do so by using only the staff from the book (or briefly from the standard staff of a universal functional analysis book)?

• @PietroMajer Ah ok, I misunderstood what you meant before. Ok, I will try to show that the function is nonexpansive. Thank you very much for your hint! – Peter Jul 29 at 6:44
• @PietroMajer But are you sure that this map is nonexpansive? In this link math.uni-frankfurt.de/~baumeist/Nonex-Kap4.pdf p.92, it is stated that the your mapping (which is called radial mapping therein) in in general not nonexpansive. The best Lipschitz constant is expected to be closed to 2. – Peter Jul 29 at 8:14
• Thankyou! I even thought I had a proof – Pietro Majer Jul 29 at 9:22
• Apologies for this gap in my book that was first pointed out to me in 2015. As a result, I added a remark to the corrections page for the 7th edition, and in the 8th edition the problem appears with $X$ being a Hilbert space, where the proof seems to be clear. -- I don't remember the argument I had in mind for the case of uniformly convex Banach spaces (mea culpa!); I'll still try to hunt it down! – Dirk Werner Aug 5 at 18:52
• Here are the references to the original papers where this result is proved (so it's true after all, but the proof is not so easy): D. O'Regan, Fixed point theorems for nonlinear operators, JMAA 202 (1996), 413-432; R. Precup, On the continuation theorem for nonexpansive maps, Studia Univ. Babes-Bolyai Math 41 (1996), 85--89. See also D. O'Regan and R. Precup, Theorems of Leray-Schauder Type and Applications, Gordon and Breach 2001, pp. 44--46. Thanks to Donal O'Regan for providing these references! – Dirk Werner Aug 6 at 21:19