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Consider the following (less general than possible) statement of Schauder's fixed point theorem:

Suppose that $X$ is a Banach space, that $B_1$ is the unit ball of $X$ and that $f: X \to X$ is a continuous function. If $f(B_1)$ is a compact subset of $B_1$, then $f$ has a fixed point in $B_1$.

Now let $B_{1+\varepsilon}$ denote the ball around $0$ of radius $1+\varepsilon$ and suppose that $f(B_1)$ is a compact subset of $B_{1+\varepsilon}$ for some $\varepsilon > 0$. Is there anything that can be said about possible fixed points of $f$? For example, is it possible to prove the existence of a point $x$ such that $\|f(x) - x\| < \varepsilon$? Can anything at all be said about this scenario? Does anyone know of any works in which such functions have been examined in some detail?

Thanks in advance.

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    $\begingroup$ Why wouldn't the fixed point theorem apply to balls of any radius, rather than just the unit balls? $\endgroup$
    – Srivatsan
    Dec 15, 2011 at 8:25
  • $\begingroup$ @Srivatsan the hypothesis does not guarantee that the entire ball $B_{1 + \epsilon}$ is mapped into a compact subset of $B_{1 + \epsilon}$. It only guarantees that $B_1$ is mapped into a compact subset of $B_{1 + \epsilon}$. So, as stated, the fixed point theorem does not apply. $\endgroup$
    – user12014
    Dec 15, 2011 at 8:35
  • $\begingroup$ unless you mean that I could ask the question for $B_k$ and $B_{k+\epsilon}$ in which case of course you are right. $\endgroup$
    – user12014
    Dec 15, 2011 at 8:36
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    $\begingroup$ You are definitely guaranteed no fixed points in $B_1$. Consider the constant map to a point in $B_{1+\epsilon}$\ $B_1$. $\endgroup$
    – bzc
    Dec 15, 2011 at 8:46

2 Answers 2

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The function $X \to X : x \mapsto \frac{1}{1+\varepsilon} f(x)$ maps $B_1$ to a compact subset of $B_1$. So by the standard fixed point theorem, there exists an $a \in B_1$ such that $\frac{1}{1 + \varepsilon}f(a) = a$, or $f(a) = (1+ \varepsilon)a$. Hence, $\| f(a) - a \| = \| \varepsilon a \| \leqslant \varepsilon$, since $\| a \| \leqslant 1$. $\quad {\small \square} $

In fact, this result is tight, in the sense that there are functions $f$ satisfying $\|f(x) - x\| \geqslant \varepsilon$ for all $x \in B_1$. [E.g., consider the constant function mapping all points to a fixed point of norm $1 + \varepsilon$.] In particular, $f$ is not guaranteed to have a fixed point inside $B_1$.

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  • $\begingroup$ @PZZ It's getting late here, so I am not sure if I am making some mistake here. =) $\endgroup$
    – Srivatsan
    Dec 15, 2011 at 8:43
  • $\begingroup$ Thank you. Since I am also waiting on any other things people might have to say about the problem I will wait a day or two to accept your answer, but if nothing comes the checkmark is yours :) $\endgroup$
    – user12014
    Dec 15, 2011 at 8:46
  • $\begingroup$ no, it doesn't look like theres a mistake here to me. quick and easy after all. $\endgroup$
    – user12014
    Dec 15, 2011 at 8:46
  • $\begingroup$ @PZZ Sure, do wait for other answers. That is the recommended practice. $\endgroup$
    – Srivatsan
    Dec 15, 2011 at 8:49
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Actually I have more of a question than an answer, but it is related to your topic. Is it possible to show in the general case that:

$ f(x)= x $ is given, with solution $x = \theta \gt 0 $

I would like to show :

$ \forall \epsilon \in (0,1), \forall x : 1/\epsilon > |x -\theta| > \epsilon => \inf_x |f(x) - x | > 0 $.

I guess it is not true in the general case, but what properties should the function f(x) have in orther this to be true?

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