Question: 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.

Answer: 
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 $B\_1$.
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?