Holomorphic map from closed convex domain in hilbert ball into itself has fixed point

Question

Let $B=\{x \in \ell_2, ||x|| \leqslant 1\}$ - Hilbert ball $X \subset B$ - open convex connected set in Hilbert ball, $\bar{X}$ - closure of $X$. $F: \bar{X} \to \bar{X}$ - continuous map that holomorphic into each point of $X$. Is it true in general, that $F$ has fixed point?

Answer 1

At first ignoring the holomorphic assumption, in general, no, since the Hilbert Ball is not compact for an infinite-dimensional Hilbert space.

Moreover, just because $X$ is open and connected does not mean that it is convex, and even though $\bar{X}$ is closed, since it is not a priori a subset of a compact set, we cannot assume that it is compact either.

So in terms of topological fixed point theorems, even the powerful Brouwer fixed point theorem will not apply in general. https://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem

As regards whether the holomorphy properties of $F$ are sufficient to remedy this situation, I cannot say -- I am not sufficiently familiar with complex analysis on Hilbert spaces. I apologize.

Answer 2

Define $f: \overline{X}\to \overline{X}$, where $X=\{x\in B: 1/2<\|x\|<1\}$, as $$ f(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots). $$ No fixed point for $f$ is $B$.

Answer 3

$X=\{x:1/2<||x||<1\}$, $f(x)=-x$. (This was before convexity was added to the hypotheses...)