On the failure of the infinite dimensional Brouwer Theorem Let $K$ be the closed unit ball of some infinite dimensional Banach
space, and let $H$ be an autohomeomorphism of $K$, having fixed
points. Can $H/2$ be fixed point free ?
Also, let ${\mathcal{F}}$ := { $S\in\mbox{C}(K,K), \mbox{Fix}(S)\neq\textrm{Ø } $}.
Let $T$ in $\mbox{C}(K,K)$  such that $TS\in\mathcal{F}$ for all $S\in\mathcal{F}$
. Must $T$ be necessarily compact ? 
 A: Indeed, $H/2$ can be fixed point free. An example for this can be constructed as follows: take $X = c_0(\mathbb{Z},\mathbb{R})$, the Banach space of bounded, null-convergent, real sequences indexed by the integers $\mathbb{Z}$. Now take an equicontinuous sequence of homeomorphisms $f_i:[-1,1] \to [-1,1], i\in \mathbb{Z}$ with the property that $f_i(0) \to 0$ as $i\to \infty$ and so that the sequences of inverses is equicontinuous as well.
We can now construct a homeomorphism of $H:K\to K$ by
$$ (H(x))_{i+1} = f_i(x_i) .$$
The map can be extended to the space of all twosided sequences with values in $[-1,1]$ and there it is obviously a bijection. Also sequences in $K$ are mapped to sequences in $K$ by the assumption of equicontinuity and the convergence $f_i(0) \to 0$. Similarly, the preimage of a sequence in $K$ is in $K$. Finally, equicontinuity gives continuity of $H$ and its inverse.
Now the fixed points of $H$ are those sequences $x=(x_i)_{i\in \mathbb{Z}}$ such that 
$$ x_{i+1} = f_i(x_i) \quad \forall i\in \mathbb{Z}. \tag{1}$$
I.e., the fixed points in $K$ are those solutions of the nonautonomous dynamical system (1) which have the property that they converge to $0$.
The trick is now to construct a nonautonomous system of the described form which has a solution converging to zero and so that the system 
$$ x_{i+1} = \frac{1}{2} \ f_i(x_i) . \tag{2}$$
does not have this property. For this it suffices to give the system (2) a globally attractive fixed point away from $0$. In this way no solution of (2) lies in $X$ and so $H/2$ is fixed point free.
A concrete example for (1) is as follows. Take $f_i=\mathrm{id}$ for $i< 0$. For $i\geq 0$ define $f_i$ by the linear interpolation of the following points:
$$ (-1,-1) \ , \quad (-1/4, -1/2) \ , \quad (2^{-(i+2)},0)\ , \quad (2^{-(i+1)},2^{-(i+2)}) ,\, \quad (1,1). $$
A solution $x$ of (1) is then given by $x_i = 1/2, i\leq 0$, $x_i = 2^{-i}, i>0$. And we have a fixed point of $H$ as $x \in K$.
The maps defining system (2) are interpolations of the points
$$ (-1,-1/2) \ , \quad (-1/4, -1/4) \ , \quad (2^{-(i+2)},0)\ , \quad (2^{-(i+1)},2^{-(i+3)}) \, \quad (1,1/2). $$
Note the nice fixed point at $(-1/4,-1/4)$ and it is not hard to see that all solutions become negative in finite positive time as the solution $x$ with the condition $x(0) = 1$ is $ ...,1,1/2,5/24,17/224,101/4480,...$ and at this stage the solution becomes negative. Also solutions are strictly decreasing if they live in the (invariant) interval $(-1/4,0)$. Thus $H/2$ has no fixed point as there are no solutions of $(2)$ that converge to $0$.
(BTW: I was worried there for a moment that I am contradicting the Banach fixed point theorem, but this is not the case, luckily. The Lipschitz constants of the $f_i$ converge to $2$, so that $H/2$ is not a contraction.)
Re question 2: The identity map would be a $T$ that works. This is not compact. If there are other maps that work I do not know.
