Fixed point theorem involving nonexpansive mapping on uniformly convex Banach space This is an exercise from the Dirk Werner's book about fixed point theorem: Let $X$ be a uniformly convex Banach space. Let $F:B_X\to X$ be a nonexpansive mapping, i.e.
$$ \forall x,y\in B_X: \|F(x)-F(y)\|\leq \|x-y\|.$$
Here denotes $B_X$ the closed unit ball in $X$. Then


*

*Either $F$ has a fixed point, or there exists some $x\in S_X$ and some $\lambda>1$ such that $F(x)=\lambda x$. Here $S_X$ denotes the unit sphere of $X$.

*If $F(S_X)\subset B_X$, then $F$ has a fixed point.


I am still working with 1. What I tried is to assume that $F$ has no fixed point. In this case, the range of $F$ is contained in the ball $B_{1+\|F(0)\|}$, and using rescaling one easily concludes that $F(x)=\lambda x$ for some $x\in B_X$ and $\lambda>1$. However, I am not able to conclude that the point $x$ is exactly a point on the sphere.
Any advice and suggestion is very welcome!

Some thoughts: To show the statement (assuming no fixed points) it is equivalent to show that the function $(F(z)-z)/\|F(z)-z\|$ has a fixed point. Notice that this function has image in $S_X$, so one might want to apply the Browder's FPT to this function stated in the Werner's textbook. However, it is no more nonexpansive now, so my goal is to construct a new function, derived from this one, which has image in $B_X$ and nonexpansive.
 A: Here I want to give an incomplete answer which is only valid for the case that $X$ is a Hilbert space. Suppose there is no FP and we want to show the second statement. We first claim that for each $x_0\in X$, there is a unique point $x\in B_X$ such that
$$ |x_0-x|=\inf_{y\in B_X}|x_0-y|.$$
Indeed, let $(x_n)$ be a minimizing sequence in $B_X$. since it is a minimizing sequence, it must be bounded and therefore it has a weakly convergence subsequence due to the reflexivity of $X$, which converges to some $x\in B_X$. Using the weakly lower semicontinuity of the norm function we immediately infer that $x$ is our sought. Suppose there are two such minimizers $x_1$ and $x_2$ with $x_1\neq x_2$, then $\frac{1}{2}(x_1+x_2)\in B_X$, since $B_X$ is convex, and
\begin{align}
&|x_0-\frac{1}{2}(x_1+x_2)|\\
<& \frac{1}{2}(|x_0-x_1|+|x_0-x_2|)\\
=&\frac{1}{2}(d+d)=d,
\end{align}
where $d$ is the distance of $x_0$ to $B_X$ and the strict inequality is due to the strict convexity of $X$. But this constradicts the minimality of $d$. Thus $x$ is uniquely determined. thus the function $P$ given by $P(x_0)=x$ for $x_0\in X$ is well-defined. In fact, $P$ is the projection onto $B_X$ and it is well known that $P$ is nonexpansive (for a proof, see for instance Prop.2 in this paper). Clearly, if $|x|\leq 1$, then $P(x)=x$. The most important observation is that 
$$ P(x)=\frac{x}{|x|}$$
for $|x|>1$. Suppose it was not the case, then using the minimal distance property of $P(x)$ one could find some $y\in B_X$ such that
$$ |x-\frac{x}{|x|}|>|y-x|\geq |x|-|y|$$
and it follows 
$$ |y|>\frac{|x|}{|x|}=1,$$
contradiction. Now we consider the function 
$$ G(x)=(P\circ F)(x).$$
From the previous mentioned claims it follows that $G$ is from $B_X$ to $B_X$ and nonexpansive. Thus using the Browder's FPT we can find a FP $x\in B_X$ of $G$, i.e.,
$$ x=G(x)=(P\circ F)(x).$$ 
If $|F(x)|\leq 1$, then it follows $(P\circ F)(x)=F(x)$, thus $x$ is a FP of $F$, which is impossible due to the given assumption. Thus $|F(x)|>1$, and therefore
$$ x=(P\circ F)(x)=\frac{F(x)}{|F(x)|}.$$
In this case we obtain that $|x|=1$ and the claim is fulfilled for this $x\in S_X$ with $\lambda=|F(x)|>1$.
In general, this method can not be adopted to the case where $X$ is a Banach space. The failure is due to the fact that the projection $P$ is not necessarily nonexpansive. Nevertheless, it seems possible to define the so called generalized metric projection (see this paper) to modify this method for Banach spaces. But this is in no meaning trivial for an exercise given in a textbook (particularly, nothing involving this topic is given in the book), so I will maybe skip this part. 
