Proof of a fixed-point lemma I'm trying to prove the following fixed-point lemma.

Let $\mathcal X$ be a Banach space and $A \neq \emptyset$ a closed, bounded and convex subset of $\mathcal X$. Further let $g: \mathbb R^+ \to \mathbb R^+$ be continuous with $g(x)<x$ for all $x>0$ and let $F: A \to A$ be a function such that
$$
\left\Vert F(x)-F(y) \right\Vert \leq g(\left\Vert x-y \right\Vert)
$$
for all $x,y\in A$. Then there exists a fixed-point for $F$.

This doesn't seem to be a special case of any fixed-point theorem I know. I've combed the proofs of some of them to find a starting point without success. Can anyone help me with this?

One route I've been exploring is to learn more about $F$. It is clearly continuous. If $\overline{F(M)}$ were compact for every subset $M$ of $A$, then the claim would directly follow from Schauder's fixed-point theorem which says that each compact $F$ in this setting has a fixed-point. However, I don't think that $F$ is compact here, at least I see no way of proving it.
Since there exists an "near-fixed-point", that is a sequence $(x_n)$ such that $\|F(x_n)-x_n\|\to 0$ as $n\to\infty$ (this, I know), it would even suffice to show compactness for $\overline{F(C)}$ to yield the claim which is a bit weaker. But again I can't show it and don't know whether this is the case.
 A: Elaboration on the hint given by janmarqz:
First, we can assume that $g$ is monotonically non-decreasing:
Otherwise we can use $\hat g(s):=\sup_{0\leq t \leq s} g(t)$ instead
to make $g$ non-decreasing.
We also define $\bar s:=\sup\{\|x-y\| \mid x,y\in A\}<\infty$.
Then we have
$$
\| F(x)-F(y)\| \leq g(\|x-y\|) \leq g(\bar s)
\qquad\forall x,y\in A.
$$
A repeated application yields
$$
\|F^n(x)-F^n(Y)\| \leq g^n(\bar s)
\qquad\forall x,y\in A,
$$
where $F^n$, $g^n$ denote the $n$-fold composition
$F\circ F\circ\cdots\circ F$ and $g\circ g\circ\cdots\circ g$.
Clearly, $\{g^n(\bar s)\}_{n\in\Bbb N}$ is a decreasing sequence.
If we denote its limit by $t$, then continuity of $g$
yields $g(t)=t$.
But this is only possible if $t=0$, i.e. $g^n(\bar s)\to0$.
Let us return to the given hint:
Let $x_0\in A$ be given and let us define
$x_{n+1}:=F(x_n) = F^n(x_0)$ as suggested by janmarqz.
Then we have
$$
\| x_{n}-x_{n+m}\|
= \| F^n(x_0)-F^n(x_m)\|
\leq g^n(\bar s) \to0 
\quad\text{as }n\to\infty.
$$
Thus the sequence $\{x_n\}_{n\in\Bbb N}$ is a Cauchy sequence.
Let $\bar x\in A$ denote its limit.
Since $F$ is continuous, it follows that $\bar x$ is a fixed point for $F$.
Remarks on generalization:
This also holds true in complete metric spaces (which is maybe a bit surprising).
Therefore this result is a strict generalization of the Banach fixed-point theorem
(which is just the case $g(s)=q s$ for some $q<1$).
A proof of this result can be found
in Theorem 1 in this article by Boyd and Wong from the 1960s.
Note that my proof above only works for bounded $A$, but the result also
holds for unbounded complete metric spaces.
A: Hint: Take $x_0\in A$ and iterate $x_1=F(x_0)$, $x_2=F(x_1)$, . . . , $x_{n+1}=F(x_n)$,...
so
$$\|x_{n+1}-x_n\|=\|F(x_n)-F(x_{n-1})\|<g(\|x_n-x_{n-1}\|)<\|x_n-x_{n-1}\|$$
then the sequence is Cauchy's.
