If a bounded function $f:\Bbb R\to \Bbb R $ and $\left|\,f(x)-f(y)\right|<\left|x-y\right|$ for $x\ne y$, then there is an $a$ s.t. $f(a)=a$. 
If a bounded function $f:\Bbb R\to \Bbb R $ and $\left|\,f(x)-f(y)\right|<\left|x-y\right|$ for $x\ne y$, then there is an $a$ s.t. $f(a)=a$. 

What I know is $f$ should be uniformly continuous. As far as I know, for the fixed point theorem to holds, the above condition is not strong enough to imply a fixed point exists as there should be a contraction,i.e.$\left|\,f(x)-f(y)\right|<\alpha\,\left|x-y\right|$ where $\alpha<1$.
This is a question I saw from a book but now I wonder if it is true .
 A: I think I have a simpler proof without usage of Banach fixed point theorem. Let's assume that the range of $f$ is inside of some closed interval $X=[-M,M]$, which is compact. Let's view $f$ as a funcion from $X$ to $X$. Then consider the function on $X$, $g(x)=|f(x)-x|$, which obtains its minimum value, say $\delta$, at $x_{0}$ on $X$. If $\delta=0$, then $x_{0}$ is the fixed point. If $\delta>0$, by the assumption, we have 
$$|f(f(x_{0}))-f(x_{0})|<|f(x_{0})-x_{0}|=\delta.$$
This is a contradiction. 
So this proof applies equally well to any compact metric space. 
A: 
Every bounded and continuous function has a fixed point. This implies the desired result.

Proof: Assume that $f$ is bounded. Then $f(x)\lt x$ for some $x$ close enough to $+\infty$, say $x=c$, and $f(x)\gt x$ for some $x$ close enough to $-\infty$, say $x=b$. Assume furthermore that $f$ is continuous. The intermediate value theorem guarantees the existence of some $a$ in $(b,c)$ such that $f(a)=a$.
A: Edit: Did's and Yunfeng's proofs make mine utterly ridiculous...Oh well, I'll leave it like this.
Take $n\geq 1$ and consider the function $f_n(x):=f(\alpha_nx)$ with $\alpha_n=1-\frac{1}{n}<1$. Then
$$
|f_n(x)-f_n(y)|\leq\alpha_n|x-y|\qquad\forall x,y\in\mathbb{R}.
$$
By Banach fixed point theorem, there exists a unique $x_n\in \mathbb{R}$ such that
$$
f_n(x_n)=x_n.
$$
Snce $f$ is bounded, we obtain a bounded sequence $x_n$ from which we can extract a converging subsequence $x_{n_k}\longrightarrow x$ as $k\rightarrow +\infty$. Now
$$
|f_{n_k}(x_{n_k})-f(x)|\leq|f_{n_k}(x_{n_k})-f_{n_k}(x)|+|f_{n_k}(x)-f(x)|.
$$
The lhs term tends to $0$ by the fact that each $f_{n_k}$ is $1$ Lipschitz. The rhs term tends to $0$ by continuity of $f$ at $x$. Hence
$$
f(x)=\lim f_{n_k}(x_{n_k})=\lim x_{n_k}=x
$$
is a fixed point of $f$. Note that it is unique, as $f$ is a contraction.
