Suppose $f:[1,\infty)\to\Bbb R$ is uniformly continuous function. Prove that there is a positive $M$ such that $\frac{|f(x)|}{x} \lt M$ for $x\ge 1$.
Every uniformly continuous function is not Lipschitz function. Thus I can't find out satisfactory proof.
Fix $\epsilon >0$
W.l.o.g assume $f$ is positive
Let $L_i$ be the supremum of the lengths of all the intervals for which $f(x)/x\geq i$
If for infinitely many $i$, $L_i>\epsilon/(2i)$ then take $\epsilon '=\epsilon/2$ then for these value, there does not exist any $\delta>0$ for which $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon '$, hence $f$ is not uniformly continuous.
Hence after some $i\geq N$, $L_i\leq \epsilon/(2i)$
Now fix $\delta>0$, so there exists an integer $M>N$ such that $\epsilon/(2M)<\delta$.
So if at any point $a$, $f(a)/a>M$, then there exists a point $b \in (a,a+\epsilon/2M)$ such that $f(b)/b<M$.
But we know that there exists a sequence of points $a_i$ such that $f(a_i)/a_i>iM$, also there exists a point $b_i \in (a_i,a_i+\epsilon/2M)$ such that $f(b_i)/b_i<M$ $\forall i\in \mathbb{N}$.
So let $S_i=sup|f(x)-f(y)|$ where $x,y\in (a_i,a_i+\epsilon/2M)$
Now let $S=sup\{S_i\}$ Clearly, $S$ goes to infinity,and since $\delta$ was arbitrary hence the function cannot be uniformly continuous.