Show uniform continuity of $f(x)/x$ I'm stuck on a question where I have the following:
Let $f$ be a function defined on $[1,\infty)$ such that there exists a constant $M>0$ with $|f(x)-f(y)| \leq M|x-y|$ for all $x,y \in [1,\infty)$. Show that the function $f(x)/x$ is uniformly continuous on $[1,\infty)$.
I understand how to solve this for $f$ to be continuous but could anyone give me hints to how I solve this for $f(x)/x$? Thanks.
 A: Without loss of generality, $f(1) = 0$ since $f$ is Lipschitz if and only if $f + a$ is Lipschitz for any $a\in\Bbb R$. Suppose that $0 < |h| < 1/2$. Then $x+h > 1/2$ no matter what $x$ we choose in $[1,\infty)$. If $x\ge 1$, then
\begin{align*}
\bigg|\frac{f(x+h)}{x+h}-\frac{f(x)}{x}\bigg| &= \bigg|\frac{[f(x+h)-f(x)]x - f(x)h}{x(x+h)}\bigg| \\
&\le \bigg|\frac{f(x+h)-f(x)}{x+h}\bigg| + \bigg|\frac{f(x)h}{x(x+h)}\bigg| \\
&\le 2M|h| + \bigg|\frac{f(x)h}{x(x+h)}\bigg| \\
&= 2M|h| + \bigg|\frac{[f(x)-f(1)]h}{x(x+h)}\bigg| \\
&\le 2M|h| + M|h|\frac{x-1}{x+h} \\
&\le 2M|h| + M|h| = 3M|h|,
\end{align*}
Where in the last inequality we used the fact that $x-1 \le x+h$. As $h\to0$ we get the desired conclusion.
A: Hint 1: The condition $|f(x) - f(y)| \leqslant M |x-y|$ shows that $f$ is uniformly continuous on $[1,\infty)$.
Hint 2: We have for $x,y \in [1,\infty)$,
$$\left|\frac{f(x)}{x} - \frac{f(y)}{y} \right| = \left|\frac{f(x)}{x} - \frac{f(y)}{x} + \frac{f(y)}{x} - \frac{f(y)}{y}\right|  $$
Hint 3: I'll leave it to you to show this leads to 
$$\left|\frac{f(x)}{x} - \frac{f(y)}{y} \right|  \leqslant M|x-y| + \frac{|f(y)|}{y}|x-y|.$$
Hint 4: If $f$ is uniformly continuous, then $|f(x)|/x$ is bounded on $[1,\infty)$. With this result,  we are finished proving $f(x)/x$ is uniformly continuous, since $|f(y)|/y \leqslant C$ implies that 
$$\left|\frac{f(x)}{x} - \frac{f(y)}{y} \right|  \leqslant (M+C) |x-y| .$$
Either try proving the claim in Hint 4 on your own or see my proof here
