Prove that $\frac{f(x)}{x}$ is uniformly continuous in $[1, +∞)$ if $f$ is Lipschitz

Question

Let $f(x)$ be a Lipschitz function on $[1, +∞)$, i.e. there exists a positive constant $C$ such that $$|f(x) − f(y)| ≤ C|x − y|, ∀x, y ∈ [1, +∞).$$

Prove that $\frac{f(x)}{x}$ is uniformly continuous in $[1,+\infty)$.

I know that a Lipschitz function is uniformly continuous. What I did so far is:

let $g(x) = \frac{f(x)}{x}$. Then I assumed $g(x)$ is Lipschitz. (Is the assumption wrong?) Then $|g(x)-g(y)| \le K|x-y|$ satisfies the Lipschitz condition.

Therefore $|\frac{yf(x)-xf(y)}{xy}| \le K|x-y|$.

How to continue from here?

Answer 1

The function $g(x) = \frac{f(x)}{x}$ is indeed Lipschitz: First of all, from $$|f(x) - f(y)| \le C|x-y|$$ putting $y=1$ gives $$\tag{1} |f(x)| = |f(x) - f(1)+ f(1)|\le C|x-1| + |f(1)|.$$ Now using $(1)$ and $x, y\ge 1$,

$$\begin{split} |g(x) - g(y)| &= \left| \frac{f(x)}{x} - \frac{f(y)}{y}\right| \\ &= \left| \frac{f(x)}{x} - \frac{f(y)}{x} + \frac{f(y)}{x} - \frac{f(y)}{y}\right| \\ &\le \left| \frac{f(x)-f(y)}{x}\right| + |f(y)| \left|\frac 1x - \frac 1y\right| \\ &\le |f(x) - f(y)| + |f(y)| \left|\frac{x-y}{xy} \right| \\ &\le C|x-y| + \frac{C|y-1| + |f(1)|}{|xy|} |x-y| \\ &\le C|x-y| + \left( C + |f(1)\right) |x-y| \\ &= K|x-y|, \end{split}$$

where $K = 2C + |f(1)|$. Thus $g$ is also Lipschitz and so is uniformly continuous.

Answer 2

Below is an incomplete proof that might not even be the right way to go.

But I'm posting it anyway (at least temporarily) so that you see one way to think about/work through these problems in general. The problem at the end of this proof is that $|f(y)|$ is not necessarily bounded by a constant, so we were unable to find a Lipschitz constant.

$\left | \dfrac{f(x)}{x} - \dfrac{f(y)}{y} \right | $

$= \left | \dfrac{yf(x) - x f(y)}{xy} \right |$

$ = \left | \dfrac{yf(x) - yf(y) + yf(y)- x f(y)}{xy} \right |$

$ \leq \left | \dfrac{yf(x) - yf(y)}{xy} \right | + \left | \dfrac{yf(y)- x f(y)}{xy} \right |$

$= \left | \dfrac{f(x) - f(y)}{x} \right | + |f(y)|\left | \dfrac{y- x }{xy} \right |$

$\leq C\left | \dfrac{x - y}{x} \right | + |f(y)|\left | \dfrac{y- x }{xy} \right |$

$\leq \left (C + \dfrac{|f(y)|}{|y|} \right ) \left | \dfrac{x - y}{x} \right |$

$\leq \left (C + \dfrac{|f(y)|}{|1|} \right ) \left | \dfrac{x - y}{1} \right |$ (since $x, y \in [1, \infty)$)

$= (C + |f(y)| ) \cdot |x - y|$