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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?

Martin Sleziak
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Ramos
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  • This reminds me of Cauchy's mean value theorem, except that here $f$ is not differentiable. You might want to check out how the theorem is proved. – hrkrshnn Nov 26 '15 at 05:09
  • @GNUEmacs I did not learn this theorem in class. But I will still check the theorem though. – Ramos Nov 26 '15 at 05:19

2 Answers2

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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.

  • @johmma why do we have to check for y=1? – Ramos Nov 26 '15 at 05:29
  • We need $(1)$ so that $f$ has at most "linear growth". You could use $y=2$ or $3$ instead. Then you get $|f(x)| = |f(x) - f(2)+ f(2)|\le |f(x) - f(2)| + |f(2)|$. $1$ is not important. Very important is the linear growth estimates of $f$ (equation (1)). This is used in the last inequlity @Ramos –  Nov 26 '15 at 05:32
  • In your second to last inequality, where did the $|y-1|$ disappear to? – layman Nov 26 '15 at 05:33
  • $\frac{C|y-1|}{|xy|}\le \frac{C|y|}{|x||y|} \le C$ @user46944 –  Nov 26 '15 at 05:34
  • neat and clean answer :) – Kushal Bhuyan Nov 26 '15 at 05:45
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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 $x, y \in [1, \infty)$)

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

layman
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