I have a homework question to which is

If $f(x)$ is diferentiable in $(0,\infty)$ and $f'(x)>\frac{1}{x}$ for every $x>0$ then $f$ is not uniformly continuous in $(0,\infty)$.

The question has a clue which asks me to prove that for $0<a<b$ then $f(b)-f(a)>\frac{b-a}{b}$ which I have also not managed to prove.

But even if I did I can't how to use that fact to solve the question.

Can someone help me out? Thanks a lot :)

  • 2
    $\begingroup$ Have you seen the mean value theorem? $\endgroup$ – Andrés E. Caicedo Jan 13 '12 at 15:15
  • $\begingroup$ Another hint: $f(x)=2\ln(x)$ satisfies the hypothesis of the theorem, so it is not uniformly continuous. Where does uniform continuity fail? Can you show that the same thing happens for any $f$ satisfying the hypothesis? $\endgroup$ – Andrés E. Caicedo Jan 13 '12 at 15:20

We'll first prove the hint.

Let $0 <a<b$.

For this, we use the mean value theorem in $[a,b]$.

Why does $f$ satisfy the hypothesis of MVT on $[a,b]$?

As $f$ is differentiable on $(0,\infty)$, $f$ is continuous on $[a,b]$. And, $f$ is, in fact, differentiable on $[a,b]$, if you define differentability at endpoints using one-sided derivatives.

So, there exists $c \in (a,b)$ $$\dfrac{f(b)-f(a)}{b-a}=f'(c) >\dfrac{1}{c}>\dfrac{1}{b}$$ Now, noting that, $b-a>0$, we can actually take $b-a$ to the other side, taking us to $$f(b)-f(a)>\dfrac{b-a}{b}$$

This proves your hint.

Now to prove that, $f$ is not uniformly continuous, we actually prove the contrapositive of the definition:

Contrapositive of uniform continuity

$f$ is NOT uniformly continuous if there exists $\epsilon>0$ such that for each $\delta >0$ there exists $x,y$ such that $|x-y|<\delta$ but $|f(x)-f(y)|\geq \epsilon$.

I think this should not be hard. So, how do we go about this? Set $b=2a$. Note that this is consistent with our assumption on $a$ and $b$ (i.e. $a < b $). There exists an $\epsilon>0$, here $\epsilon =\dfrac{1}{2}$, such that for any $\delta>0$, here any $a>0$, the claimed inequality holds.

So, we are through.

  • $\begingroup$ To satisfy the MVT conditions wouldn't $f$ have to be continuous at $[0,\infty)$? This is not ensured by the given arguments $\endgroup$ – Jason Jan 13 '12 at 15:17
  • $\begingroup$ I'll edit my answer to add what you're missing. Please Wait. $\endgroup$ – user21436 Jan 13 '12 at 15:19
  • $\begingroup$ @Jason Note that $a\neq 0$, so $[a,b]\subseteq (0,\infty)$. $\endgroup$ – M Turgeon Jan 13 '12 at 15:21
  • $\begingroup$ @MTurgeon Ah thanks your right $\endgroup$ – Jason Jan 13 '12 at 15:27


1) Use the Mean Value Theorem to obtain the inequality.

2) Set $b=2a$. In this case $f(b)-f(a)>1/2$. So, can you choose $a$ and $b$ as closely as you like and still have $|f(b)-f(a)|$ big?

  • $\begingroup$ Thanks :) Your 2'nd hint combined with Kannappan's answer helped me complete the question :) $\endgroup$ – Jason Jan 13 '12 at 15:33

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