Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been stuck on this one for a while. Comes from an analysis qual question.

Let f be a function that is continuous on $\left[0,1\right]$ and differentiable on $(0,1)$. Show that if $f(0)=0$ and $|f'(x)| \leq |f(x)|$ for all $x \in (0,1)$, then $f(x)=0$ for all $x \in \left[0,1\right]$.

What I've tried doing so far is see if there was anything I could do with MVT. I didn't really see anything to do with definitions which I have a feeling I'll be playing around with them. Drawing a picture was a little difficult with these conditions as well

Any hints/suggestions?

share|cite|improve this question
MVT actually does work here, but you have to use it many times in succession... – Potato Aug 22 '13 at 5:31
up vote 10 down vote accepted

It suffices to show that $f=0$ on every interval $[0,b]\subsetneq [0,1]$ (because of continuity). To show this, let $x_0$ be the maximum of $|f|$ on $[0,b]$. Then,

$$|f(x_0)|=\left|\int_0^{x_0}f'(t)\, dt\right|\leqslant \int_0^{x_0}|f'(t)|\, dt\leqslant\int_0^{x_0}|f(t)|\, dt\leqslant x_0 |f(x_0)|$$

Since $x_0<1$, this implies that $|f(x_0)|=0$, and so $f=0$.

share|cite|improve this answer
(+1) nice answer. – Mhenni Benghorbal Aug 22 '13 at 5:23
This is nice, thanks. I also found an alternate solution that I don't quite understand, I'll try to digest it, maybe post it if people are interested. – DaveNine Aug 22 '13 at 6:12

There is also a proof using the MVT:

Let $x_{0} \in (0,1)$. The MVT implies that $\left | f(x_{0}) - f(0) \right | = \left| f'(x_{1}) \right| \left| x_{0} - 0 \right|$ for some $x_{1} \in (0,x_{0}).$ Also, $$ |f(x_{0})| = |x_{0}||f'(x_{1})| \leq |x_{0}| |f(x_{1})|$$ by assumption. Applying the MVT again gives $$ |f(x_{0})| \leq |x_{0}| |x_{1}| |f(x_{2})| $$ for some $x_{2} \in (0,x_{1}).$ Continuing in this way, we obtain a sequence of points $(x_{n})$ such that $x_{n} \to 0$ as $n \to \infty$ and $$ 0 \leq |f(x_{0})| \leq |x_{0}| ...|x_{n}| |f(x_{n+1})|.$$ But $f(x_{n}) \to 0$ as $n \to \infty$ (by continuity), so the squeeze theorem gives that $f(x_{0})=0.$ Thus $f(x)=0$ on $[0,1).$

Continuity then implies that $f(1)=0,$ completing the proof.

share|cite|improve this answer
Just a tip, you can edit your previous post instead of deleting it. That way you can also keep the upvotes on the post :). – Dan Rust Feb 8 '14 at 12:38

Here's an approach that's a little ad-hoc, but it works. Let $A=\max |f|$. Use MVT to bound $|f|$ under $Ax$, then look at $f(1/2)$ and show that it's too small...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.