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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 either..to 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?

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MVT actually does work here, but you have to use it many times in succession... –  Potato Aug 22 '13 at 5:31

3 Answers 3

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 $f=0$.

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(+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

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...

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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.

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Just a tip, you can edit your previous post instead of deleting it. That way you can also keep the upvotes on the post :). –  Daniel Rust Feb 8 at 12:38

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