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Today we had a probational exam in analysis. I wasn't able to solve one of the exercises and I have no idea what theorem to apply in order to solve it:

Let $I=[0,1]$ and $f: I \rightarrow \mathbb{R}$ be continuously differentiable. Assuming that $f$ has a root. Show: $$\max_{x \in I} |f(x)| \leq \max_{x \in I} |f'(x)|$$

Does this theorem have a name? What other theorem will I need in order to prove it? I'm sure the fact that the function has a root is important, but I don't see why and how to make use of it...

Thanks for your help!

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3 Answers 3

up vote 9 down vote accepted

The example $f(x)\equiv 1$ shows that having a root is important.

Let $x_0\in I$ be such that $|f(x_0)|=\max_{x\in I}|f(x)|$, and let $r\in I$ be such that $f(r)=0$. Then $\left|\frac{f(x_0)-f(r)}{x_0-r}\right|\geq |f(x_0)|$, and you can apply the Mean Value Theorem to finish. Continuity of the derivative isn't necessary, just continuity of $f$ on $I$ and differentiability on $(0,1)$, except that with continuity of $f'$ you can really say that the right-hand side of the inequality is a max rather than a sup.

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This is so far the most inspiring answer to this question, thanks. –  awllower Feb 21 '11 at 14:01

I don't know if this theorem has a name, but you're guessing right that it is important that $f$ has a root (otherwise adding a constant allows you to make the values of $f$ arbitrarily large without changing $f'$). One proof is a straightforward application of the fundamental theorem of analysis and a standard estimate on the integral. Indeed, if $x_{0}$ is such that $f(x_{0}) = 0$ then $f(x) = \int_{x_{0}}^{x} f'(t)\,dt$. Now combine this with the fact that $|\int_{a}^{b} h| \leq \int_{a}^{b}|h|$ for all $a,b$ and all integrable $h$.

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Hints: Mean value theorem between a root and any point + the fact that the diameter of $I$ is $1$.

By the way, continuity on the closed interval $I$ and differentiability on the interior of $I$ are sufficient.

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