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I came across the following exercise in Spivak's book: Suppose that you have a function $f$ such that $f'(x) \ge M > 0$, for any $x \in [0,1]$. I would like to show that there is an interval of length $1/4$ contained in $[0,1]$, on which $|f(x)| \ge M/4$.

My feeling is that this can be shown using the Mean Value Theorem: Applying this yields, for any $x \in (0,1]$, that $$ \frac{f(x) - f(0)}{x-0} \ge M. $$ However, I don't see how to proceed from here, as I don't have a lower bound for $f(0)$.

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up vote 1 down vote accepted

Here's a proof that doesn't use the MVT. (Well, perhaps indirectly.) Suppose to the contrary no interval of length 1/4 satisfies the property. Hence, there must be a $t_0 \in [0,\frac{1}{4}]$ so that $|f(t_0)| < \frac{M}{4}$. We then have $\frac{-M}{4} \leq f(t_0)$, from which we obtain the inequality $$f(t + t_0) \geq Mt + f(t_0) \geq M\bigg(t - \frac{1}{4}\bigg).$$ We now see that if $t \geq \frac{1}{2}$, then $f(t + t_0) \geq \frac{M}{4}$. But $t_0 + \frac{1}{2} \leq \frac{3}{4}$, and so $f(x) \geq \frac{M}{4}$ on the interval $[t_0 + \frac{1}{2},1] \supset [\frac{3}{4},1]$.

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Nice answer, thanks! :) – somebody Oct 23 '12 at 4:58

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