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Let $f\in C([0,1])$ be differentiable on $(0,1)$. Suppose that $f(0)=f(1)=0$ and that there is an $x_0\in(0,1)$ where $f(x_0)=1$. I have to prove that there is a $c\in(0,1)$ satisfying $|f^\prime(c)|>2$.

This is what I have done:

$f(x)=\int^x_0f^\prime(t)dt=-\int^1_xf^\prime(t)dt$. Now suppose in the seek of a contradiction that $|f^\prime(x)|\leq2$ for every $x\in(0,1)$. Then we have $|f(x)|\leq\int^x_0|f^\prime(t)|dt\leq 2x$ and $|f(x)|\leq\int^1_x|f^\prime(t)|dt\leq 2(1-x)$. And so $|f(x)|\leq\mathrm{min}\{2x,2(1-x)\}$. Evaluating this inequality in $x_0$ we obtain $1\leq2\mathrm{min}\{x_0,1-x_0\}$ and this is a contradiction if $x_0\neq1/2$. Now, if $x_0=1/2$ we can easily reduce to the case $|f(x)|<1$ if $x\neq1/2$. But I can't continue the proof in this case. Could you help me ,please?

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up vote 2 down vote accepted

For the case where $x_0=1/2$, assume that $|f'(x)|\le 2$ for all $x$. Then we can conclude by the Mean Value Theorem that $$f(x)\le 2x \qquad\text{for}\qquad 0 \le x \le 1/2.$$ Similarly, or by symmetry, $f(x)\le 2-2x$ for $x\ge 1/2$.

Now we examine the difference quotient $\dfrac{f(x)-1}{x-1/2}$ that is used in the definition of $f'(1/2)$. If $x>1/2$, then $$\frac{f(x)-1}{x-1/2}\le \frac{2-2x - 1}{x-1/2}=-2.$$ If $x<1/2$, then $f(x)-1<2x-1$. But $x-1/2$ is negative. Therefore $$\frac{f(x)-1}{x-1/2} \ge \frac{2x-1}{x-1/2}=2,$$ since dividing by the negative number $x-1/2$ reverses inequalities. Thus the right difference quotient and the left difference quotient cannot have a common limit, contradicting the hypothesis that $f$ is differentiable at $x=1/2$.

Comment: Let's interpret the formal work above informally, since after all that's where the intuition for the argument came from. The bound $|f'(x)|\le 2$ would force a sharp peak at $x=1/2$ of the same general type as the one that the function $-|x|$ has at $0$.

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I thought I understood your answer, but now I think I didn't. Could you explain me why $f^\prime(1/2)$ doesn't exist? – Alex M Nov 19 '11 at 2:26
probabily I understood, it'd because for $x\in[0,1/2]$ the limit of the difference quotient is greater than 2, instead for $x\in[1/2,1]$ it's less than -2. Now I have an observation, in my proof I proved that $|f(x)|\leq 2x$ if $x\in[0,1/2]$ that is $-2x\leq f(x)\leq2x$. If we now make the difference quotient and then take the limit we obtain that this limit should be greater than 2 and less than -2 and this is a contradiction, right? – Alex M Nov 19 '11 at 2:44

One good reason to use the Mean Value Theorem instead of integration is that not all derivatives are integrable. That is, the step $f(x)=\int_0^x f'(t)dt$ is not true in full generality. See the MathOverflow question "Integrability of derivatives" for more on this.

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