Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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 $0$.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

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.