Let $f$ be continuous on $[0,1]$ and differentiable on $(0,1)$, with $f(0) = f(1) = 0$ and $f(k)=1$ for some $k\in (0,1)$. Prove that there exist $c \in (0,1)$ such that $|f'(c)|>2$

I tried the taylor expansion of $f$ at $x=k$,ie $f(x) = f(k) + f'(\xi)(x-k) = 1+f'(\xi)(x-k)$.

$f(0) = 1+ f'(\xi_0)(-k) = 1+ f'(\xi_1)(1-k) = f(1) = 0$

But i am not sure how to show the desired inequality. Any hints?

  • $\begingroup$ You might use the mean value theorem. $\endgroup$ – Brian Borchers Nov 16 '17 at 17:04
  • $\begingroup$ I would try MVT $\endgroup$ – Rodrigo Dias Nov 16 '17 at 17:05
  • $\begingroup$ hmm im not sure what function i should use MVT on, its probably not on f and the intervals [0,k] and [k,1], right? $\endgroup$ – Timothy Nov 16 '17 at 17:10
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    $\begingroup$ Do you mean $|f'(c)| \geq 2$? $\endgroup$ – Gibbs Nov 16 '17 at 17:17
  • $\begingroup$ @Gibbs I don't think so, perhaps you have some counterexamples in mind? $\endgroup$ – Timothy Nov 16 '17 at 17:19

Use the mean value theorem:

$$f'(c_0)=\frac{f(k)-f(0)}{k-0}=\frac{1}{k}\\f'(c_1)=\frac{f(1)-f(k)}{1-k}=\frac{-1}{1-k}$$ Now, there are 3 cases: $\begin{cases}k\in(0,\frac12)\\k=\frac12\\k\in(\frac12,1)\end{cases}$

In case 1, $f'(c_0)>\frac1{\frac12}=2$

In case 3, $f'(c_1)<-\frac1{1-\frac12}=-2\implies |f'(c_1)|>2$

In case 2, both $f'(c_0),|f'(c_1)|$ are equal to 2, if every point between $0$ to $k$ and between $k$ to $1$ has slope of $2$ in one side and $-2$ in the other then $f'(k)$ is undefined​, so I know that it can't be, but I also know that the average is $2$ in one side and $-2$ in the other, I know that because the average value of a function is: $$\frac1{b-a}\int_a^b g(x) dx$$ put in this $b=\frac12,a=0$ or $b=1, a=\frac12$, and $g(x)=f'(x)$ and you will get $2$ or $-2$

So there exist at leas one point between $0$ and $k$ that has slope is greater than $2$ and at least one point between $k$ and $1$ that has slope that is less than $-2$

  • $\begingroup$ I can see that "the average is or −2−2" is pretty intuitive but is there a more precise way of phrasing it? $\endgroup$ – Timothy Nov 16 '17 at 17:36
  • $\begingroup$ @Timothy I added it to the answer $\endgroup$ – Holo Nov 16 '17 at 17:51
  • $\begingroup$ We can handle the second case without using integrals. See my answer. $\endgroup$ – Paramanand Singh Nov 16 '17 at 18:41
  • $\begingroup$ @ParamanandSingh i understand this but i still think that using average like this is easier, the integral may look like it require some work but all it really says is that the average slope is the slope of the line that pass through the endpoints. computing it is really easy $\endgroup$ – Holo Nov 16 '17 at 18:53
  • $\begingroup$ There is nothing wrong with the integral approach but it requires the derivative $f'$ to be integrable. $\endgroup$ – Paramanand Singh Nov 16 '17 at 19:41

It is easy to dispose of the cases $k<1/2$ and $k>1/2$ using the equations $$f(0)+kf'(c_{1})=f(k)=1=f(k)=f(1)+(k-1)f'(c_{2})$$ The case when $k=1/2$ requires a bit more analysis. If $f'(1/2)>0$ then there is a point $k'>1/2$ such that $f(k') >f(1/2)=1$ and then $$1<f(k')=f(1)+(k'-1)f'(c_{3})$$ so that $|f'(c_{3})|>2$. Similarly we can deal with the case $f'(1/2)<0$.

Let's consider the case when $f'(1/2)=0$ and $f(1/2)=1$ is the maximum value of $f$ in $[0,1]$. If $f'(c) \leq 2$ for all $c\in(0,1/2)$ then we can see that $$f(x)=f(0)+xf'(c)\leq 2x,f(x)=f(1/2)+(x-1/2)f'(d)\geq 1+2(x-1/2)=2x$$ so that $f(x) =2x$ for all $x\in[0,1/2]$ and this contradicts $f'(1/2)=0$. Thus we must have some point $c\in(0,1/2)$ for which $f'(c) >2$.

  • $\begingroup$ Sorry, I don't see how $f(x) \ge 2x$ since $f'(d) /le 2$. Can you elaborate on it please? $\endgroup$ – Timothy Nov 17 '17 at 7:07
  • $\begingroup$ @Timothy: $f'(d) \leq 2$ and multiply this with negative number (actually non-positive number) $(x-1/2)$ then the inequality is reversed. $\endgroup$ – Paramanand Singh Nov 17 '17 at 7:15

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