Proving that the function is bounded by the function of its derivative on the given interval

Question

Suppose $h:R \longrightarrow R$ is differentiable everywhere and $h'$ is continuous on $[0,1]$, $h(0) = -2$ and $h(1) = 1$. Show that:

$|h(x)|\leq max(|h'(t)| , t\in[0,1])$ for all $x\in[0,1]$

I attempted the problem the following way: Since $h(x)$ is differentiable everywhere then it is also continuous everywhere. $h(0) = -2$ and $h(1) = 1$ imply that h(x) should cross x-axis at some point (at least once). Denote that point by c to get $h(c) = 0$ for some $c\in[0,1]$.

$h'(x)$ continuous means that $lim[h'(x)] = h'(a)$ as $x\rightarrow a$ but then I am stuck and I don't see how what I have done so far can help me to obtain the desired inequality.

Thank you in advance!

Answer 1

If $\max_{x\in[0,1]}|h(x)|=|h(0)|=2$ we have by the mean value theorem

$h'(\xi)=\frac{h(1)-h(0)}{1-0}=3$ for some $\xi\in(0,1)$ and You´re done.

If $\max_{x\in[0,1]}|h(x)|=|h(x_0)|$ for some $x_0\in(0,1)$ and $h(x_0)>0$ we have by the mean value theorem

$h'(\xi)=\frac{h(x_0)-h(0)}{x_0-0}\geq h(x_0)$ for some $\xi\in(0,x_0)$ and You´re done. If $h(x_0)<0$ we have by the mean value theorem

$|h'(\xi)|=|\frac{h(1)-h(x_0)}{1-x_0}|\geq 1+|h(x_0)|>|h(x_0)|$ for some $\xi\in(x_0,1)$.

Answer 2

Since $h(c) = 0$, then $h(x) = \int_c^x h'(t)\, dt$. Thus, for all $x\in [0,1]$,

$$\lvert h(x)\rvert \le \lvert x - c\rvert \max_{0\le t \le 1} \lvert h'(t)\rvert \le \max_{0 \le t \le 1} \lvert h'(t)\rvert$$

Answer 3

We can do this without assuming $h'$ is continuous, if we replace $\max |h'|$ with $\sup |h'|.$

Because $h$ is differentiable, it is continuous. The intermediate value theorem then gives $x_0\in (0,1)$ such that $h(x_0)=0.$ Let $x\in [0,1].$ Then by the mean value theorem,

$$|h(x)| = |h(x)-h(x_0)| = |h'(c_x)(x-x_0)| \le \sup_{[0,1]}|h'||x-x_0| \le \sup_{[0,1]}|h'|.$$

Note that if the supremum is finite, we actually get $|h(x)| < \sup_{[0,1]}|h'|,$ since $|x-x_0| < 1.$