I was working through Strichartz's "The Way of Analysis" and got stuck on this problem (6.1.5 #14):
"If $f$ is $\mathcal{C}^2$ on $[a,b]$ with $|f''(x)| \leq M_2$ and $f(a)=f(b)=0$, prove that $|f(x)| \leq M_2|x-a||x-b|$."
So far I've done the following:
By the Mean Value Theorem, there exists a $y \in [a,x]$ such that $f(x)-f(a)=f'(y)(x-a)$.
By Rolle's Theorem, there exists a $z \in [a,b]$ such that $f'(z)=0$.
By the Mean value Theorem, there exists a $w \in [y,z]$ (or possibly the other order, just between the two points) such that $f'(y)-f'(z)=f''(w)(y-z)$.
Substituting this and using that $f(a)=0$ and $f'(z)=0$ we get that: $f(x)=f''(w)(y-z)(x-a)$.
I would be done if I could show that $|y-z| \leq |x-b|$ but I don't see why this would necessarily have to be true. Does anyone have advice, or am I overlooking something simple?
As you have already observed, there exists $c\in (a,b)$ such that $f'(c) = 0$. Hence $$ |f'(t)| = \left| \int_c^t f''(s)\, ds\right| \leq M_2 |t-c|. $$
Let $x\in [a,b]$.
If $x\in [a,c]$, then $|t-c| \leq |x-c| \leq |x-b|$ for every $t\in [x,c]$, hence $$ |f(x)| = \left|\int_a^x f'(t)\, dt\right| \leq M_2 \int_a^x |t-c|\, dt \leq M_2 |x-a|\, |x-b|. $$
Similarly, if $x\in [c,b]$, then $|t-c| \leq |x-a|$ for every $t\in [c,x]$, hence $$ |f(x)| = \left|\int_b^x f'(t)\, dt\right| \leq M_2 \int_x^b |t-c|\, dt \leq M_2 |x-a|\, |x-b|. $$