Let $f(x)$ be twice-differentiable and $f(a)=f(b)=0$ with $a<b$
How to prove that
$$\max|f(x)| \le \frac 18(b-a)^2 \max|f''(x)|$$
(With $\max|f(x)|$ beeing maximum (or minimum) between a and b.)
How would I use Taylor-Expansion and Mean Value Theorem here?
Remark: $\frac{1}{8}$ is the best constant. For example, $f(x)=x(1-x)$, $f''(x) = -2$ and $\max_{x\in [0,1]} f(x) = \frac{1}{4} = \frac{1}{8}(1-0)^2 \max_{x\in [0,1]} |f''(x)|.$
Assume that $f(x)$ is not identically zero. Clearly, there exists $c\in (a,b)$ such that $\max_{x\in [a,b]} |f(x)| = |f(c)| > 0$.
Let $$g(x) = f(x) - f(c)\frac{(x-a)(x-b)}{(c-a)(c-b)}.$$ We have $g(a) = g(b) = g(c) = 0$. By MVT, we have $g'(\xi_1) = 0$ for some $\xi_1\in (a, c)$ and $g'(\xi_2) = 0$ for some $\xi_2\in (c, b)$. By MVT again, we have $g''(\xi_3) = 0$ for some $\xi_3 \in (\xi_1, \xi_2)$ which results in $$f''(\xi_3) = \frac{2f(c)}{(c-a)(c-b)}.$$ Thus, we have $$|f(c)| = \frac{1}{2}(c-a)(b-c)|f''(\xi_3)| \le \frac{1}{2}\frac{(c-a + b-c)^2}{4}|f''(\xi_3)| = \frac{1}{8}(b-a)^2 |f''(\xi_3)|.$$ We are done.
I also did not get any 1/8, or used Taylor expansion, but...
Assume there is only 1 maximum. since that's the most extreme case for the bound, it should cover all other cases. Define $c$ the value in $[a,b]$ that achieves this. since $f$ is smooth, then if $c\neq a$ and $c\neq b$ then $f'(c)=0$.
Now we use MVT to find a bound on the second derivative.
$\exists z_1\in [a,c]$ where $f'(z_1)=f(c)/(c-a)$
$\exists z_2\in [c,b]$ where $f'(z_2)=-f(c)/(b-c)$
Therefore $\exists z_3\in [z_1,z_2]$ where
$$f''(z_3)=\frac{f'(z_2)-f'(z_1)}{z_2-z_2}\\ =\frac{f(c)/(c-a)+f(c)/(b-c)}{z_2-z_1}\\ =f(c)\frac{b-a}{(b-c)(c-a)(z_1-z_2)}\\ $$
To get a bound, we want to pick $c,z_1,z_2$ to minimize the right hand side, so we pick $c=(b+a)/2$, $z_2=b$, $z_1=a$. Simplifying gives
$f''(z_3)\geq 4f(c)/(b-a)^2\iff f(c) \leq \frac{(b-a)^2}{4}f''(z_3)$
I'm not quite sure where that extra factor of 2 is coming in since I think without more assumptions MVT is tight...
