Prove that $|f''(\xi)|\geqslant\frac{4|f(a)-f(b)|}{(b-a)^2}$

Let ${\rm f}:\left[a, b\right]\to\mathbb{R}$ be twice differentiable, and suppose

$$\lim_{x\to a^{+}} \frac{{\rm f}\left(x\right) - {\rm f}\left(a\right)}{x - a} = \lim_{x\to b^{-}}\frac{{\rm f}\left(x\right) - {\rm f}\left(b\right)}{x - b} =0 $$

Show that there exists $\xi \in \left(a, b\right)$ such that $\displaystyle{% \left\vert\vphantom{\Large A}\,{\rm f}''\left(\xi\right)\right\vert \geq \frac{4\left\vert\vphantom{\Large A}% \,{\rm f}\left(a\right) - {\rm f}\left(b\right)\right\vert} {\left(b - a\right)^{2}}}$.

I don't know how to start. Any hints ?.

Solution 1:

let $g(x) = f'(x)$, then with $g(a) = g(b)= 0$ we are going to prove

$$|\int_{a}^{b}g(x)dx| \leq\left(\sup_{\xi \in (a,b)}|g'(\xi)|\right)\frac{(b-a)^2}{4}$$

\begin{align} |\int_{a}^{\frac{a+b}{2}}g(x)dx| &= |\int_{a}^{\frac{a+b}{2}}g(x)-g(a)dx| \\ &\leq \left(\sup_{\xi \in (a,b)}|g'(\xi)|\right)\int_{a}^{\frac{a+b}{2}}(x-a)dx \\ &= \left(\sup_{\xi \in (a,b)}|g'(\xi)|\right) \frac{(b-a)^2}{8} \end{align}

Similarly we could also get $|\int_{\frac{a+b}{2}}^{b}g(x)dx| \leq\left(\sup_{\xi \in (a,b)}|g'(\xi)|\right) \frac{(b-a)^2}{8}$, then the result follows by adding the two inequalities

Solution 2:

An alternative, more enlightening proof that follows what has been done here https://math.stackexchange.com/a/835185/66096

For variational purposes, we assess the integral $\displaystyle \int_a^bf''(t)(t+\beta)$dt where $\beta$ is arbitrary.

By integration by parts, and since $f'(a)=f'(b)=0$, this rewrites as $\displaystyle \int_a^bf''(t)(t+\beta)dt =f(a)-f(b)$

Hence $\displaystyle \max|f''|\int_a^b|t+\beta|dt\geq \left|\int_a^bf''(t)(t+\beta)dt \right| =|f(b)-f(a)| $

And $\displaystyle \max|f''|\geq \frac{|f(b)-f(a)|}{\int_a^b|t+\beta|dt}$

Choosing $\displaystyle \beta=-\frac{a+b}{2}$ yields the result.