# If $f(0) = f(1)=0$ and $|f'' | \leq 1$ on $[0,1]$, then $|f'(1/2)|\le 1/4$

Let $f : [0,1] \rightarrow \mathbb{R}$ be a function whose second order derivative $f''(x)$ is continuous on $[0,1]$. Suppose that $f(0) = f(1)=0$ and that $|f''(x)| \leq 1$ for any $x \in [0,1]$.

Prove that $|f'(\frac{1}{2})| \leq \frac{1}{4}$.

I think we need to use the mean value theorem, and I have proven that $|f(x)| \leq \frac{1}{8}$ for any $x \in [0,1]$. I'm not sure how to proceed, though. Could someone please help? Thanks!

• How did you prove $|f(x)|\leq 1/8$? Commented Dec 16, 2014 at 13:46
• @Behaviour okay ! Done :)
– r9m
Commented Dec 16, 2014 at 17:22
• @r9m Not a duplicate at all...? This question asks $|f'(1/2)| \le 1/4$, that other question asks $|f'(x)| \le 1/2 \forall x$. Commented Dec 16, 2014 at 18:00
• it's not a duplicate since in that question integrals are forbidden Commented Dec 16, 2014 at 18:04
• @Zarrax Noted ! thanks ! I retracted my close vote ! :)
– r9m
Commented Dec 16, 2014 at 18:12

As Najib Idrissi pointed out in chat, the argument given by the answers in this question solves your question as well ! Substitute $x = \frac{1}{2}$ in the final inequality.

Integrate by parts to verify that for $f(0) = f(1) = 0$,

$$f'(x) = \int_{0}^{x}tf''(t)\,dt - \int_{1}^{x}(t-1)f''(t)\,dt$$

Take absolute value on both sides and use triangle inequality:

\begin{align}|f'(x)| &= \left|\int_{0}^{x}tf''(t)\,dt - \int_{1}^{x}(t-1)f''(t)\,dt\right| \\ &\le \left|\int_{0}^{x}tf''(t)\,dt\right| + \left|\int_{1}^{x}(t-1)f''(t)\,dt\right| \\ &\le \max_{x \in (0,1)} |f''(x)|\left(\left|\int_{0}^{x}t\,dt\right| + \left|\int_{1}^{x}(t-1)\,dt\right|\right) \\& \le \frac{1}{2}(x^2 + (1-x)^2) \end{align}

Thus, $\displaystyle f'\left(\frac{1}{2}\right) \le \frac{1}{4}$.

• I downvote! (-1) Commented Dec 16, 2014 at 18:12
• @Venus
– r9m
Commented Apr 7, 2015 at 6:30