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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!

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

1 Answer 1

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

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

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