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Let $f$ is two differentiable function on $[0,1]$ such that $f(0)=1$, $f(1)=0$, $f'(x)\le 0$, $f''(x)\ge0$

Show that

$$\int_{0}^{1}x \sqrt{1+\{f'(x)\}^2}dx\le\ \frac{1}{\sqrt{2}}$$

How do have it? Thank you.

JW Dark
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1 Answers1

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By integration by parts the given integral equals

$$\left[\frac{x^2}{2}\sqrt{1+f'(x)^2}\right]_{0}^{1}-\int_{0}^{1}\frac{x^2}{2}\cdot\frac{-2f'(x)f''(x)}{2\sqrt{1+f'(x)^2}}\,dx $$ or $$ \frac{1}{2}\sqrt{1+f'(1)^2}+\frac{1}{2}\int_{0}^{1}\frac{x^2 f'(x) f''(x)}{\sqrt{1+f'(x)^2}}\,dx\leq \frac{1}{2}\sqrt{1+f'(1)^2} $$ since $f'(x)\leq 0$ and $f''(x)\geq 0$.
$|f'(1)|\leq 1$ holds by convexity, so we are done.

Jack D'Aurizio
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  • Why is there a minus sign in the numerator of the integral after the derivation of $\sqrt{1+f'(x)^2}$? I don't really get it, maybe is trivial but I don't see it. Thanks. – Bernkastel Mar 12 '20 at 17:01
  • @Dunkelheit: integration by parts gives $$\int_{a}^{b}f(x)g(x),dx = \left[F(x)g(x)\right]{a}^{b}-\int{a}^{b}F(x)g'(x),dx.$$ – Jack D'Aurizio Mar 16 '20 at 12:38
  • Thanks for you answer, so if $f(x)=x$ and $g(x)=\sqrt{1+f'(x)^2}$ shouldn't we get $$\left[\frac{x^2}{2}\sqrt{1+f'(x)^2}\right]_0^1-\int_0^1 \frac{x^2}{2} \frac{1}{2}\frac{2f'(x)f''(x)}{\sqrt{1+f'(x)^2}}\text{d}x=\frac{1}{2}\sqrt{1+f'(1)^2}-\frac{1}{2}\int_0^1 \frac{x^2f'(x)f''(x)}{\sqrt{1+f'(x)^2}}\text{d}x$$ Since $$\left(\sqrt{1+f'(x)^2}\right)'=\frac{1}{2}\frac{2f'(x)f''(x)}{\sqrt{1+f'(x)^2}}$$ So I still don't get why there is a minus sign in the numerator of the first integrand function you've written, that leads to a plus sign in front of the second integral you've written. – Bernkastel Mar 16 '20 at 15:18