How to prove this integral-inequality.

Suppose $f$ is twice differentiable and satisfies $f(0)=0$. Prove the inequality. $$\int_0^1 |f(x)f'(x)| dx \le\ \frac{1}{2} \int_0^1 |f'(x)|^2 dx$$

This is a problem from undergraduate math competition in Korea.

I may use the Cauchy-Schwarz inequality, but it is difficult for me to rearrange this inequality to make form for the Cauchy-Schwarz inequality.

How to prove?

• As written, it's not true, consider e.g. $f(x) = 2x^2$. Is it maybe $\int_0^1 \lvert f(x) f'(x)\rvert\,dx$ on the left hand side? – Daniel Fischer Dec 7 '14 at 10:21
• Why did two people upvote this question I wonder? – Suzu Hirose Dec 7 '14 at 10:23
• @SuzuHirose This is an excellent question. Be ready for more if you continue on the site. – Did Dec 7 '14 at 11:49
• yes $\int_0^1 |f(x)f'(x)| dx \le\ \frac{1}{2} \int_0^1 |f'(x)|^2 dx$ – Younghoon Jeon Dec 7 '14 at 12:02
• @Did Like Michael Knight, I am here to prove that one person can make a difference. – Suzu Hirose Dec 7 '14 at 14:13

$$\int f f' dx=\int f df=1/2f^2$$ Thus $$\int_0^1f(x)f'(x)dx=f(x)|_0^1=1/2f(1)^2$$ since $f(0)=0$. Let $g(x)=1$ then by Cauchy's theorem $$\left|\int (f')^2 dx\right|\left|\int g^2 dx\right|\geq\left(\int f'g\,dx\right)^2=(f)^2$$ Thus since $\int_0^1 g(x)dx=1$ $$1/2\int_0^1 (f')^2 dx\geq1/2f(1)^2=\int_0^1 f f' dx$$ The only part I did not get was the absolute value sign around $ff'$, could this be another typo?
• $\int_0^1|\sin'2\pi x|dx=2\pi\int_0^1|\cos2\pi x|dx=8\pi>\int\sin'2\pi x dx=2\pi\int_0^1\cos2\pi x=0$ – Suzu Hirose Dec 7 '14 at 14:35
• To complete this, consider the function $$h(x)=\int_0^x|f'(t)|\,\mathrm dt,$$ and note that $h\geqslant|f|$ and $h'=|f'|$ hence the inequality you proved shows that $$\int(f')^2=\int(h')^2\geqslant2\int hh'\geqslant2\int|f|\,|f'|.$$ – Did Dec 8 '14 at 9:13