# Integral Inequality $\int\limits_0^1f^2(x)dx\geq12\left( \int\limits_0^1xf(x)dx\right)^2.$

Let $f:[0,1]\to\mathbb{R}$ be a continuous function such that $\int\limits_0^1f(x)dx=0$. Prove that $$\int\limits_0^1f^2(x)dx\geq12\left( \int\limits_0^1xf(x)dx\right)^2.$$

My approach as follow

Let $F(x)=\int\limits_0^xf(t)dt$. Integrate by part we have $$\int\limits_0^1F(x)dx=-\int\limits_0^1xf(x)dx.$$ By Cauchy-Schwarz inequality \begin{aligned} \left( \int\limits_0^{1/2}xf(x)dx\right)^2&\leq\left( \int\limits_0^{1/2}x^2dx\right)\left( \int\limits_0^{1/2}f^2(x)dx\right)=\frac{1}{24}\int\limits_0^{1/2}f^2(x)dx\\ \left( \int\limits_{1/2}^1(x-\frac{1}{2})f(x)dx\right)^2&\leq\left( \int\limits_{1/2}^1(x-\frac{1}{2})^2dx\right)\left( \int\limits_{1/2}^1f^2(x)dx\right)=\frac{1}{24}\int\limits_{1/2}^1f^2(x)dx \end{aligned} Hence $$\left( \int\limits_0^{1/2}xf(x)dx\right)^2+\left( \int\limits_{1/2}^1(x-\frac{1}{2})f(x)dx\right)^2\leq \frac{1}{24}\int\limits_{0}^1f^2(x)dx.$$

Since $f$ has mean zero, $$\int_{0}^{1} x\,f(x)\,dx = \int_{0}^{1}\left(x-\frac{1}{2}\right)\,f(x)\,dx,\tag{1}$$ and by the Cauchy-Schwarz/Buniakowski inequality we have: $$\left(\int_{0}^{1}\left(x-\frac{1}{2}\right)\,f(x)\,dx\right)^2 \leq \int_{0}^{1}f(x)^2\,dx \int_{0}^{1}\left(x-\frac{1}{2}\right)^2\,dx \tag{2}$$ so we just have to check that: $$\int_{0}^{1}\left(x-\frac{1}{2}\right)^2\,dx = \frac{1}{12},\tag{3}$$ that is straightforward.

• Where does the first line come from? – nbubis Apr 14 '15 at 15:47
• @nbubis: if $f$ has mean zero, $\int_{0}^{1}\frac{1}{2}\,f(x)\,dx = 0$. – Jack D'Aurizio Apr 14 '15 at 15:49
• Must.. get.. more.. sleep :) – nbubis Apr 14 '15 at 15:51
• One could also notice that the $1/2$ is the optimal choice of constant. – mickep Apr 14 '15 at 15:54

A "mechanical" way to deal with this question is to follow the proof of Cauchy Inequality and write

\begin{aligned} 0&\leq \int_0^1 (x+\lambda f+\mu)^2\\ &=E(f^2)\lambda^2+2\lambda E(xf) +\frac{1}{3}+\mu+\mu^2 \end{aligned}

Here $E(g)=\int_0^1 g,\;E^2(g)=(E(g))^2$. Thus

$$\Delta_1=4 E^2(xf)-4 E^2(f)\left(\frac{1}{3}+\mu+\mu^2\right)\leq 0$$

Now treat $\Delta_1\leq 0$ as an inequality of $\mu$,

$$\Delta_2=E(f^2)\left (-\frac{1}{3}E(f^2)+4 E^2(xf)\right)\leq 0$$

If $E(f^2)=0$ then $f=0$ almost everywhere, so done. Otherwise, we have

$$E^2(xf)\leq \frac{1}{12} E(f^2)$$