8
$\begingroup$

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 $$\left( \int\limits_0^{1/2}xf(x)dx\right)^2+\left( \int\limits_{1/2}^1(x-\frac{1}{2})f(x)dx\right)^2\leq \frac{1}{24}\int\limits_{0}^1f^2(x)dx.$$

But I can't deduce the result. Please help me. Thank in advanced.

$\endgroup$
12
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Where does the first line come from? $\endgroup$ – nbubis Apr 14 '15 at 15:47
  • 3
    $\begingroup$ @nbubis: if $f$ has mean zero, $\int_{0}^{1}\frac{1}{2}\,f(x)\,dx = 0$. $\endgroup$ – Jack D'Aurizio Apr 14 '15 at 15:49
  • 2
    $\begingroup$ Must.. get.. more.. sleep :) $\endgroup$ – nbubis Apr 14 '15 at 15:51
  • 4
    $\begingroup$ One could also notice that the $1/2$ is the optimal choice of constant. $\endgroup$ – mickep Apr 14 '15 at 15:54
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.