# For all integrable $f:[-1,1]\mapsto \mathbb{R}$ prove that $\int_{-1}^1f^2(x)\ge\frac12(\int_{-1}^1f(x))^2+\frac32(\int_{-1}^1xf(x))^2$

For all integrable $f:[-1.1]\mapsto \mathbb{R}$ peove that $$\int_{-1}^1f^2(x)dx\ge\frac12\left(\int_{-1}^1f(x)dx\right)^2+\frac32\left(\int_{-1}^1xf(x)dx\right)^2$$

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If $f$ is even then the second term on the right hand side vanishes, if it is odd the first term vanishes. The decomposition of $f$ in its even and odd parts could perhaps help. – Sebastien B Jan 26 '13 at 18:37

Apply The Cauchy-Schwarz Inequality for Integrals on the right side.

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thanks i know The Cauchy-Schwarz Inequality for Integrals – Maisam Hedyelloo Jan 26 '13 at 18:51
Christ's sister: i appreciate you . you helped me to solve it. i knew it but you remember me to use it, – Maisam Hedyelloo Jan 26 '13 at 19:09
@MaisamHedyelloo: So, it deserves a +1 for remembering you. – Babak S. Jan 26 '13 at 19:16

If $f$ is even, then $xf(x)$ is odd so its integral over $[-1,1]$ is zero and the inequality boils down to $$\int_{-1}^1 f^2(x)dx \geq \frac{1}{2}\left( \int_{-1}^1 f(x)dx\right)^2$$ and the result follows from Cauchy-Schwarz.

If $f$ is odd, then its integral over $[-1,1]$ is zero and the inequality becomes $$\int_{-1}^1 f^2(x)dx \geq \frac{3}{2}\left( \int_{-1}^1 xf(x)dx\right)^2$$ and the result follows again from Cauchy-Schwarz.

Now write $f=g+h$ with $g$ even and $h$ odd.

We find $$\int_{-1}^1 (g+h)^2=\int g^2+\int h^2 + 2\int gh= \int g^2+\int h^2$$ since $gh$ is odd.

Also $$\left( \int g+h\right)^2=\left( \int g\right)^2+ \left( \int h\right)^2+ 2 \int g\int h=\left( \int g\right)^2$$ since $h$ is odd and $\int h=0$.

Finally, we have $$\left( \int x(g+h)\right)^2= \left( \int xg\right)^2 + \left( \int xh\right)^2+ 2\int xg\int xh=\left( \int xh\right)^2$$ since $xg(x)$ is odd and $\int xg=0$.

It only remains to add the two inequalities for $g$ and $h$, to obtain the one for $f$.

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Nice attempt +1 – Babak S. Jan 26 '13 at 19:16
@Babak Thank you! – 1015 Jan 26 '13 at 20:13

By Cauchy-Schwarz we have

$$\left(\int_{-1}^1f(x)dx\right)^2 \leq 2 \left(\int_{-1}^1f^2(x)dx\right)$$ $$\left(\int_{-1}^1xf(x)dx\right)^2 \leq \left(\int_{-1}^1x^2dx\right) \left(\int_{-1}^1f^2(x)dx\right)=\frac{2}{3} \left(\int_{-1}^1f^2(x)dx\right)$$

This proves the inequalities for odd and even functions.

Then use SebastienB's hint.

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Please spell Cauchy-Schwarz correctly. – Phira May 30 '14 at 5:49