Suppose $f(x)$ is a nonnegative convex function in $[0,1]$, prove an inequality.

Question

Suppose $f(x)$ is a nonnegative convex function in $[0,1]$. Prove:

$$\displaystyle \int_0^1f^2(x)\,\mathrm dx\leqslant\frac43\left(\int_0^1f(x)\,\mathrm dx\right)^2$$

I have tried Cauchy Mean Value Theorem: Construct $\displaystyle F(x)=\frac{\displaystyle \int_0^1 f^2(x)\,\mathrm dx}{\displaystyle \left(\int_0^1f(x)\,\mathrm dx\right)^2}$... But it doesn't work :-(
Any tips would be appreciated!

Not sure if it will work, but you're in a probability space and you have a convex function, so try Jensen's inequality. — Rick Sanchez, Apr 24 '16 at 04:03

score 5 · Accepted Answer · answered Apr 24 '16 at 04:59

5

Let $f(x)=x^2$. Then $$ \int_0^1 f^2(x)\,dx=\int_0^1x^4\,dx=0.2, $$ while $$ \frac43\,\left(\int_0^1 f(x)\,dx\right)^2=\frac43\,\left(\int_0^1x^2\,dx\right)^2=\frac4{27}=0.\overline{148}. $$

answered Apr 24 '16 at 04:59

Martin Argerami

205,756

All to easy! +1 – Mark Viola Apr 24 '16 at 05:01
Excuse me, but $f(x)=x^2$ should be concave function in $\mathbb R$? – Shine Mic Apr 24 '16 at 05:17
For any $x,y\in\mathbb R$, and $t\in[0,1]$, you have $$ (tx+(1-t)y)^2\leq tx^2+(1-t)y^2.$$ That makes it convex. – Martin Argerami Apr 24 '16 at 05:31

Suppose $f(x)$ is a nonnegative convex function in $[0,1]$, prove an inequality.

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