Measure inequality in $L^2$

Question

Let $f\in L^2([0,1])$ such that $\Vert f \Vert^2=1$ and $\displaystyle \int_{[0,1]} fdx>\alpha>0.$ If $E_{\beta}=\{x\in [0,1]: f(x)\geq \beta\}$ and $0<\beta<\alpha$, prove that $$m(E_{\beta})\geq (\alpha-\beta)^2.$$ I’ve been stuck in this problem for hours and I don’t know what to do. Please help.

Answer 1

Let $g = 1_{E_{\beta}}$, by holder

$||fg ||_1 \leq || f||_ 2 ||g||_2 \Leftrightarrow \int_{E_{\beta}} f \leq m(E_{\beta})^{1/2} $

Notice that

$\alpha< \int_{[0,1]} f = \int_{E_{\beta}}f + \int_{E_{\beta}^c}f $

but then

$\int_{E_{\beta}}f > \alpha - \int_{E_{\beta}^c}f > \alpha -\beta$

so

$m(E_{\beta})^{1/2} \geq \alpha -\beta \Leftrightarrow m(E_{\beta}) \geq (\alpha -\beta)^{2}$