real analysis problem.

liyi

Suppose $1<p<\infty$ . Define
$\mathcal{F} = \left\{ f\in L([0,1]): \int_0^1 |f(x)|dx = 1, \int_0^1 |f(x)|^p dx = 2 \right\},$
Prove that $\forall 0<\varepsilon<1$ , $\exists \delta>0$ , such that
$m(\{ x\in[0,1]: |f(x)|>\varepsilon \}) \geq \delta, \quad f\in \mathcal{F}.$

**Posted:** Mon Jan 31, 2005 5:26 pm

Myth

Let $E=\{x\mid |f(x)|>\epsilon\}$ , $f\in \mathcal{F}$ . Suppose WLOG $f\geq 0$ .
We have $\int_{E}f(x)dx>1-\epsilon(1-m(E))$ . Moreover, form Holder inequality $\int_{E}f(x)dx=\int_{E}1\cdot f(x)dx \leq m(E)^{1/q}||f||_p\leq 2m(E)^{1/q},$ where $1/p+1/q=1$ .
Define $m=m(E)$ . Then we have obtained $1-\epsilon(1-m)<2m^{1/q}$ . It follows $1-\epsilon<2m^{1/q}$ , i.e. $m>\left(\frac{1-\epsilon}{2}\right)^q$ for all $f\in\mathcal F$ .

**Posted:** Tue Feb 01, 2005 6:01 am

_________________
Myth is out of here