# L^p norm

• Mar 13th 2009, 12:51 PM
math8
L^p norm
(X, sigma, measure) is a finite measure space. Show that as p--> infty, lim |f|_p = |f|_infty, where f lies in L^infty.

I know that
|f|_infty= ess sup |f|= smallest number k st measure ({x:|f||>k})=0

I would note that given e>0, measure({x in X: |f(x)|> |f|_infty - e} ) > 0.

But from here, I am kind of lost.
• Mar 14th 2009, 02:53 AM
Opalg
Quote:

Originally Posted by math8
(X, sigma, measure) is a finite measure space. Show that as p--> infty, lim |f|_p = |f|_infty, where f lies in L^infty.

I know that
|f|_infty= ess sup |f|= smallest number k st measure ({x:|f||>k})=0

I would note that given e>0, measure({x in X: |f(x)|> |f|_infty - e} ) > 0.

But from here, I am kind of lost.

Since $|f(x)|\leqslant\|f\|_\infty$ for almost all x, it follows that $\textstyle\|f\|_p = \left(\int|f|^p\right)^{1/p} \leqslant \left(\int\|f\|_\infty^p\right)^{1/p} = \left(\|f\|_\infty^p\mu(X)\right)^{1/p} \to \|f\|_\infty$ as p→∞. Hence $\limsup_{p\to\infty}\|f\|_p\leqslant \|f\|_\infty$.

Let $m<\|f\|_\infty$, and let $S_m = \{x\in X:|f(x)|\geqslant m\}$. Then $\mu(S_m)>0$, and $\textstyle\|f\|_p \geqslant \left(\int_{S_m}\!\!m^p\right)^{1/p} = \left(m^p\mu(S_m)\right)^{1/p} \to m$ as p→∞. Since this holds for all $m<\|f\|_\infty$, it follows that $\liminf_{p\to\infty}\|f\|_p\geqslant \|f\|_\infty$.