1

I 'm not getting the solution by David Giraudo for Limit of $L^p$ norm completely. The step in which i'm having problem is

As $|f(x)|\leqslant\lVert f\rVert_\infty$ for almost every $x$, we have for $p>q$, $$ \lVert f\rVert_p\leqslant\left(\int_X|f(x)|^{p-q}|f(x)|^qd\mu\right)^{1/p}\leqslant \lVert f\rVert_\infty^{\frac{p-q}p}\lVert f\rVert_q^{q/p},$$

Please give the detailed explanation for this completely.

Styles
  • 3,609

1 Answers1

5

The first inequality is actually an equality since $|f(x)|^p=|f(x)|^{p-q}|f(x)|^q$.

To get to the second step, replace $|f(x)|^{p-q}$ by $\|f\|_{\infty}^{p-q}$, which you can do since $|f(x)| \leq \|f\|_{\infty}$, and factor it out of the integral. Also note that $\left(\int_X |f(x)|^q \ d\mu\right)^{1/p} =\left( \left(\int_X |f(x)|^q \ d\mu \right)^{1/q} \right)^{q/p} = \|f\|_q^{q/p}$.

kccu
  • 21,188