How is $\|f\|_p \geq \bigg( \int_A |f|^p d \mu \bigg)^{1/p}$?

Question

How is $\|f\|_p \geq \bigg( \int_A |f|^p d \mu \bigg)^{1/p}$?

This is presented as "trivial inequality". However since this is essentially the definition of $\|f\|_p$, then I don't see how $>$ is trivial?

https://ocw.mit.edu/courses/mathematics/18-125-measure-and-integration-fall-2003/lecture-notes/18125_lec17.pdf

Answer 1

Because the set $A$ may be a subset of the "whole" space, say $\mathbb{R}$, where you define the $L^p$ norm, and the right-hand side of your inequality is $$ \left(\int_{\mathbb{R}} |f|^p1_A\ d\mu\right)^{1/p}. $$ But $$ \|f\|_p=\bigg(\int_{\mathbb{R}}|f|^p\ d\mu\bigg)^{1/p}. $$

[Added: On the other hand, note that it is trivially true that $|f|^p\ge |f|^p1_A$. But Lebesuge integrals have the property that $\int |f|\leq \int |g|$ if $|f|\le |g|$.]

Consider for instance $f(x)=e^{-x^2}$, $A=[0,1]$ and $\mu$ the Lebesgue measure on $\mathbb{R}$. In this case you have the strict inequality.

An even simpler example: let $f(x)=1_{[0,2]}(x)$ and $A=[0,1]$.