How to show that if $\int_a^b|{f(x)}|^p\,d\alpha = 0$ then $\int_a^b|{f(x)}|\,d\alpha = 0$

Question

How can I prove that if $\int_a^b|{f(x)}|^p\,d\alpha = 0$ then $\int_a^b|{f(x)}|\,d\alpha = 0$? (Assuming that $\alpha$ is increasing and $p > 1$). Intuitively this is clear but I'm having trouble proving it rigorously.

Answer 1

Let $q$ such that $\dfrac{1}{p}+\dfrac{1}{q}=1$ so $q>1$ and $$\int_a^b|f|d\alpha\leq\Big(\int_a^b|f|^pd\alpha\Big)^\frac1p\Big(\int_a^b|f|^qd\alpha\Big)^\frac1q=0$$

Without Holder, we say for all step function $\phi$ $$\int_a^b|f|^pd\alpha\leq\sup_{|f|^p\leq\phi}\int_a^b\phi d\alpha=0$$ this shows $|f|=0$ almost-everywhere for $\alpha$ (measure) on $[a,b]$. That is $$\int_a^b|f|d\alpha=0$$

Answer 2

Edit: Here is another proof without using Holder's inequality.

First we prove that if $$ \int_A|{g(x)}|\:d\mu=0 $$ then $g(x)=0$ almost everywhere on $A$.

Let $B_n=\{x:x\in A, \:|g(x)|\geqslant 1/n\}$. Then by Chebychev's inequality $$ \mu(B_n)\leqslant n\int_A|{g(x)}|\:d\mu=0 $$ Hence $$ \mu(\{x:x\in A, \:|g(x)|\ne0\})=\mu\left(\bigcup_{n=1}^{\infty}B_n\right)\leqslant \sum_{n=1}^{\infty}\mu(B_n)=0 $$ i.e. $g(x)=0$ almost everywhere on $A$.

Now since $$ \int_a^b|{f(x)}|^p\,d\alpha=0 $$ $f^p(x)=0$ a.e. on $[a,b]$ and so $f(x)=0$ a.e. on $[a,b]$. Thus $$ \int_a^b|{f(x)}|\,d\alpha=0\tag*{$\blacksquare$} $$

This is a hint that use Holder's inequality. $$ \int_a^b|{f(x)}|\cdot 1\,d\alpha\leqslant \left(\int_a^b|{f(x)}|^p\,d\alpha\right)^{1/p}\left(\int_a^b1\,d\alpha\right)^{1-1/p}=0 $$

Answer 3

If $|f(x)| > \epsilon$ then $|f(x)|^p > \epsilon^p$.

If $A_{\epsilon} := \{|f| > \epsilon\}$,

$0 = \int_{[a,b]} |f|^p d\alpha \ge \int_{A_{\epsilon}} |f|^p d\alpha \ge \epsilon^p \alpha(A_{\epsilon})$, from which it follows that $\alpha(A_{\epsilon})=0$.

Since $[a,b] = \{|f|=0\} \cup \bigcup_n A_{\frac{1}{n}}$, we have that $|f| = 0$ a.e., and the result follows.