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.
How to show that if $\int_a^b|{f(x)}|^p\,d\alpha = 0$ then $\int_a^b|{f(x)}|\,d\alpha = 0$
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$\begingroup$
real-analysis
integration
measure-theory
lebesgue-integral
riemann-integration
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0@MyGlasses Yes. – 2017-01-29
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1Can you edit the question to include 1) the non-obvious fact that you want to avoid depending on Hölder's inequality, and 2) exactly which type of integral you're referring to? The question feels rather different for a Riemann integral versus a Lebesgue integral. Also, you specifically mention $\alpha$ is increasing so perhaps you are referring to a Stieljtes integral instead. So, which one is it? – 2017-01-29
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0User1006 and MyGlasses mentioned Hölder's inequality. You could also use the convexity of the map $\varphi(z)=z^p$. By [Jensen's inequality](https://en.wikipedia.org/wiki/Jensen's_inequality), $\frac{1}{b-a}\int_a^b |f(x)|^p dx\geq \left(\frac{1}{b-a}\int_a^b |f(x)|\, dx\right)^p$. – 2017-01-29
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1@ErickWong I am referring to a Stieljtes integral. Basically, Holder's inequality says $|\int_a^bf(x)g(x)\,d\alpha| \leq \left[\int_a^b|{f(x)}|^p\,d\alpha\right]^{1/p}\left[\int_a^b|{g(x)}|^q\,d\alpha\right]^{1/q}$ and I am trying to prove using Young' s inequality, but I need to prove the case when $\int_a^b|{f(x)}|^p\,d\alpha = 0$ separately. – 2017-01-29
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0@ihmth Thanks, these details need to be part of the question :). Not too late to edit. – 2017-01-29
3 Answers
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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.
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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. $$ \int_a^b|{f(x)}|\cdot 1\,d\alpha\leqslant \left(\int_a^b|{f(x)}|^p\,d\alpha\right)^{1/p}\left(\int_a^b1\,d\alpha\right)^{1-1/p}=0 $$
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0I am using this to prove Holder's inequality, so this won't work for me – 2017-01-29
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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 $$\int_a^b|f|d\alpha=0$$
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0Actually, I am using this to prove Holder's inequality – 2017-01-29
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1@ihmth Is $f$ continuous.? – 2017-01-29
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0How do you know with Holder's that $f$ is in $L^q$? If the integral is infinite the bound is invalid – 2017-01-29
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0@pasta My first question (I erased it, was on the top) was: Is it for every $p$ with positive answer. – 2017-01-29