Let $f:[a,b]\to\mathbb{R}$ be a nonnegative Riemann integrable function. If $\int_a^bf^p=0$ for some $p>0$, is it true that $\int_a^bf=0$ ? It is obviously true if $p=1$, and the Hölder's inequality
$$\left|\int_a^bfg\right|\leq\left(\int_a^b|f|^p\right)^{1/p}\left(\int_a^b|g|^p\right)^{1/p}$$
implies that it is true if $p>1$. But what if $0<p<1$ ?
I would prefer a solution which does not use Lebesgue integral or measure zero sets if possible, please.
Does $\int_a^bf^p=0$ imply $\int_a^bf=0$
real-analysisriemann-integration
Related Question
- Is there a simple proof for $\int_a^bf^p=0\implies\int_a^bf=0$ when $p>1$
- Elementary proof of $\int f=0\implies f=0$ a.e. for Riemann integrals
- Prove that $ \int_a^x f\,dx=0$ for all $x\in [a,b]$ implies $ \int_a^b fg\,dx=0$ for any integrable $g$.
- If $\int_a ^b g^2 = 0$, then $\int_a ^b fg = 0$ (edit: without Cauchy-Schwarz)
Best Answer
Suppose $ 0<p<1$. Let $M=\sup \{f(x): a \leq x \leq b\}$. Let $g=\frac f M$. Then $\int g^{p}=0$ and $0\leq g \leq 1$. Hence $0 \leq g \leq g^{p}$ which gives $0\leq \int g \leq \int g^{p}=0$ which gives $\int g=0$. This implies $\int f=0$.