Lemma: Let $f,g\in\mathscr{R}[a,b]$ and $f,g\geq0$ ($\mathscr{R}[a,b]$ is the set of all Riemann integrable functions in $[a,b]$) such that $F=\left\{\int_{a}^{b}f(x)^pdx\right\}^{\frac{1}{p}}=0$, where $p\in[1,\infty)$. Prove that $\int_{a}^{b}f(x)g(x)dx=0$.
Proof:
Let $F=\left\{\int_{a}^{b}f(x)^pdx\right\}^{\frac{1}{p}}=0$. Since $f,g$ are Riemann intrgrable in $[a,b]$, $f,g$ are bounded on this closed interval. Let $A:=\sup\{g(x):x\in[a,b]\}$ and $B:=\sup\{f(x)^{1-p}:x\in[a,b],f(x)>0\}$. Since $f,g$ are non-negative, $f(x)g(x)=f(x)^pf(x)^{1-p}g(x)\leq f(x)^pAB$ for all $x\in[a,b]$ such that $f(x)>0$. For $f(x)=0$, this inequality is trivially true (in fact, equality holds in that case). So $$f(x)g(x)\leq f(x)^pAB$$ for all $x\in[a,b]$. Integrating both sides of the inequality we get, $$0\leq\int_{a}^{b}f(x)g(x)dx\leq AB\int_{a}^{b}f(x)^pdx=0$$ Therefore $$\int_{a}^{b}f(x)g(x)dx=0$$
Is my proof correct? Any suggestion will be appreciated. (This is basically a step in proving the Holder's inequality for integrals)
The proof looks fine.
Another proof:
Since $\int f^p = 0$ you must have $f(x) = 0 $ ae. $x$.
And another:
Suppose $\int f \cdot g > 0$. Then there is some partition $P$ such that $L(f \cdot g,P) >0$ and hence there is some non trivial interval $I \subset [a,b]$ and $\delta>0$ such that $(f \cdot g)(x) \ge \delta >0$ for $x \in I$.
If we let $B= \sup g$ then this gives $B f(x) \ge \delta$ for $x \in I$ and so $B^p f^p(x) \ge \delta^p$. Then integrating gives $0 = B^p\int f^p \ge \delta^p m(I) >0$ which is a contradiction.