Suppose $f(x) \in C[a,b]$ and $\varphi(x)$ is Riemann integrable and satisfy : $\int_a^b\varphi(x)\mathscr{dx}=0$. $\int_a^bf(x)\varphi(x)\mathscr{dx}=0$ can we conclude that $f(x)\equiv c$?
My approach:
for $f$ is continuous on $[a,b]$, then there exists $M$ and $m$ in $\Bbb{R}$,satisfy: $m \leq f(x) \leq M$,so we have $M-f(x)\geq0$,so I can use the mean value theorem: $\int_a^b(M-f(x))\varphi(x)\mathscr{dx}=\mu\int_a^b(M-f(x))\mathscr{dx}=0 \Rightarrow \int_a^b f(x)\mathscr{dx} =M(b-a)$ where $\inf\varphi(x)\leq\mu \leq \sup \varphi(x)$
use MVT to : $\int_a^b(f(x)-m)\varphi(x)\mathscr{dx}$we get $\int_a^bf(x)\mathscr{dx}=m(b-a)$ this implies $m=M$,then $f$ is constant on $[a,b]$
Am I right?