# can we conclude that $f$ is a constant?

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?

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What you've actually shown is that $\mu \int_a^b f(x) dx = \mu M(b-a)$. You cannot conclude that $\mu \neq 0$, and so your argument does not follow. –  anonymous Dec 28 '12 at 16:48

No. For example, you could have $\varphi=0$, in which case you will have $$\int_a^b f(x)\varphi(x)\,dx=0$$ for any $f\in C[a,b]$ whatsoever.
In fact, any function $\varphi$ which is $0$ on some open interval in $[a,b]$ will admit non-constant functions $f\in C[a,b]$ such that $\int_a^b f(x)\varphi(x)\,dx=0$.
A slightly less trivial example: for any $t>0$, we could choose $\varphi(x)=x$ on $[-t,t]$, which satisfies $\int_{-t}^t\varphi(x)\,dx=0$, and any even function $f\in C[-t,t]$ will have \begin{align}\int_{-t}^t f(x)\varphi(x)\,dx&=\int_{-t}^0xf(x)\,dx+\int_0^t xf(x)\,dx\\\\ &=-\int_0^{-t}xf(x)\,dx+\int_0^txf(x)\,dx\\\\ &=-\int_0^{t}xf(x)\,dx+\int_0^txf(x)\,dx\\\\ &=0.\end{align}
As an addendum, you will sometimes see people using the fact that the equality holds for all test functions $\phi$ to conclude that $f$ is constant. –  user7530 Dec 28 '12 at 18:31