Let $f\in C([a,b])$, $\int_{a}^{b}f(x)\phi'(x) \ dx=0$ for all continuous and differentiable functions $\phi$. Show $f$ is constant

Question

Let $f\in C([a,b])$ and $\int_{a}^{b}f(x)\phi'(x) \ dx=0$ for all continuous and differentiable functions $\phi$ on $[a,b]$ with $\phi(a)=\phi(b)=0$. Show that $f$ is constant on $[a,b]$.

I would like to know if my proof holds, please. My attempt is to pass by primitive of $f$.

First of all, as $f\in C([a,b])$, $f$ has a primitive $F\in C^1([a,b])$ such that $F'(x)=f(x) \ \forall x \in [a,b]$. Then, as $\int_{a}^{b}f(x)\phi'(x) \ dx=0$ holds for all continuous and differentiable functions $\phi$ on $[a,b]$, we can choose $\phi=F$. So we have to show that $\int_{a}^{b}f(x)F'(x) \ dx=\int_{a}^{b}f^2(x)=0 \implies f$ constant.

Let $G$ be a primitive of $f^2$. We have that $G'(x)=f^2(x)\ge 0 \ \forall x \in [a,b]$. So, $G$ is increasing on $[a,b]$. Moreover, $G(b)-G(a)=0 \iff G(a)=G(b)$. So, we conclude that $G$ is constant as $a\le x\le b\iff G(a)\le G(x)\le G(b)=G(a)$. Thus,

$G'=f^2=0 \Rightarrow f=0$. So we showed that $f$ is constant.

I'm curious as well if there is a maneer to prove the statement using upper and lower Darboux sum considering by absurd that $f\neq 0$ so there exists $c\in[a,b]: f(c)\neq0$. I tried to prove the statement using it, but I feel lost in the inequalities. I tried to pass by integrability of th functions implies integrability of absolute value of functions and by Upper Darboux sum.

Answer 1

Of course you can drop the requirement $\phi(a) = \phi(b) = 0$ to make your proof work. However, it turns out that you only need to tweak your approach a little bit to prove the given statement:

Define the mean $\overline{f} = \frac{1}{b-a} \int_a^b f(x) \, \mathrm{d} x$ of $f$. Instead of $\phi = F$ choose $$\phi(x) = F(x) - F(a) - (x-a) \overline{f}$$ for $x \in [a,b]$. You can check that $\phi$ satisfies all hypotheses and $\phi' = f - \overline{f}$. Therefore, $$ 0 = \int \limits_a^b f(x) \phi'(x) \, \mathrm{d} x - \bar{f} \int \limits_a^b \phi'(x) \, \mathrm{d} x = \int \limits_a^b \left[f(x) - \overline{f}\right] \phi'(x) \, \mathrm{d} x = \int \limits_a^b \left[f(x) - \overline{f}\right]^2 \, \mathrm{d} x \, .$$ You have already shown that this implies $f - \overline{f} = 0$, so $f$ must be constant.