Integration by parts to prove a function is constant a.e. Let $(a,b)$ be an interval on $\mathbb{R}$. Let $f \in L^1(a,b)$. Assume that $$
\int_a^b f(x)g'(x)\, dx =0
$$
for all $C^1$ functions $g$ with support compactly contained in $(a,b)$. Prove that there is a constant $c$ such that $f(x)=c$ for almost every $x \in (a,b)$. 
My thought was to use integration by parts so as to have $$
\int_a^b g(x)df(x)=0
$$
but since $f(x)$ is only integrable, it does not seem to work. 
Any help/hint is appreciated! 
 A: This is a standard result in calculus of variations (it is a version of the so-called "fundamental lemma").
First of all, by density argument, if the relation holds for every $g\in C^1$ with compact support in $(a,b)$, then it holds also for every $g\in AC$ with $g(a) = g(b) = 0$.
Let $c := \frac{1}{b-a} \int_a^b f$ be the integral mean of $f$, and consider the absolutely continuous function
$$
g(x) := \int_a^x [f(t) - c]\, dt,
\qquad x\in [a,b].
$$
Clearly $g(a) = 0$ and $g(b) = \int_a^b f - (b-a) c = 0$, so that $g$ can be used as a test function.
We have that
$$
\int_a^b (f-c) g' = \int_a^b f g' - c \int_a^b g' = 0.
$$
On the other hand
$$
\int_a^b (f-c) g' = \int_a^b (f-c)^2,
$$
so we can conclude that $f = c$ a.e. in $[a,b]$.
A: Hint: If you happen to know that 
$$ \int_a^b f(x) h(x) dx = 0 , \ \ \forall h\in C_c((a,b)): \int_a^b h\; dx=1$$
implies that $f$ vanish
then you may reduce to this situation by considering the difference 
of two such $h$'s. One of them will give rise to the constant $c$.
