Prove Corollary of the Fundamental lemma of calculus of variations From the Fundamental lemma of calculus of variations we know the following:
Let $u \in L_{loc}(a,b)$ and for every $\phi \in C_0^\infty$,
$$\int_a^b u(x) \phi(x) dx = 0$$ 
holds, then $u(x) = 0$ almost everywhere. 
From that I need to prove the following Corollary:
Let $u \in L_{loc}(a,b)$ and for every $\phi \in C_0^\infty$,
$$\int_a^b u(x) \phi'(x) dx = 0$$ 
holds, then $u(x) = C$ almost everywhere, for a constant $C$.
My start was to performe integration by parts,
$$0 = \int_a^b u(x) \phi'(x) dx = \int_a^b u'(x) \phi(x) dx \ \forall \phi \in C_0^\infty \ .$$ 
Now the Fundamental lemma gives that $u'(x) = 0$ almost everywhere and therefore $u(x)$ is constant almost everywhere.
But this idea has two shortcommings:
First, $u(x)$ doesn't even need to have a weak derivative, nor a normal derivative, therefore it is unclear why an integration by parts should be possible at all.
Second, it only shows that $u(x)$ is constant almost everywhere, not that it is equal to (the same) constant almost everywehre.
So I think I am on the wrong track, anyone got a hint?
 A: My suggestion in the comments was not a productive one. Here is a workable approach:
I'm not exactly sure what you mean by $C_u^\infty$. I imagine it is the same as $K= \{ \phi: (a,b) \to \mathbb{R} | \phi \text{ is smooth}, \overline{\operatorname{supp}} \phi \subset (a,b) \}$. If not, the proof below may need to be tweaked.
Suppose $u \in L_{\text{loc}}(a,b)$ such that $\int u \phi' = 0$ for all $\phi \in K$.
Lemma (similar to Lemma 1 in Section 21.4 of Kolmogorov & Fomin's "Introductory Real Analysis"): Let $\phi_1 \in K$ such that $\int \phi_1 = 1$ and $\phi \in K$.
Then we can write $\phi = \phi_0' + \alpha \phi_1$, where $\alpha$ is a constant, and $\phi_0 \in K$.
Proof: 
Let $\overline{\operatorname{supp}} \phi_0 \cup \overline{\operatorname{supp}} \phi \subset [\sigma_0, \sigma_1] \subset (a,b)$.
Let $\alpha = \int \phi$ and $\phi_0(t) = \int_a^t (\phi(\tau)-\alpha \phi_1(\tau)) dt$.
Then $\phi_0$ is smooth and $\phi_0(t) = 0$ for $t \in (a,\sigma_0) \cup (\sigma_1,b)$, so $\phi_0 \in K$, and furthermore $\phi_0'(t) = \phi(t) - \alpha \phi_1(t)$, hence $\phi = \phi_0'+ \alpha \phi_1$. 
Now choose any $\phi_1 \in K$ such that $\int \phi_1 = 1$, let $c = \int u \phi_1$, and let $\tilde{u} = u-c$. Choose $\phi \in K$, and using the above lemma, write $\phi = \phi_0' + \alpha \phi_1$. Then we have
$$\int \tilde{u} \phi = \int (u -c)(\phi_0' + \alpha \phi_1) = \int u \phi_0' - c \phi_0' + \alpha(\int u \phi_1 - c\int \phi_1) = 0$$
hence the fundamental lemma shows $\tilde{u}(t) = 0$ a.e. $t \in (a,b)$, and hence $u(t)=c$ a.e. $t \in (a,b)$.
It is not immediate to me how to extend the lemma to domains in $\mathbb{R}^n$.
