I am reading an expository article about Variational Methods. There is a point where he talks about weak solutions, and gives me a functional $\Phi$ associated to a differential equation. There is a passage which is not clear where he puts:
"...we see that, if $u_0$ is a minimum of $\Phi$ in $C_0^1[a,b]$, then $$\Phi(u_0) \leq \Phi(u_0 + tv), \forall t \in \mathbb{R}, \forall v \in C_0^1[a,b]$$, and from there we have $\Phi'(u_0)v = 0$, that is, $u_0$ is a weak solution of the problem..."
I'm having difficulty with this passage because I can clearly see that $\Phi'(u_0)v \geq 0$, but I don't see how we conclude that $\Phi'(u_0)v = 0$. Is there something about passing the limit on both sides of the inequality that I'm not seeing? Thanks in advance.
Actually, this gives $$t\Phi'(u_0)v\ge 0,\,\forall t\in\Bbb R$$ So take $t=\pm 1$, and you get $$\Phi'(u_0)v\ge 0\\\Phi'(u_0)v\le 0\\\therefore\Phi'(u_0)v= 0$$