The following is a part of the "line of thought" on how we obtain the "weak" formulation of the classical form of the one-dimensional problem for the Finite Element Method, but there is one small part I don't quite get.
"Supposing we have a given $f\in L_2(0,1)$, and we look for a "minimizing function", $u \in H^1_0(0,1)$ such that, $$F(u)\le F(v),\,\forall v\in H^1_0(0,1),$$ where $F:H^1_0(0,1)\to\mathbb R$ is the energy functional: $$F(v)=\frac12(v',v')-(f,v),$$ where $(\cdot,\cdot)$ denotes the inner product function. Now, assuming that some $u$ satisfies the first inequality, then given a function $v\in H^1_0(0,1)$ and $\epsilon\in\mathbb R$, then $u+\epsilon v\in H^1_0(0,1)$, so, $$F(u)\le F(u+\epsilon v)$$ If we define $g(\epsilon):=F(u+\epsilon v)$ this function is minimized when $\epsilon=0$, so that...
And then follows the part I don't quite get,
... $\frac{dg}{d\epsilon}|_{\epsilon=0}=0$"
How is it we can say that the entire expression, once evaluated at $\epsilon=0$, equals zero?
What I get when working it out by hand is,
$$\frac{dg}{d\epsilon}=\epsilon(v',v')+(u',u')-(f,v)$$ $$\implies\frac{dg}{d\epsilon}|_{\epsilon=0}=(u',u')-(f,v)$$
How do we know that this is equal to zero?
Cheers!
You are trying to find extreme points of the energy functional $F$ (i.e. trying to find the function that minimizes total energy). Opening up your condition gives $$ 0=\frac{\mathrm{d}g}{\mathrm{d}\epsilon}\Big|_{\epsilon=0} = \frac{\mathrm{d}}{\mathrm{d}\epsilon}F(u+\epsilon v)\Big|_{\epsilon=0} =: DF(u)v, $$ where on the right hand side you have the definition of Gâteaux differential of $F$ at $u$ into direction $v$ (generalization of the directional derivative to Banach spaces)—here denoted by $DF(u)v$ emphasizing that $DF(u)$ defines an operator acting on $v$. Now analogously to real valued functions, you want your extreme point $u$ to satisfy $DF(u)v=0$ for every direction $v$.
Since your energy is of such form, you can actually show that the resulting $u$ corresponds to the global minimum.