I have been struggling with this for a while. Here, as you can see, they define the weak formulation of the Poisson equation as:
$-\int_{\Omega}\nabla u\cdot\nabla v\,ds = \int_{\Omega}fv\,ds \equiv -\phi(u,v)$ (not sure what the $\equiv$ symbol means here(?)), where they specify that the ''." denotes the dot product, $v$ is a test function, $u$ the unknown and $\phi$ an operator (or map). Few lines later; they explain: "Such functions are (weakly) once differentiable and it turns out that the symmetric bilinear map $\phi$ then defines an inner product which turns $H_0^1(0,1)$ into a Hilbert space"
Does this mean that the weak formulation in this case is by default an inner product that uses the dot product?
EDIT: I checked this pdf file that studies the weak formulation of Poisson's equation, apparently the weak formulation defines an inner product.
I just need a confirmation if what I am doing is correct...
The reason it's an inner product is that the inner product axioms are satisfied for $\phi(u,v)=\int_\Omega\nabla u\cdot\nabla v.$
The easy ones to show are symmetry and linearity, i.e., $\phi(u,v)=\phi(v,u)$ and $\phi(au,v)=a\phi(u,v)$ ($a$ constant). One has to also show that $\phi(u,u)\ge0$ for all $u$, and $\phi(u,u)=0$ if and only if $u=0$.
One has that $\phi(u,u)=\int_\Omega|\nabla u|^2\ge0$. Then, if $u=0$, then $\phi(u,u)=\int_\Omega0=0$, and if $\phi(u,u)=0$, then $\int_\Omega|\nabla u|^2=0$, and so $u$ must be constant, but also $u\in H^1_0(\Omega)$, so the constant must in fact be zero, so $u=0$.