Let $I(u) = \int_{\Omega}\frac{1}{2}|\nabla u|^2+\frac{1}{2}|u|^2-fu dx$, where $f\in L^2(\Omega), \Omega$ is a bounded open set in $R^n$. Prove that $I(u)$ achieves its minimum over $v\in H^1(\Omega)$.
I can prove the $v\in H^1_0(\Omega)$ case: $I(u)$ is convex and we can prove that $I(u)$ has a local minimum by considering $F(t) = I(u+tv)$, and thus has a global minimum.
How about the $H^1(\Omega)$ case?
This is the consequence of the direct method in the Calculus of Variation. For instance, you have $$ \int_\Omega fu \,dx\leq \frac12\int_\Omega f^2dx+\frac12\int_\Omega u^2dx \tag 1 $$ (think the algebra fact that $a^2+b^2\geq 2ab$)
Equation $(1)$ implies that $$ I(u)\geq \frac12\int_\Omega |\nabla u|^2dx-\frac12\int_\Omega f^2dx\geq -\frac12\int_\Omega f^2dx>-\infty. \tag 2 $$
Hence, you may have a minimizing sequence $u_n$ in $H^1$ such that $$ I(u_n)\to \inf_{v\in H^1}I(v)=:m $$ where the right hand side is not $-\infty$ because of $(2)$. Therefore, you have for $n$ large enough, $$ I(u_n)\leq m+1. $$ Hence, you have $u_n$ is bounded in $H^1$ norm. The rest is as @Jeff's comment. you have weakly convergence in $H^1$ and by using strictly convexity, you done.