Let $f \in L^2(\Omega)$ for $\Omega$ an open bounded set of $\mathbb{R}^N$. We consider $$ J(u) := \int_{\Omega} \left( \left| \nabla_x u(x) \right|^2 - f(x) \cdot u(x) \right) {\rm d} x $$ Prove that there exists a unique minimizer.
My attempt: By using the definition, I can check this functional is convex. That means if i can prove that this functional exists a minimizer, then it will be unique. But i don't know how to prove that. Can you help me to solve this problem? Any help is appreciated.
We have $$J(u) = \frac{1}{2}a(u,u) - L(u)$$ Where $$a(u,v) = 2 \int_\Omega \nabla u(x) \nabla v(x) dx$$ and $$ L(u) = \int_\Omega f(x)dx $$ We can verify these functions fullfill all the Lax-Milgram hypotheses. Then by Lax-Milgram theorem there exists a unique $v$ such that $ \forall u, a(u,v)=L(u)$. We show that this $v$ is the also the minimizer of $v$ :
Let $w$ be an arbitrary function, then \begin{align} J(v+w) &= J(v) + ( a(v,w)-L(w) ) + \frac{1}{2} a(w,w) \\ &= J(v) + \frac{1}{2} a(w,w) \\ &\geq J(v) + \frac{\alpha}{2} \|w\|^2 \text{ since $a(.|.)$ is coercive}\\ \end{align}
Therefore $\forall w, J(v) \leq J(w)$ and $v$ is unique.