Let $D$ be in $\mathbb R^n $. Let $f\in L^2(D)$. I proved that for all $\epsilon >0$ there exists a unique $\bar g\in H^1(D)$ such that $\int_D \nabla \bar g \nabla h+\int_D\epsilon \bar g h=\int_D hf$ for all $h\in H^1(D)$.
There exists a unique $g\in H^1(D)$ such that $\int_Dg=0$ and $\int_D \nabla g \nabla h=\int_D hf$ for all $h\in H^1(D)$ if and only if $\int_Df=0$.
I want to show that if $\int_Df=0$ then $||\bar g-g||$ converges to $0$.
If $\int_D f=0$ then by choosing $h=\bar g -g$ we can write $||\bar g -g||_{H^1(D)}=\int _D(\epsilon \bar gg−ϵ\bar g^2+\bar g^2−2\bar gg+g^2)$. Is there a way to show that the limit of this integral is $0$? Is that even true?
Hint: Test your very first variational equation with the constant function $h \equiv 1$.