I want prove that the following problems are equivalent under addtional assumption.
Problem 1 Let $a\colon X\times X\to \mathbb{R}$ and $j\colon X\to(-\infty,+\infty]$. Given $f\in X$ (Hilbert space) find an element $u\in X$ such that $$a(u,v-u)+j(v)-j(u)\ge(f,v-u)_X.\qquad (1)$$ Problem 1 has a unique solution under assumptions:
a) $a$ bilinear symetric form,
b) $\exists_{M>0}\,\forall_{u,v\in X}\,|a(u,v)|\le M\|u\|_{X}\|v\|_{X},$
c) $\exists_{m>0}\,\forall_{v\in X}\,a(v,v)\ge m\|v\|^{2} $
d) $j$ is a proper convex l.s.c. function.
Problem 2 Find $u$ such that
$$Au+\nabla j(u)=f,\qquad (2)$$ where $A\colon \,X\to X$ is the linear continuous operator given by $$(Au,v)_X=a(u,v)$$ and $\nabla j \colon X\to X$ is the gradient operator of $j$ i.e., $$(\nabla j(u),v)_{X}=\lim_{h\to 0}\frac{j(u+hv)-j(u)}{h}.$$ Now, suppose that $j$ is in addition Gateaux differentiable.
Proof: Suppose problem 2 has a solution. Multiplying (2) by $v-u$ and taking the inner product we have $$(Au,v-u)_X+(\nabla j(u), v-u)_X=(f,v-u)_X.$$ Since $j$ is convex then $$j(u+h(v-u))\le j(hv+(1-h)u)\le hj(v)+(1-h)j(u)=j(u)+h(j(v)-j(u)).$$ The last one is valid for $h\in (0,1)$. Hence $$\frac{j(u+h(v-u))-j(u)}{h}\le j(v)-j(u).\qquad (3)$$ Now, I can pass to the limit and finish the first part of the proof. What worries me is that (3) is valid for $h\in (0,1)$. So, can really pass to the limit? From my point of view, I can pass to the limit but only one-sided so that (3) holds.
What would be the second part of the proof?