Let $H$ be a Hilbert space, $A(\cdot,\cdot):H\times H\to\mathbb{R}$ be a symmetric coercive bi-continuous bi-linear form on $H$ and $F:H\to\mathbb{R}$ be a continuous linear functional. Then the function $J:H\to\mathbb{R}$ defined by $$J(v)=\frac{1}{2} A(v,v)-F(v)$$ is continuous and convex.
Question 1 Are the conditions sufficient for the result to hold? Doesn't one need some condition on $F$?
Question 2 It is hinted that if $A(v,v)\geq 0$ for all $v\in H$, then there is a simple proof for the above proposition. Does the result hold even without the positivity condition?
Also, any hints how to tackle problems regarding convexity in Hilbert spaces in general?
(Note: This operator arises from the Lax-Milgram theorem, which gives a solution to the weak formulation A(u,v)=F(v) of some partial differential equation. A solution $u$ to that equation minimizes $J(v)$).
We show that for any $u,v\in H$ and $t\in[0,1]$, $$ J(tu+(1-t)v)\leq tJ(u)+(1-t)J(v).$$ Indeed we have, \begin{align*} J(tu+(1-t)v)&=\frac{1}{2}A(tu+(1-t)v,tu+(1-t)v)-F(tu+(1-t)v)\\ &=\frac{t^2}{2}A(u,u)+\frac{(1-t)^2}{2}A(v,v)+t(1-t)A(u,v)-tF(u)-(1-t)F(v), \end{align*} and \begin{align*} tJ(u)+(1-t)J(v)&=\frac{t}{2}A(u,u)+\frac{1-t}{2}A(v,v)-tF(u)-(1-t)F(v). \end{align*} Therefore, \begin{align*} J(tu+(1-t)v)-\bigg(tJ(u)+(1-t)J(v)\bigg)&=\frac{t}{2}(t-1)(A(u,u)+A(v,v))\leq 0, \end{align*} since $A(v,v)\geq C\|v\|_H\geq 0$ for all $v\in H$ and $t\leq 1$.