Equality of two properties for $J(x):=B(x,x)-2l(x)$ 
Let H be a real Hilbert space, $l\in H'$, which is the dual space of H, and $B:H\times H\rightarrow \mathbb{R}$ a symmetric, bounded and coercive Bilinear form. We define $J:H\rightarrow \mathbb{R}$ as $J(x):=B(x,x)-2l(x)$.
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For $u\in H$: $$i)J(u)\le J(x) \forall x\in H \leftrightarrow ii)B(u,x)=l(x) \forall x\in H$$

So $u$ is the minimum of $J$. I don't know how to start either way. I guess ii)$\rightarrow$i) uses the derivative. Can someone help me?
 A: Let us calculate the first variation of $J$ at a point $u\in H$ applied to a vector $v\in H$:
$$\begin{align}[\delta J(u)](v)&:=\frac{d}{dt}\bigg( J(u+tv) \bigg)\bigg|_{t=0}=\frac{d}{dt}\bigg( B(u,u)+t^2B(v,v)+2t(Bu,v)-2l(u+tv)\bigg)\bigg|_{t=0} =\\&=2(B(u,v)-l(v)). \end{align}$$
Since $l$ is linear, by the hypotheses on $B$ we have that $J$ is strictly convex, in fact for $t\in[0,1]$ we have:
$$\begin{align} B(tx+&(1-t)y,tx+(1-t)y)= t^2B(x,x)+(1-t)^2B(y,y)+2t(1-t)B(x,y) \le \\ &  \le t^2B(x,x)+(1-t)^2B(y,y)+2t(1-t)\sqrt{B(x,x)}\sqrt{B(y,y)} =\\&=\bigg(t\sqrt{B(x,x)}+(1-t)\sqrt{B(y,y)}\bigg)^2\le \\& \le tB(x,x)+(1-t)B(y,y), \end{align}$$
where the first inequality is the Cauchy-Schwarz on the symetric bilinear positive definite form $B(\cdot,\cdot)$ (it is positive by the coercivity hypothesis), while the second inequality is the strict convexity of $(\cdot)^2$. Together with the linearity of $l$ we have the strict convexity of $J$.
So by convexity we have that for all $u,v\in H$ it holds:
$$J(u+v)\ge J(u)+[\delta J(u)](v)=J(u)+2(B(u,v)-l(v)).$$
Now if $u\in H$ is such that i) holds, then by minimality the first variation at $u$ is such that $[\delta J(u)](x)=0$ for all $x\in H$, that is $B(u,x)=l(x)$ for all $x\in H$ by the first calculation.
While if $u\in H$ is such that ii) holds, then using the convexity of $J$ for all $x\in H$ we have:
$$J(x)=J(u+x-u)\ge J(u)+[\delta J(u)](x-u)=J(u)+2(B(u,x-u)-l(x-u))=J(u),$$
where in the last equality we used the hypothesis ii).
