I was studying the proof of Stampacchia's theorem from Haim Brezis's book, attached the theorem with the proof:
Theorem: Assume that $a(u,v)$ is a continuous coercive bilinear form on $H$. Let $K\subseteq H$ be a nonempty closed and convex subset. Then, given any $\phi\in H^{\star}$, there exists a unique element $u\in K$ such that \begin{equation} a(u,v-u) \geq \langle\phi,v-u\rangle\quad \forall v\in K \end{equation} Moreover, if $a$ is symmetric, then $u$ is characterized by the property: \begin{equation*} u\in K\text{ and }\frac{1}{2}a(u,v)-\langle\phi , u\rangle = \min_{v\in K}\{a(v,v)/2 - \langle\phi , u\rangle\} \end{equation*}
I can't understand how the same $ u \in K $ fulfills both properties, is it the same? and where in the demo do I see that the u element of the first property is used for the second. As I see in the demonstration, nothing ensures that the element $ u $ that fulfills the first property, when adding the new hypotheses, is the same that fulfills the second property
The proof is: (I add a few steps to make it easier to read):
By the Riesz-Fréchet representation theorem, for $\phi\in H^{\star}$ exists $f\in H$ such that \begin{equation*} \langle\phi,v\rangle = (f,v) \quad \forall v\in H \end{equation*} On the other hand, taking $u\in H$, we have the function $v\mapsto a(u,v)$ a continuous linear function on $H$. Using again the Riesz-Fréchet representation theorem, there is a single representative in $H$, denoted by $Au$, such that: \begin{equation} \label{BilinealStampacchia} a(u,v) = (Au,v) \quad \forall v\in H \end{equation}
Let's see next that $u\mapsto Au$ is a continuous linear function on $H$: Note that $Au\in H$, then considering $v = Au$ on $a(u,v) = (Au,v) \quad \forall v\in H$, we have \begin{eqnarray} a(u,Au) = (Au,Au) = |Au|^{2} \end{eqnarray} and since $a$ is continuous we limit obtaining \begin{equation*} \begin{array}{rrcl} & |Au| = \sqrt{a(u,Au)} & \leq & \sqrt{C|u||Au|} \quad\forall u\in H\\ \Rightarrow & \dfrac{|Au|}{\sqrt{|Au|}} & \leq & \sqrt{C|u|} \quad\forall u\in H\\ \Rightarrow & \sqrt{|Au|} & \leq & \sqrt{C|u|} \quad\forall u\in H\\ \Rightarrow & |Au| & \leq & C|u| \quad\forall u\in H\\ \end{array} \end{equation*}
Let's see what $(Au,u) \geq C'|u|^{2}$: We know that $a(u,v) = (Au, v)$ for all $v\in H$, in particular if we consider $u \in H$ we have to \begin{equation*} a(u,u) = (Au,u) \geq C'|u|^{2}\quad\forall u\in H \end{equation*}
Let's see what exists $u\in K$ such that $(Au,v-u) \geq (f,v-u)$ for all $v\in K$: Note that the desired is equivalent to: \begin{equation*} \begin{array}{rrcl} \Leftrightarrow & (Au,v-u) & \geq & (f,v-u)\\ \Leftrightarrow & 0 & \geq & (f,v-u)-(Au,v-u)\\ \Leftrightarrow & 0 & \geq & (f-Au,v-u)\\ \Leftrightarrow & 0 & \geq & \rho(f-Au,v-u)\\ \Leftrightarrow & 0 & \geq & (\rho f-\rho Au,v-u)\\ \Leftrightarrow & 0 & \geq & (\rho f-\rho Au+u-u,v-u) \quad\forall v\in K\\ \Leftrightarrow & u & = & P_{K}(\rho f-\rho Au+u) \end{array} \end{equation*} Let $S(v):= P_{K}(\rho f - \rho Av+v)$, choosing some $\rho>0$, $S$ is a strict contraction and we can use the Fixed Point Theorem (Banach). Then, how $P_{K}$ does not increase distances (that is, it is Lipschitz with constant $ 1 $), then: \begin{eqnarray*} |S(v_{1}) - S(v_{2})| \leq |(v_{1}-v_{2}) - \rho(Av_{1}-Av_{2})| \end{eqnarray*} So we can finally get a bound using the properties (1) and (2) tested for $A$ \begin{eqnarray*} |S(v_{1}) - S(v_{2})|^{2} & = & S(v_{1})^{2} - 2S(v_{1})S(v_{2}) + S(v_{2})\\ & = & |v_{1}-v_{2}|^{2} +\rho^{2}|Av_{1}-Av_{2}|^{2} - 2\rho(Av_{1}-Av_{2},v_{1}-v_{2})\\ &\leq& |v_{1}-v_{2}|^{2}\cdot(1-2\rho C'+\rho^{2} C) \end{eqnarray*} and finally, choosing $\rho$ such that $0<\rho <\frac{2C'}{C^{2}}$, we have to $S$ has a single fixed point, with which we can conclude that \begin{equation*} S(u):= P_{K}(\rho f - \rho Au+u) = u \end{equation*} equivalent to: \begin{equation*} a(u,v-u) \geq \langle\phi , v-u\rangle\ \forall v\in K \end{equation*}
To prove the second property of the theorem, we now assume that $a(u,v) $ is symmetric, then $a(u,v) $ defines a new dot product over $H$ and thus defines a Hilbertian norm $\sqrt{a(u,u)} $ which turns out to be equivalent to the original norm of $H$. Since the norms are equivalent then $H$ remains a Hilbert space with the norm $\sqrt{a (u,u)}$. Using the Riesz-Fréchet representation theorem, for $\phi\in H^{\star} $ there is a single representative $g\in H$ such that
\begin{equation*} \langle\phi,v\rangle = a(g,v) \quad v\in H \end{equation*} Now let's find some $u\in K$ such that \begin{equation*} a(g-u,v-u)\leq0\quad\forall v\in K \end{equation*} The $u$ we are looking for is simply the projection of $g$ onto $K$ for the inner product $a(\cdot,\cdot)$. By the projection theorem on closed convexes there is a unique $u\in K$ such that it reaches \begin{equation*} \min_{v\in K}\sqrt{a(g-v,v-v)} \end{equation*} so it is enough to minimize the function over $K$ \begin{eqnarray*} v \mapsto a(g-v,g-v) & = & a(v,v)-2a(g,v)+a(g,g)\\ & = & a(v,v)-2\langle\phi,v\rangle+a(g,g) \end{eqnarray*} or equivalently the function \begin{equation*} v\mapsto\frac{1}{2}a(v,v)-\langle\phi,v\rangle \end{equation*}
From the first part of the theorem we have $$a(u,v-u)\geq \langle \phi,v-u\rangle, $$ and then $$a(u,v)-\langle \phi,v\rangle\geq a(u,u)-\langle \phi,u\rangle.\;\;\;\;\; (*) $$
Also, by bilinearity, coercivity and now symmetry, from $$a(v-u,v-u)\geq \alpha\|v-u\|^2\geq 0 $$ we have $$\frac{1}{2}a(v,v)\geq a(v,u)-\frac{1}{2}a(u,u).$$
Then, $$\frac{1}{2}a(v,v)-\langle \phi,v\rangle\geq a(v,u)-\frac{1}{2}a(u,u)-\langle \phi,v\rangle,$$ using the inequality (*) $$\frac{1}{2}a(v,v)-\langle \phi,v\rangle\geq a(u,u)-\langle \phi,u\rangle-\frac{1}{2}a(u,u),$$ and that is it !