English Translation of a French Theorem by Brezis

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I am interested in an English translation of the French "THEOREME 24" in "Équations et inéquations non linéaires dans les espaces vectoriels en dualité" by Haïm Brézis.

It states

THEOREME 24. — Soient $X$ un convexe compact de $E$, $A$ une application pseudo-monotone de $X$ dans $F$ et $\phi$ une fonction convexe S.C.I. de $X$ dans $]-\infty, +\infty]$ non identiquement égale à $+\infty$. Alors pour tout $f\in F$, il existe $u\in X$ tel que $$(f, v - u) - (Au, v - u) \leq \phi(v) - \phi(u) \quad\forall v \in X.\quad(4.1)$$ L'ensemble des solutions de (4.1) est compact ; il est convexe compact lorsque $A$ est monotone hémicontinu.

Si de plus, $(Ax - A y , x - y ) > 0, \forall x, y \in X , x \neq y$, alors la solution de (4.1) est unique.

Prior to THEOREME 24, he states

Soient $E$ et $F$ deux espaces vectoriels sur $\mathbf R$ mis en dualité par une forme bilinéaire $(f, u), f\in F , u \in E$. On supposera dans la suite que $E$ est muni d'une topologie d'espace vectoriel topologique, plus fine que $\sigma(E , F)$ et que $F$ est muni de la topologie $\sigma(F , E)$.

My best attempt at translating this:

Let $E$ and $F$ be two vector spaces on $\mathbb R$ that are brought into duality by a bilinear form $\langle f, u \rangle$ for $f\in F, u\in E$. We suppose, that $E$ is equipped with a finer topology than $\sigma(E,F)$ and $F$ is equipped with the topology $\sigma(F,E)$.

Theorem 24. — Let $X$ be a compact, convex subset of $E$, let $A: X\to F$ be a pseudo-monotone operator and let $\phi: X \to (-\infty, \infty]$ be convex and lower semi-continuous with $\phi(x)=\infty$ does not hold for all $x\in X$. Then for all $f\in F$ there exists some $u\in X$ such that $$\langle f, v - u\rangle - \langle A(u), v - u\rangle \leq \phi(v) - \phi(u) \quad\forall v \in X.\quad(4.1)$$ The set $S$ of solutions to (4.1) is compact; if $A$ is monotone and hemicontinuous, $S$ is even convex and compact.

If $\langle Ax - A y , x - y \rangle > 0$ holds for all $x, y \in X , x \neq y$, then the solution is unique.

Edit1: I adjusted the translation as proposed in a comment.

Edit2: I just realized, that the definition of pseudo-monotone is also not straight-forward to me and a translation would be great.

Definition F. — Soit $X$ un sous-ensemble de $E$. On dit qu'une application $A$ de $X$ dans $F$ est pseudo-monotone si elle vérifie les deux propriétés suivantes:

($\text{PM}_1$) Pour tout filtre $u_i$ porté par un ensemble compact de X tel que $u_i \to u$ dans X et $\lim \sup (Au_i , u_i - u) \leq 0$ on a $$(Au , u - v) \leq \lim \inf(Au_i, u_i - v) \quad\forall v \in X .$$

($\text{PM}_2$) Pour tout $v\in X$, l'application $u \to (A u, u - v)$ est bornée intérieurement sur les sous-ensembles compacts de $X$.

Remarque. — Si la topologie induite par E sur les sous-ensembles compacts de X est métrisable, alors la propriété ($\text{PM}_1$) est equivalente à la même propriété formulée pour des suites.

My best translation to this would be:

Definition F. — Let $X\subseteq E$. An operator $A: X \to F$ is called pseudo-monotone if it fulfills the following two conditions:

($\text{PM}_1$) For any filter $u_i$ on a compact subset of $X$ with $u_i\to u$ in $X$ and $\lim\sup\langle A(u_i), u_i - u \rangle \leq 0$ we have $$\langle A(u) , u - v\rangle \leq \lim \inf\langle A(u_i), u_i - v\rangle \quad\forall v \in X .$$ ($\text{PM}_2$) For all $v\in X$, the mapping $u \mapsto \langle A(u), u - v\rangle$ is bounded from below on all compact subsets of $X$.

Remark. — If the topology induced on $E$ by the compact subsets of $X$ is metrizable, the property ($\text{PM}_1$) is equivalent to the same properties formulized for sequences.

Additional questions:

Q1: What does $A(u_i)$ mean, if $u_i$ is a filter? How can we calculate with it as if it was an element of $F$?

Q2: A "filter $u_i$ on a compact subset of $X$" means that all elements of the filter are subsets of one compact subset of $X$, right?

Q3: Is ($\text{PM}_2$) equivalent to the following? For all $v\in X$ and all compact subsets $C\subseteq X$ there is some $M\in\mathbb R$ with $\langle A(u),u-v\rangle\geq M$ for all $u\in C$.