Why is a weak variational inequality *equivalent* to a strong variational inequality for a continuous, monotone operator? Consider a closed set $Q$, and a continuous monotone operator $g: Q \rightarrow E^*$. Consider the "weak" variational inequality: $$\mbox{find } x^* \in Q: \, g(x)^T (x-x^*) \geq 0, \, \forall x \in Q. $$  Another problem is the "strong " variational inequality: $$\mbox{find } x^* \in Q: \, g(x^*)^T (x-x^*) \geq 0, \, \forall x \in Q. $$  By monotonicity of $g$, $$ (g(x) - g(y))^T(x - y) \geq 0, \, \forall x, y, \in Q.$$  So it is clear to me that the strong variational inequality implies the weak. But this paper I'm reading claims that the two are equivalent, so the weak also implies the strong. Specifically, it says this is because of continuity of $g$. I'm not able to see how this is the case. Can someone provide any ideas? Thank you! 
 A: This is a famous results which is sometimes called Minty lemma or Linearization lemma. (See also any book about variational inequalities, for example the well-known book An Introduction to Variational Inequalities and Their Applications of Kinderlehrer and Stampacchia, p. 84). 
Let us denote the variational inequalities by (1) and (2). If $Q$ is a non-empty, closed and convex subset of a finite-dimensional Euclidean space $E$, then the continuity of $g: Q \to E$ is sufficient to show that (1) implies (2):
Let $x^* \in Q$ be a solution of (1), and put $x= x^* + t(w-x^*)$, $t\in (0,1]$, where $w\in Q$ is arbitrarily chosen. Since $Q$ is convex (!), $x\in Q$. Now, inserting this element into (1) yields
$$\langle g(x^*+t(w-x^*)), t(w-x^*) \rangle \geq 0.$$
Dividing this inequality by $t$ and then passing to the limit $t\downarrow 0$ yields 
$$ \langle g(x^*), w-x^*\rangle \geq 0, \qquad \text{for every } w\in Q,$$
where we used the fact that $g$ is continuous.
If $E$ is not finite-dimensional, then one has to replace the continuity condition for $g$ by one of the two following (equivalent) conditions:


*

*$g:Q \to E^*$ is continuous on finite-dimensional subspaces of $E$, that is, for every finite-dimensional subspace $F\subset E$, the restriction of $g$ on $Q\cap F$ is weakly continuous.

*The real valued function $\mathbf R \to \mathbf R$, $t\mapsto \langle g(u+tv),w\rangle$ is continuous at $0^+$ for every fixed elements $u,v,w\in X$.

