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For a monotone operator $A:K\to X^{\ast }$ on a nonempty convex set $K$ in a real Banach space, assume hemicontinuity. Then $u\in K$ solves $⟨Au,v-u⟩\ge 0$ for every $v\in K$ if and only if $⟨Av,v-u⟩\ge 0$ for every $v\in K$. The quantifier applies to the entire inequality on each side.

The forward implication is operator monotonicity. For the reverse implication, choose $v=u+t(w-u)$ with $w\in K$ and $t>0$. Divide the assumed inequality by $t$ and pass to $t\downarrow 0$ using hemicontinuity; the result is $⟨Au,w-u⟩\ge 0$. For $K=X$, applying this to $A-f$ shows that $Au=f$ is equivalent to $⟨Av-f,v-u⟩\ge 0$ for every $v\in X$.

A useful form of Minty's lemma is the following. Let $X$ be a real Banach space, $K\subset X$ a nonempty convex set, and $A:K\to X^{\ast }$ a monotone operator that is hemicontinuous. Here operator monotonicity means $⟨Ax-Ay,x-y⟩\ge 0$ for $x,y\in K$, and hemicontinuity means that $t\mapsto ⟨A(x+t(y-x)),z⟩$ is continuous on $[0,1]$ for $x,y\in K$ and $z\in X$. For $u\in K$,The forward implication follows from operator monotonicity:For the reverse implication, fix $w\in K$ and take $v_t=u+t(w-u)\in K$ for $0<t\le 1$. The assumed inequality givesDivide by $t$ and send $t\downarrow 0$. Hemicontinuity gives $⟨Au,w-u⟩\ge 0$, as required.

In the whole-space version, for a monotone operator $A:X\to X^{\ast }$ that is hemicontinuous,Apply the just-proved result to $A-f$ and $K=X$; the resulting inequality $⟨Au-f,v-u⟩\ge 0$ for all $v$ forces $Au=f$ by choosing $v=u\pm z$. This is the variational inequality form of Minty's lemma.