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For a real Banach space $X$ and a convex set $K\subset X$, a map $A:K\to X^{\ast }$ is hemicontinuous if $t\mapsto ⟨A(x+t(y-x)),z⟩$ is continuous on $[0,1]$ for all $x,y\in K$ and $z\in X$. This tests continuity only along line segments and only through pairings with fixed vectors. It is enough for the limiting step in Minty's lemma.

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.