= Solution
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 $A:K\to X^*$ a <monotone operator> that is <hemicontinuous>. Here <operator monotonicity> means $\langle Ax-Ay,x-y\rangle\geq0$ for $x,y\in K$, and <hemicontinuity> means that $t\mapsto\langle A(x+t(y-x)),z\rangle$ is continuous on $[0,1]$ for $x,y\in K$ and $z\in X$. For $u\in K$,
$$
\boxed{\langle Au,v-u\rangle\geq0\ \forall v\in K
\quad\Longleftrightarrow\quad
\langle Av,v-u\rangle\geq0\ \forall v\in K.}
$$
The forward implication follows from <operator monotonicity>:
$$
\langle Av,v-u\rangle
=\langle Au,v-u\rangle+\langle Av-Au,v-u\rangle\geq0.
$$
For the reverse implication, fix $w\in K$ and take $v_t=u+t(w-u)\in K$ for $0<t\leq1$. The assumed inequality gives
$$
0\leq\langle Av_t,v_t-u\rangle=t\langle A(u+t(w-u)),w-u\rangle.
$$
Divide by $t$ and send $t\downarrow0$. <Hemicontinuity> gives $\langle Au,w-u\rangle\geq0$, as required.
In the whole-space version, for a <monotone operator> $A:X\to X^*$ that is <hemicontinuous>,
$$
\boxed{Au=f\quad\Longleftrightarrow\quad
\langle Av-f,v-u\rangle\geq0\quad\forall v\in X.}
$$
Apply the just-proved result to $A-f$ and $K=X$; the resulting inequality $\langle Au-f,v-u\rangle\geq0$ for all $v$ forces $Au=f$ by choosing $v=u\pm z$. This is the <variational inequality> form of <Minty's lemma>.
Back to article page
