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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 $v\in X$.

For a real Banach space $X$, a map $A:X\to X^{\ast }$ has operator monotonicity when $⟨A(u)-A(v),u-v⟩\ge 0$ for all $u,v\in X$. This is the monotone operator property, with the pairing between $X$ and its continuous dual space $X^{\ast }$. In a Hilbert space one can identify that pairing with the inner product.

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.

Take the functionalThe factor $1/8$ is essential: the derivative of $|s{|}^8/8=s^8/8$ is $s^7$, while the gradient with respect to $\xi$ of $|\xi {|}^4/4$ is $|\xi {|}^2\xi$. As in part (c), finite polynomial expansions and Hölder's inequality justify differentiation along every direction $v\in X$, giving $DJ(u)[v]=⟨A(u),v⟩$.

We prove the global and strict minimum directly. The strictly convex functions $\Phi (\xi )=|\xi {|}^4/4$ and $\Psi (s)=|s{|}^8/8$ satisfy their tangent inequalitieswith strict inequality when the respective arguments differ. For example, strict convexity of $\Phi$ follows from strict operator monotonicity of its gradient, proved by the midpoint identity in part (b)(iii); strict convexity of $\Psi$ follows because its derivative $s^7$ is strictly increasing.

Apply these inequalities pointwise to $\nabla u_{\ast },\nabla v$ and $u_{\ast },v$, then integrate. The weak formulation at $u_{\ast }$ with test function $v-u_{\ast }$ givesIf $v\ne u_{\ast }$ in $X$, the Poincaré inequality implies $\nabla v\ne \nabla u_{\ast }$ on a set of positive Lebesgue measure. The first tangent inequality is strict on that set and the second is nonnegative everywhere. Their integrable remainders therefore have strictly positive total integral. HenceThus $u_{\ast }$ is the unique global minimizer of $J$. Conversely, every minimizer has zero directional derivative in every direction and so satisfies $A(u)=0$, which also identifies it with the unique weak solution from part (c).

For real vectors $\xi ,\eta$, put $m=(\xi +\eta )/2$ and $d=\xi -\eta$. Expansion givesThis proves strict operator monotonicity of the gradient of $\xi \mapsto |\xi {|}^4/4$. Integrating the inequality controls the difference of two weak gradients in the fourth power of their Lp norm, which yields uniqueness for suitable nonlinear Dirichlet boundary conditions.

A strictly monotone operator $A:X\to X^{\ast }$ on a real Banach space satisfies strict operator monotonicity for every pair of distinct inputs. It is injective: $Au=Av$ would make the pairing zero and hence force $u=v$. In particular $Au=f$ has at most one solution for each $f\in X^{\ast }$.