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A mapping $T$ on an inner product space is firmly nonexpansive whenEvery firmly nonexpansive mapping is nonexpansive, and the resolvent of a maximal monotone operator is firmly nonexpansive.

A map $A:X\to X^{\ast }$ on a real Banach space is a monotone operator exactly when the displayed scalar function is nondecreasing on $[0,1]$ for every $u,v\in X$. Applying operator monotonicity to $u+tv$ and $u+sv$, $s<t$, gives $(t-s)(\phi (t)-\phi (s))\ge 0$. Conversely take $u=y$, $v=x-y$ and compare the endpoints to recover $⟨A(x)-A(y),x-y⟩\ge 0$. No continuity or differentiation of $A$ is needed.

A monotone operator is maximal when its graph has no proper enlargement that is still monotone. In Euclidean or Hilbert space, its resolvent of a monotone operator is single-valued and everywhere defined for every positive step. Subdifferentials of proper lower semicontinuous convex functions are fundamental examples.

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 $v\in X$.

Adding the two subgradient inequalities for $p\in \partial f(x)$ and $q\in \partial f(y)$ givesThus the subdifferential is a monotone operator. This inequality proves uniqueness of its resolvent: two solutions of $z\in x+\tau \partial f(x)$ must coincide when $\tau >0$.

For a map $A:M\to X^{\ast }$ on a subset of a real Banach space, coercivity with respect to $u_0\in M$ means the displayed limit as $u\in M$ and $\Vert u-u_0\Vert \to \infty$. Here the brackets denote the dual pairing. This growth condition is useful in existence theorems for monotone operators. A linear Hilbert-space estimate $⟨Au,u⟩\ge c\Vert u{\Vert }^2$, $c>0$, implies the condition at zero. Monotonicity alone does not: the zero operator is monotone and fails this growth condition on an unbounded space.

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.

Using the dual pairing, a monotone operator satisfies $⟨A(u)-A(v),u-v⟩\ge 0$ for all $u,v\in M$. It is a strongly monotone operator if some $m>0$ givesA nonlinear coercive operator with respect to $u_0\in M$ satisfies $⟨A(u),u-u_0⟩/\Vert u-u_0\Vert \to +\infty$ as $\Vert u-u_0\Vert \to \infty$ in $M$.

Strong monotonicity, with $v=u_0$, givesAfter division, the lower bound is $m\Vert u-u_0\Vert -\Vert A(u_0){\Vert }_{X^{\ast }}\to +\infty$, proving coercivity. Reflexivity and separability are not needed for this particular estimate.

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.

The first-order subgradient inequality for a differentiable convex function givesand, after interchanging $v$ and $w$,Adding and rearranging yieldsThus the gradient $F=\nabla f$ is a monotone operator.

If $T(w)=w$, the defining equation becomes $M(w)+F(w)=M(w)$, so $\boxed{F(w)=0}$.

A map is firmly nonexpansive in the $M$-inner product whenPut $p=T(v)$ and $q=T(w)$. The two implicit equations giveBecause $F$ is a monotone operator,Thereforewhich is precisely firm nonexpansiveness. In particular, the preconditioned proximal point algorithm map $T=(M+F{)}^{-1}M$ is nonexpansive in the norm induced by the positive-definite matrix $M$.

Use the sign conventionfor the Lagrangian function in constrained optimization. The Lagrangian dual problem iswhere $f^{\ast }$ is the convex conjugate. For this convex problem with affine equality constraints, the stationarity and feasibility parts of the Karush-Kuhn-Tucker conditions areThey say exactly that the displayed operator satisfiesThus its zeros are precisely the primal-dual optimal points, subject to the usual attainment assumptions.

For $u=(x,z)$ and $\tilde{u}=(y,s)$, the Euclidean inner product givesThe last two terms cancel by the defining property of the matrix transpose, and the first is nonnegative by part a. Hence $F$ is a monotone operator.

Using the dual pairing, monotonicity meansStrict monotonicity requires strict positivity whenever $u\ne v$. Coercivity with respect to zero, in the growth sense used for monotone-operator existence theory, meansThis is a condition on all sequences escaping to infinity, not just each fixed direction with a bound depending on that direction. Strict monotonicity concerns pairs of distinct arguments; it does not by itself imply this uniform large-norm growth.

Using the dual pairing between $X^{\ast }$ and $X$, a monotone operator satisfiesA nonlinear coercive operator with respect to $u_0$ satisfiesThe directional derivative is ${\partial }_{v}F(u)={\lim }_{t\to 0}[F(u+tv)-F(u)]/t$, when this finite limit exists through admissible points. At a boundary point of $M$ only an inward direction may be admissible; then one uses the corresponding one-sided limit.

If $A:X\to X^{\ast }$ is monotone and $s<t$, apply its defining inequality at $u+tv$ and $u+sv$. It becomes $(t-s)[\phi (t)-\phi (s)]\ge 0$, so $\phi$ is nondecreasing. Conversely choose $u=y$, $v=x-y$. Nondecrease from zero to one gives $⟨A(x)-A(y),x-y⟩\ge 0$. This proves the line criterion for a monotone operator.

The graph area functional is finite on $H^1$, since $\sqrt{1+|\nabla u{|}^2}\le 1+|\nabla u|$ and $\Omega$ has finite volume. For $p=\nabla u$, $q=\nabla v$, the difference quotient of $\sqrt{1+|p+tq{|}^2}$ is bounded in absolute value by $|q|$. This is integrable by the Cauchy-Schwarz inequality. The dominated convergence theorem therefore givesThis defines a bounded linear functional of $v$, since $|⟨A(u),v⟩|\le |\Omega {|}^{1/2}\Vert \nabla v{\Vert }_2$.

For $a(p)=p/\sqrt{1+|p{|}^2}$ the Jacobian matrix isThe inequality follows from the Cauchy-Schwarz inequality. Integrating this derivative along the segment between any two gradients shows $(a(p)-a(r))\cdot (p-r)\ge 0$. Integrating over $\Omega$ proves that $A$ is a monotone operator.

It is not coercive on $H^1(\Omega )$. The constant functions $u_k\equiv k$ satisfy $\Vert u_k{\Vert }_{H^1}=k|\Omega {|}^{1/2}\to \infty$ but $A(u_k)=0$. Thus $⟨A(u_k),u_k⟩/\Vert u_k{\Vert }_{H^1}=0$.

The proximal point algorithm seeks a zero of a monotone operator $F$ by repeatedly applying its resolvent:For $F=\partial f$, this is iteration of a proximal operator.