Monotone operator

An operator $F$ on an inner product space is monotone whenfor every $v,w$. The gradient of every differentiable convex function is monotone.

A map from a subset of a real Banach space to its continuous dual space is strongly monotone if the displayed inequality holds for every pair in its domain, with one constant $m>0$. The brackets denote the dual pairing. This implies strict monotonicity for distinct arguments. Fixing $v=u_0$ and applying the dual norm inequality gives $⟨A(u),u-u_0⟩\ge m\Vert u-u_0{\Vert }^2-\Vert A(u_0)\Vert \Vert u-u_0\Vert$, so it is a nonlinear coercive operator with respect to $u_0$.

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.

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 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 $f\in X^{\ast }$.

Let $\Omega$ be a nonempty bounded Lipschitz domain, and define $A:H^1(\Omega )\to H^1(\Omega {)}^{\ast }$ byThis operator is strictly monotone: the difference pairing is the sum of $\Vert \nabla (v-w){\Vert }_2^2$ and the nonnegative integral $(\tanh v-\tanh w)(v-w)$, which is strictly positive wherever $v\ne w$. The bounded-domain hypothesis also makes nonzero constant functions admissible in $H^1$.

For constant $v=t$,Thus $A$ is not a nonlinear coercive operator in the large-norm sense. On $H_0^1(\Omega )$, the Poincaré inequality instead giveswhich proves coercivity on that zero-boundary space. The example separates being a strictly monotone operator from coercivity and identifies the uncontrolled constant direction.

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.

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.

An operator $G$ is $\beta$-cocoercive when $⟨Gz-Gw,z-w⟩\ge \beta \Vert Gz-Gw{\Vert }^2$ with $\beta >0$. This strengthens monotonicity and implies a Lipschitz constant $1/\beta$ by the Cauchy-Schwarz inequality. Gradient maps of smooth convex functions provide an important converse through the Baillon–Haddad theorem.

On the whole Euclidean space, an $L$-Lipschitz gradient of a convex function is $1/L$-cocoercive. Subtract the supporting affine function at one point, apply the descent lemma to a gradient step at the other point, then interchange the points and add. The conclusion needs convexity but no second derivative.

For an operator $A$, its resolvent with step $\tau >0$ is $J_{\tau A}=(I+\tau A{)}^{-1}$. A maximal monotone operator has an everywhere-defined single-valued firmly nonexpansive mapping as its resolvent. For $A=\partial f$, the resolvent is the proximal map of $\tau f$. This monotone-operator resolvent uses a different convention from the spectral resolvent of a linear operator.

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.

A mapping $T$ on a normed vector space is nonexpansive when $\Vert T(v)-T(w)\Vert \le \Vert v-w\Vert$ for every $v,w$.

A strictly distance-decreasing map decreases the distance between every pair of distinct points. Unlike a contraction mapping, it need not have a single Lipschitz constant below one. Two distinct fixed points would contradict strict decrease, so there is at most one. Existence can fail even on a complete metric space: on $[0,\infty )$, $T(x)=x+1/(1+x)$ has no fixed point, while the mean value theorem and $0\le T^{\prime }(x)<1$ show strict distance decrease.

An operator is $\alpha$-averaged for $0<\alpha <1$ when $T=(1-\alpha )I+\alpha R$ for a nonexpansive mapping $R$. Averaging controls the update residual and enables fixed-point iteration. A firmly nonexpansive mapping is exactly a $1/2$-averaged operator.

In finite-dimensional Euclidean space, iterating an averaged operator with a nonempty fixed-point set converges in norm to a fixed point from every start. The averaged-operator inequality proves boundedness and a vanishing residual; compactness produces a fixed cluster point and Fejér monotonicity turns subsequential convergence into convergence of the full sequence. A fixed point must exist.

The defining averaging representation is equivalent to $\Vert Tz-Tw{\Vert }^2\le \Vert z-w{\Vert }^2-(1-\alpha )\Vert (I-T)z-(I-T)w{\Vert }^2/\alpha$. Expanding the norm of the convex combination proves both directions. Substituting a fixed point for $w$ gives Fejér monotonicity and summability of squared residuals.

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.

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.

For two proximal maps $P_f,P_h$, define reflected maps $R_f=2P_f-I$ and $R_h=2P_h-I$. The Douglas--Rachford fixed-point map isBecause reflected proximal maps are nonexpansive, $T$ is firmly nonexpansive.

Finding a point in ${\bigcap }_{j=1}^{\ell }{C}_j$ is equivalent to intersecting the product $C_1\times \cdots \times C_{\ell }$ with the diagonal subspace in $({\mathbb{R}}^n{)}^{\ell }$. Projection onto the product is componentwise, while projection onto the diagonal replaces every component by their average.

Let $M$ be a real symmetric positive-definite matrix and $F$ a maximal monotone operator on a finite-dimensional Euclidean space, possibly set-valued. The preconditioned proximal point algorithm usesMaximal monotonicity guarantees that $T$ is defined everywhere. It is a firmly nonexpansive mapping in the weighted inner product $⟨x,y{⟩}_M=⟨x,My⟩$.

Indeed, if $x=Tu$ and $y=Tv$, operator monotonicity givesThis proves the firm bound and single-valuedness wherever the inverse exists; maximality supplies the full domain. With mere monotonicity, existence can fail. On $\mathbb{R}$, $F(x)=\mathrm{sign}x$ with $F(0)=0$ is monotone, but $x+F(x)=1/2$ has no solution. Its maximal monotone completion instead assigns the interval $[-1,1]$ at zero.