Dissipative operator

Let $X$ be a Banach space over $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$ and let $A:D(A)\subseteq X\to X$ be a linear operator with domain $D(A)$ (Unbounded linear operators: domain, graph and extension). $A$ is dissipative if $\mathrm{\parallel }(\lambda I-A)x\mathrm{\parallel }\ge \lambda \mathrm{\parallel }x\mathrm{\parallel }\text{for all }\lambda >0\text{ and all }x\in D(A);$ equivalently $\mathrm{\parallel }x-\alpha Ax\mathrm{\parallel }\ge \mathrm{\parallel }x\mathrm{\parallel }$ for all $\alpha >0$ and $x\in D(A)$. A dissipative operator has injective $\lambda I-A$ for every $\lambda >0$ and $\mathrm{\parallel }(\lambda I-A{)}^{-1}z\mathrm{\parallel }\le {\lambda }^{-1}\mathrm{\parallel }z\mathrm{\parallel }$ on the range of $\lambda I-A$; no surjectivity, closedness or density is implied. If $X$ is a Hilbert space (Hilbert space), then $A$ is dissipative if and only if $\mathrm{R}\mathrm{e}⟨Ax,x⟩\le 0$ for every $x\in D(A)$: from the norm dissipativity inequality, squaring gives $\mathrm{R}\mathrm{e}⟨Ax,x⟩\le \mathrm{\parallel }Ax{\mathrm{\parallel }}^2\mathrm{/}(2\lambda )$ for every $\lambda >0$, so letting $\lambda \to \mathrm{\infty }$ yields $\mathrm{R}\mathrm{e}⟨Ax,x⟩\le 0$. Conversely, if this real-part inequality holds, expanding $\mathrm{\parallel }(\lambda I-A)x{\mathrm{\parallel }}^2={\lambda }^2\mathrm{\parallel }x{\mathrm{\parallel }}^2-2\lambda \mathrm{R}\mathrm{e}⟨Ax,x⟩+\mathrm{\parallel }Ax{\mathrm{\parallel }}^2$ gives the norm dissipativity inequality. Finally, under the Hahn-Banach extension principle HB (The real dominated-extension principle as an additional hypothesis over ZF) dissipativity is equivalent to the norm-duality form: for every $x\in D(A)$ there exists $x^{\ast }\in X^{\ast }$ with $\mathrm{\parallel }{x}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$, $x^{\ast }(x)=\mathrm{\parallel }x{\mathrm{\parallel }}^2$ and $\mathrm{R}\mathrm{e}{x}^{\ast }(Ax)\le 0$; the equivalence is the two-dimensional argument of [T] Lemma 11.19, where HB produces the norming functionals and supplies the extension from $\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\{x,Ax\}$ (Relative dual norming, point separation, and recovery of the norm), and the extraction of the limit uses compactness of the finite-dimensional dual unit ball (For $n\ge 1$ every bounded sequence in ${\mathbb{R}}^n$ has a convergent subsequence).

Two normalisations of the defining inequality. Putting $\alpha =1\mathrm{/}\lambda$ shows that the displayed inequality is equivalent to $\mathrm{\parallel }x-\alpha Ax\mathrm{\parallel }\ge \mathrm{\parallel }x\mathrm{\parallel }$ for all $\alpha >0$ and $x\in D(A)$. Since $\lambda x=(\lambda I-A)x+Ax$, the inequality $\mathrm{\parallel }(\lambda I-A)x\mathrm{\parallel }\ge \lambda \mathrm{\parallel }x\mathrm{\parallel }$ is in turn equivalent to the one-sided estimate $\lambda \mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }(\lambda I-A)x\mathrm{\parallel }$ used below.

Injectivity and the inverse bound. If $(\lambda I-A)x=0$ for some $\lambda >0$ and $x\in D(A)$, then $\lambda \mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }(\lambda I-A)x\mathrm{\parallel }=0$, so $x=0$: each $\lambda I-A$ is injective. If $z=(\lambda I-A)x$ lies in the range, then $\mathrm{\parallel }x\mathrm{\parallel }\le {\lambda }^{-1}\mathrm{\parallel }z\mathrm{\parallel }$, so the inverse defined on the range satisfies $\mathrm{\parallel }(\lambda I-A{)}^{-1}z\mathrm{\parallel }\le {\lambda }^{-1}\mathrm{\parallel }z\mathrm{\parallel }$. No surjectivity onto $X$, no closedness of $A$ and no density of $D(A)$ is asserted, and none is implied.

The real-part form on a Hilbert space. Let $X$ be a Hilbert space. If $A$ is dissipative and $x\in D(A)$, then for every $\lambda >0$ the expansion $\mathrm{\parallel }(\lambda I-A)x{\mathrm{\parallel }}^2={\lambda }^2\mathrm{\parallel }x{\mathrm{\parallel }}^2-2\lambda \mathrm{Re}⟨Ax,x⟩+\mathrm{\parallel }Ax{\mathrm{\parallel }}^2$ gives $2\lambda \mathrm{Re}⟨Ax,x⟩\le \mathrm{\parallel }Ax{\mathrm{\parallel }}^2$, that is $\mathrm{Re}⟨Ax,x⟩\le \mathrm{\parallel }Ax{\mathrm{\parallel }}^2\mathrm{/}(2\lambda )$; letting $\lambda \to \mathrm{\infty }$ yields $\mathrm{Re}⟨Ax,x⟩\le 0$. Conversely, if $\mathrm{Re}⟨Ax,x⟩\le 0$ for all $x\in D(A)$, the same expansion gives $\mathrm{\parallel }(\lambda I-A)x{\mathrm{\parallel }}^2\ge {\lambda }^2\mathrm{\parallel }x{\mathrm{\parallel }}^2$ for every $\lambda >0$, so $A$ is dissipative. This real-part form is choice-free; the Hilbert-space vocabulary comes from Hilbert space.

The norm-duality form under HB. Assume the Hahn-Banach extension principle (The real dominated-extension principle as an additional hypothesis over ZF) and let $x\in D(A)$. We claim that $A$ is dissipative if and only if for every $x\in D(A)$ there is $x^{\ast }\in X^{\ast }$ with $\mathrm{\parallel }{x}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$, $x^{\ast }(x)=\mathrm{\parallel }x{\mathrm{\parallel }}^2$ and $\mathrm{Re}{x}^{\ast }(Ax)\le 0$.

Sufficiency. If such $x^{\ast }$ is given and $\lambda >0$, then $\lambda \mathrm{\parallel }x{\mathrm{\parallel }}^2=\lambda \mathrm{Re}{x}^{\ast }(x)=\mathrm{Re}{x}^{\ast }((\lambda I-A)x)+\mathrm{Re}{x}^{\ast }(Ax)\le \mathrm{Re}{x}^{\ast }((\lambda I-A)x)\le \mathrm{\parallel }{x}^{\ast }\mathrm{\parallel }\mathrm{\parallel }(\lambda I-A)x\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }(\lambda I-A)x\mathrm{\parallel }\mathrm{.}$ For $x\mathrm{\ne }0$ this is $\lambda \mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }(\lambda I-A)x\mathrm{\parallel }$; for $x=0$ it is trivial. Hence $A$ is dissipative.

Necessity. Fix $x\in D(A)$; for $x=0$ take $x^{\ast }=0$, so assume $x\mathrm{\ne }0$ and put $M:=\mathrm{span}\{x,Ax\}$, a subspace of finite dimension at most two over the scalar field $\mathbb{K}$. For each positive integer $\lambda =n$ the vector $(\lambda I-A)x$ is nonzero, because $(\lambda I-A)$ is injective. Construct a sequence $y_n\in M^{\ast }$ without simultaneously choosing functionals on $X$. Fix a basis of the finite-dimensional space $M$. The coordinate vectors of functionals of norm at most one form a closed bounded subset of a finite real coordinate space: $\mathrm{\mid }y(u)\mathrm{\mid }\le \mathrm{\parallel }u\mathrm{\parallel }$ for every $u\in M$ is an intersection of closed conditions, and each basis evaluation is bounded. For each $n$, intersect this set with $y((nI-A)x)=\mathrm{\parallel }(nI-A)x\mathrm{\parallel }$. The intersection is nonempty by Relative dual norming, point separation, and recovery of the norm applied to $M$, and compact by Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line. Select its lexicographically least coordinate vector by minimizing its real coordinates successively; each minimum exists because the corresponding projected compact set is nonempty. This finite deterministic procedure defines $y_n$ for all $n$, with norm one and the required norming identity, without Countable Choice. Write $y_{\lambda }=y_n$, $\lambda =n$, below. Then $\mathrm{Re}{y}_{\lambda }(Ax)=\lambda \mathrm{Re}{y}_{\lambda }(x)-\mathrm{\parallel }(\lambda I-A)x\mathrm{\parallel }\le \lambda \mathrm{\mid }{y}_{\lambda }(x)\mathrm{\mid }-\lambda \mathrm{\parallel }x\mathrm{\parallel }\le 0,$ and likewise $y_{\lambda }(x)=\mathrm{\parallel }(\lambda I-A)x\mathrm{\parallel }\mathrm{/}\lambda +y_{\lambda }(Ax)\mathrm{/}\lambda$, where $\mid \mathrm{\parallel }(\lambda I-A)x\mathrm{\parallel }\mathrm{/}\lambda -\mathrm{\parallel }x\mathrm{\parallel }\mid \le \mathrm{\parallel }Ax\mathrm{\parallel }\mathrm{/}\lambda \to 0$ and $\mathrm{\mid }{y}_{\lambda }(Ax)\mathrm{\mid }\le \mathrm{\parallel }Ax\mathrm{\parallel }$; hence $y_{\lambda }(x)\to \mathrm{\parallel }x\mathrm{\parallel }$.

Passing to a subsequence. The restrictions $y_n$ lie in the unit ball of the dual of the finite-dimensional space $M$, which is sequentially compact: after choosing coordinates for $M^{\ast }$, the coordinates of a functional amount to a bounded sequence in a Euclidean space, and For $n\ge 1$ every bounded sequence in ${\mathbb{R}}^n$ has a convergent subsequence extracts a convergent subsequence. Take ${\lambda }_n\to \mathrm{\infty }$ along a subsequence on which $y_{{\lambda }_n}$ converges to some $y\in M^{\ast }$. Then $\mathrm{\parallel }y\mathrm{\parallel }\le 1$, $\mathrm{Re}y(Ax)\le 0$ (a closed condition), and $y(x)={\lim }_n{y}_{{\lambda }_n}(x)=\mathrm{\parallel }x\mathrm{\parallel }$; consequently $\mathrm{\parallel }y\mathrm{\parallel }=1$ because $\mathrm{\mid }y(x)\mathrm{\mid }=\mathrm{\parallel }x\mathrm{\parallel }$ with $x\mathrm{\ne }0$.

Extension to $X$. The real part $g:=\mathrm{Re}y$ is a real-linear functional on the real vector space $M$ with $\mathrm{\mid }g(u)\mathrm{\mid }\le \mathrm{\parallel }u\mathrm{\parallel }$ for $u\in M$; here the real structure of $X$ is the one underlying the complex case as well. Apply HB, with the sublinear functional $p(u)=\mathrm{\parallel }u\mathrm{\parallel }$, to extend $g$ to a real-linear $G:X\to \mathbb{R}$ satisfying $G(u)\le \mathrm{\parallel }u\mathrm{\parallel }$ for all $u\in X$; then $\mathrm{\mid }G(u)\mathrm{\mid }\le \mathrm{\parallel }u\mathrm{\parallel }$ by applying the inequality to $-u$. In the real case set $\tilde{x}:=G$. In the complex case set $\tilde{x}(u):=G(u)-iG(iu)$; then $\tilde{x}(iu)=G(iu)+iG(u)=i\tilde{x}(u)$; together with real linearity this proves complex linearity, and its real part is $G$. For $\mathrm{\parallel }u\mathrm{\parallel }\le 1$ put $z=\tilde{x}(u)$. If $z\mathrm{\ne }0$, take $c=\bar{z}\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$, so $\mathrm{\mid }c\mathrm{\mid }=1$ and $\tilde{x}(cu)=cz=\mathrm{\mid }z\mathrm{\mid }$ is real. Thus $\mathrm{\mid }z\mathrm{\mid }=G(cu)\le \mathrm{\parallel }cu\mathrm{\parallel }=\mathrm{\parallel }u\mathrm{\parallel }\le 1$; if $z=0$ the same bound is immediate. Thus $\mathrm{\parallel }\tilde{x}\mathrm{\parallel }\le 1$ in either case, and $x^{\ast }:=\mathrm{\parallel }x\mathrm{\parallel }\tilde{x}$ satisfies $\mathrm{\parallel }{x}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$ because $\tilde{x}(x)$, having real part $G(x)=g(x)=\mathrm{\parallel }x\mathrm{\parallel }$ and modulus at most $\mathrm{\parallel }x\mathrm{\parallel }$, equals the positive real number $\mathrm{\parallel }x\mathrm{\parallel }$. Finally $\mathrm{Re}{x}^{\ast }(Ax)=\mathrm{\parallel }x\mathrm{\parallel }G(Ax)=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{Re}y(Ax)\le 0$, as required. The definition and both elementary forms are choice-free; only the norm-duality form uses HB and the finite-dimensional compactness above.