Original text: CC0 1.0. Prerequisite proofs and component terms.

Banach holomorphy, generators, and resolvent limits

OA-MOD-SG-04 — Stone's theorem with the derivative domain

Theorem. Let $U:\mathbb R\to B(H)$ be a strongly continuous unitary group. There is a unique self-adjoint operator $A$ such that

Its domain and action are exactly

Conversely, the spectral group of every self-adjoint $A$ is strongly continuous and has (SG.9) as its derivative domain.

Proof. Define $A$ by (SG.9). For $\varepsilon>0$, set

A difference quotient and translation of the integral give

Strong continuity gives $T_\varepsilon x\to x$, so $D(A)$ is dense. For $x\in D(A)$, the group law gives

If $x_j\to x$ and $Ax_j\to y$, integrate (SG.12) and pass to the limit:

The derivative at zero puts $x\in D(A)$ and gives $Ax=y$. Thus $A$ is closed. Differentiating $\langle U_tx,U_ty\rangle$ at zero shows

so $A$ is symmetric.

The two norm integrals

exist and have norm at most one. Changing variables in the difference quotients gives

Therefore

Both $A-i$ and $A+i$ are onto. If $y\in D(A^*)$, choose $x\in D(A)$ with $(A-i)x=(A^*-i)y$. Then $y-x\in\ker(A^*-i)=\operatorname{ran}(A+i)^\perp=0$. Hence $D(A^*)\subseteq D(A)$, and symmetry gives $A=A^*$.

Let $V_t=e^{itA}$, using the spectral calculus. Both $U_t$ and $V_t$ preserve $D(A)$, commute there with $A$, and have derivative $iA$. For $x\in D(A)$, differentiation of $U_{t-s}V_sx$ with respect to $s$ gives zero. Thus $U_tx=V_tx$; density proves (SG.8) on all of $H$.

Conversely, scalar dominated convergence proves strong continuity of $e^{itA}$. It also gives the derivative $iAx$ for $x\in D(A)$. If the difference quotient converges for an arbitrary $x$, its norms are bounded along a sequence $h\to0$. Fatou's lemma applied to

puts $x$ in the spectral domain of $A$. This proves the converse domain in (SG.9). Uniqueness follows from that formula. $\square$

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