Unbounded Self Adjoint Operators and Stones Theorem — Examples

Multiplication by $m$ is well defined on null classes: two representatives that agree off a null set have products agreeing there, and the squared-integrability condition is unchanged. It is linear on its domain, since $\mathrm{\mid }m(u+v){\mathrm{\mid }}^2\le 2\mathrm{\mid }mu{\mathrm{\mid }}^2+2\mathrm{\mid }mv{\mathrm{\mid }}^2$. For $u\in H$, $u_n=u{\mathbf{1}}_{\{\mathrm{\mid }m\mathrm{\mid }\le n\}}$ belongs to the domain and tends to $u$ in $L^2$ by domination by $\mathrm{\mid }u{\mathrm{\mid }}^2$, so the domain is dense. All multiplications and norms below use the complex Hilbert structure in [A1].

A bounded measurable multiplier $a$ gives a bounded operator with $\mathrm{\parallel }au{\mathrm{\parallel }}_2\le (\sup \mathrm{\mid }a\mathrm{\mid })\mathrm{\parallel }u{\mathrm{\parallel }}_2$. Define $E(B)u={\mathbf{1}}_{m^{-1}(B)}u$ for Borel $B\subseteq \mathbb{R}$. It is idempotent and self-adjoint by the integral pairing in [A1]. Preimages show normalization and multiplicativity. For disjoint Borel $B_j$, the difference between $E({\bigcup }_j{B}_j)u$ and its first $n$ summands has squared norm the integral of $\mathrm{\mid }u{\mathrm{\mid }}^2$ over the remaining preimages, tending to zero by dominated convergence. Thus $E$ is a PVM.

If $H=\{0\}$, sigma-finiteness forces $\mu =0$: otherwise some finite-measure set in a countable finite-measure cover would have positive measure, and its indicator would be a nonzero $L^2$ vector. All operators then have full zero-space domain and zero action, and the spectrum and essential range are empty; [A4] and [A5] give the direct zero-space conventions. All claims follow in this case. Henceforth assume $H\mathrm{\ne }\{0\}$ before using the bounded calculus or spectral uniqueness in [A4]–[A5].

Its scalar measure is $E_u(B)={\int }_{m^{-1}(B)}\mathrm{\mid }u{\mathrm{\mid }}^{2}d\mu$, a finite Borel measure of mass $\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$. The real line is Hausdorff (disjoint small intervals separate distinct points), locally compact by compact closed bounded intervals, and second-countable by the rational base in [A3]. Therefore the regularity theorem in [A3] applies to every $E_u$: $E$ is regular. Moreover, for any nonnegative Borel $q$, $\int qd{E}_u=\int (q\circ m)\mathrm{\mid }u{\mathrm{\mid }}^{2}d\mu$. This holds first for indicators by the displayed scalar measure, then finite nonnegative simple sums, then all nonnegative Borel functions by increasing simple approximation and monotone convergence.

For a complex Borel simple function $s$ on $\mathbb{R}$, complete its disjoint representation with the zero-coefficient complement. By [A4], its integral against $E$ is multiplication by $s\circ m$. Given bounded Borel $h$, partition a square containing its complex range into finitely many Borel cells of diameter tending to zero, taking a fixed corner as each coefficient; these give complex simple $s_n$ with $\sup \mathrm{\mid }{s}_n-h\mathrm{\mid }\to 0$. The multiplier norm bound from step 1.2 and the operator-norm approximation in [A4] imply ${\mathrm{\Phi }}_E(h)=M_{h\circ m}$.

For arbitrary Borel $g:\mathbb{R}\to \mathbb{C}$, step 2.1 with $q=\mathrm{\mid }g{\mathrm{\mid }}^2$ identifies the domain of $g(E)$ with $\{u\in H:\int \mathrm{\mid }g\circ m{\mathrm{\mid }}^2\mathrm{\mid }u{\mathrm{\mid }}^{2}d\mu <\mathrm{\infty }\}$. Its bounded truncations act by $g(m){\mathbf{1}}_{\{\mathrm{\mid }g(m)\mathrm{\mid }\le n\}}u$ by step 2.2, and these tend in $L^2$ to $(g\circ m)u$ by dominated convergence. In particular $\int \lambda dE$ equals $M_m$ with exactly the stated domain. Regularity proved in step 2.1 licenses the converse spectral theorem in [A5], so $M_m$ is self-adjoint and $E$ is its spectral PVM, unique among regular representing PVMs. Thus the calculation for general $g$ is indeed its functional calculus.

For any measurable set $A$, the indicator multiplier $M_{{\mathbf{1}}_A}$ vanishes if $\mu (A)=0$. Conversely, let $(X_n)$ be a countable cover by finite-measure sets. If $\mu (A)>0$, some $A\cap X_n$ has positive finite measure, since a countable union of null sets is null. Then $u={\mathbf{1}}_{A\cap X_n}$ is a nonzero $L^2$ vector fixed by that multiplier. Therefore $E(B)=0$ exactly when $\mu (m^{-1}(B))=0$. Apply the spectral essential-range formula in [A5] to the identity function: this gives precisely the stated real essential range of $m$. Nonreal points are outside that range since the PVM is on the real line. This completes the domain, self-adjointness, spectral measure, calculus and spectrum claims.

$Q$ is the multiplication operator of the previous example for the measure space $(\mathbb{R},\text{Borel},\lambda )$ and $m(x)=x$: the domain, the self-adjointness, the spectral PVM $E(B)f={\mathbf{1}}_{B}f$ and the calculus $g(Q)f=(g\circ m)f$ are those results.

The essential range of $x$ is $\mathbb{R}$, since every interval $(t-\epsilon ,t+\epsilon )$ has positive Lebesgue measure, so $\sigma (Q)=\mathbb{R}$.

By the generation theorem applied to the self-adjoint operator $Q$, the formula $U(t)=\int e^{itx}dE(x)$ is a strongly continuous unitary group with generator $iQ$, and the calculus of [A1] identifies $U(t)f(x)=e^{itx}f(x)$.

$D(Q)$ is proper and dense: it is dense by the previous example, and the function $f(x)=(1+\mathrm{\mid }x\mathrm{\mid }{)}^{-1}$ for $\mathrm{\mid }x\mathrm{\mid }\ge 1$, extended by $1$ on $[-1,1]$, lies in $L^2(\mathbb{R})$ but not in $D(Q)$, because ${\int }_1^{\mathrm{\infty }}{x}^2(1+x{)}^{-2}dx$ diverges.

The claims are steps 1.1, 1.2, 1.3 and 1.4. ∎

The domain is linear and its representatives and derivatives are well defined by [A1]. Since D(T) is dense and contained in D(P), P is densely defined. For periodic f,g in D(P), integration by parts and the first-variable convention give $⟨Pf,g⟩=-i[f\bar{g}{]}_0^1+i{\int }_0^{1}f\overline{g^{\mathrm{\prime }}}=i{\int }_0^{1}f\overline{g^{\mathrm{\prime }}}=⟨f,Pg⟩\mathrm{.}$ Periodicity of both endpoints cancels the boundary term. Thus P is symmetric.

For real t let r be its representative in [0,1) modulo integers. Splitting the x integral at 1-r and translating on the two intervals gives ${\int }_0^1\mathrm{\mid }f((x+r)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>1){\mathrm{\mid }}^{2}dx={\int }_r^1\mathrm{\mid }f(y){\mathrm{\mid }}^{2}dy+{\int }_0^r\mathrm{\mid }f(y){\mathrm{\mid }}^{2}dy=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2\mathrm{.}$ Endpoints have measure zero. The same computation for indicators of null sets proves independence of the measurable representative; periodic extension from a Lebesgue-measurable representative is measurable, and translations preserve null modifications. V(t) is linear, V(0)=I and V(s)V(t)=V(s+t) on classes by addition modulo 1. Its inverse is V(-t), so it is unitary.

Let $\epsilon \in \{1,-1\}$ and $g\in H$. Cauchy--Schwarz gives $g\in L^1(0,1)$. Put $c_{\epsilon }=\frac{-i{e}^{\epsilon }}{e^{\epsilon }-1}{\int }_0^1{e}^{-\epsilon s}g(s)ds,u(x)=e^{\epsilon x}(c_{\epsilon }+i{\int }_0^x{e}^{-\epsilon s}g(s)ds)\mathrm{.}$ The denominator is nonzero for either sign. By [A4], u is absolutely continuous and $u^{\mathrm{\prime }}=\epsilon u+ig$ almost everywhere. For completeness, the product with the smooth exponential is absolutely continuous: the integral factor is bounded and absolutely continuous, the exponential is bounded with bounded derivative, and the increment product formula verifies the defining AC estimates. Thus u is bounded, belongs to $L^2$, and u' belongs to $L^2$. The displayed constant gives $u(1)=u(0)=c_{\epsilon }$. Finally $-i{u}^{\mathrm{\prime }}+\epsilon iu=g$, so $(P+\epsilon i)u=g$. Both shifts are onto; [A2] and step 1.1 make P self-adjoint.

Every f in D(T) has a continuous periodic extension. Its restriction to [-1,2] is uniformly continuous by [A5]; hence $\mathrm{\parallel }V(t)f-f{\mathrm{\parallel }}_2\le {\sup }_{x\in [0,1]}\mathrm{\mid }f(x+t)-f(x)\mathrm{\mid }\to 0$ as t tends to zero through either sign. Given arbitrary f in H and eta>0 choose g in D(T) with $\mathrm{\parallel }f-g{\mathrm{\parallel }}_2<\eta$ by [A1]. Isometry gives $\mathrm{\parallel }V(t)f-f{\mathrm{\parallel }}_2\le 2\eta +\mathrm{\parallel }V(t)g-g{\mathrm{\parallel }}_2$. First send t to zero and then eta to zero. The group law and isometry transfer continuity to every real time. Thus V is a strongly continuous unitary group.

For f in D(P), its periodic extension is absolutely continuous on every compact interval: finitely many translates of the AC representative join with matching endpoint values, and the AC estimates combine across finitely many joins. Its a.e. derivative is the periodic extension of f'. By [A4], for positive or negative t, $\frac{V(t)f-f}{t}-f^{\mathrm{\prime }}=\frac{1}{t}{\int }_0^t(V(s)f^{\mathrm{\prime }}-f^{\mathrm{\prime }})ds$ in the scalar pointwise integral sense for almost every x. Let J_t be the interval between 0 and t. Cauchy--Schwarz in s and Tonelli yield ${\parallel \frac{V(t)f-f}{t}-f^{\mathrm{\prime }}\parallel }_2^2\le \frac{1}{\mathrm{\mid }t\mathrm{\mid }}{\int }_{J_t}\mathrm{\parallel }V(s)f^{\mathrm{\prime }}-f^{\mathrm{\prime }}{\mathrm{\parallel }}_2^{2}ds\le {\sup }_{\mathrm{\mid }s\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }}\mathrm{\parallel }V(s)f^{\mathrm{\prime }}-f^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2⟶0,$ by step 2.2 applied to the $L^2$ class f'. For joint measurability use the explicit periodic Borel representative of f' obtained as the finite limit of $(n+1)(f(x+1\mathrm{/}(n+1))-f(x))$, assigning zero where no finite limit exists. Each difference quotient is continuous, its finite-convergence set is Borel by the countable Cauchy criterion, and the limit equals f' wherever f is differentiable. Composition with addition modulo 1 is jointly Borel. Since the representative differs from f' only on a null set, translation invariance and Tonelli leave the displayed estimates unchanged. Consequently D(P) is contained in D(G) and Gf=f'=iPf.

By [A3], S=-iG is self-adjoint. Step 3.1 gives P contained in S, with equal values on D(P). P is itself self-adjoint by step 2.1, so [A2]'s maximality applies to P and its symmetric extension S, and gives P=S. Equivalently the adjoint inclusions read $P\subseteq S=S^{\ast }\subseteq P^{\ast }=P$. Thus D(G)=D(P), G=iP and Stone's uniqueness gives $V(t)=e^{itP}$.

The zero function and every constant function are in D(P); constants are fixed by V and annihilated by P and G. V(0)=I, integer translations are I, and negative times are included in both the group and derivative calculations. No division by t occurs at t=0, only a two-sided limit, and neither $e-1$ nor $e^{-1}-1$ vanishes. Endpoint values belong to the unique AC representative, while the translation action belongs to $L^2$ classes. Full AC has the uses in [A6]; the two resolvent solutions and the translation are explicit.

By A symmetric closed operator that is not self-adjoint, $T$ is densely defined, closed, symmetric and not self-adjoint, and $D(T^{\ast })=\{g\in AC[0,1]:g^{\mathrm{\prime }}\in L^2(0,1)\}$ with $T^{\ast }g=-i{g}^{\mathrm{\prime }}$.

$\ker (T^{\ast }-i)=\mathbb{C}{e}^{-x}$ and $\ker (T^{\ast }+i)=\mathbb{C}{e}^x$: the equations $T^{\ast }g=ig$ and $T^{\ast }g=-ig$ read $g^{\mathrm{\prime }}=-g$ and $g^{\mathrm{\prime }}=g$.

Hence $d_{+}=d_{-}=1$ and $K_{+}\cap K_{-}=\{0\}$; by the von Neumann parameterization the self-adjoint extensions of $T$ correspond bijectively to the unitaries $V:\mathbb{C}{e}^{-x}\to \mathbb{C}{e}^x$, that is, to the numbers $\lambda$ with $V(e^{-x})=\lambda e^x$ and $\mathrm{\mid }\lambda \mathrm{\mid }=\mathrm{\parallel }{e}^{-x}\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }{e}^x\mathrm{\parallel }=1\mathrm{/}e$.

Domains: by the parameterization, $D(T_V)=D(T)\oplus \{u+Vu:u\in K_{+}\}$. For $u=c{e}^{-x}$ the element is $x_T+c(e^{-x}+\lambda e^x)$ with $x_T(0)=x_T(1)=0$, so its endpoint values are $c(1+\lambda )$ at $0$ and $c(e^{-1}+\lambda e)$ at $1$; hence $D(T_V)$ consists exactly of the absolutely continuous $f$ with $f^{\mathrm{\prime }}\in L^2$ and $f(1)=\mu f(0)$, where $\mu =(e^{-1}+\lambda e)\mathrm{/}(1+\lambda )$.

In particular there are infinitely many self-adjoint extensions, so $T$ is not self-adjoint; the case of $\lambda$ equal to a suitable value reproduces the periodic operator of Periodic derivative and its unitary translation group.

The map $\lambda \mapsto \mu$ is a bijection from $\{\lambda :\mathrm{\mid }\lambda \mathrm{\mid }=1\mathrm{/}e\}$ onto the unit circle: for $\mathrm{\mid }\lambda \mathrm{\mid }=1\mathrm{/}e$ one computes $\mathrm{\mid }\mu \mathrm{\mid }=1$, and for $\mathrm{\mid }\mu \mathrm{\mid }=1$ the formula $\lambda =(\mu -e^{-1})\mathrm{/}(e-\mu )$ inverts it and satisfies $\mathrm{\mid }\lambda \mathrm{\mid }=1\mathrm{/}e$.

Therefore the self-adjoint extensions of $T$ are exactly the operators $T_{\mu }$ of the statement, one for each $\mu$ on the unit circle; $T$ itself is not among them because it is not self-adjoint.

The closed graph theorem gives that $T$ is bounded: an everywhere defined linear map between Banach spaces with closed graph is bounded.

Let $H$ be a Hilbert space and let $S$ be a self-adjoint operator on $H$. A self-adjoint operator is closed, being equal to the adjoint of a densely defined operator; if in addition $D(S)=H$, then step 1.1 applied to $S:H\to H$ shows that $S$ is bounded.

Therefore a self-adjoint operator that is unbounded must have $D(S)\mathrm{\ne }H$, and its domain is dense by the definition of self-adjointness; the claimed impossibility of an unbounded everywhere-defined self-adjoint operator follows.

Both conclusions are steps 1.1 and 2.1, and the hypothesis used is exactly Dependent Choice, through the closed graph theorem. ∎

Multiplication by $e^{itx}$ is well defined on almost-everywhere classes, preserves the $L^2$ norm, and has inverse multiplication by $e^{-itx}$. The scalar exponential identities give $U(s)U(t)=U(s+t)$ and $U(0)=I$, so $U$ is a unitary group.

By the position-operator example $U(t)=e^{itQ}$, and by the generation lemma its generator is $iQ$ on $D(Q)=\{f:\int x^2\mathrm{\mid }f(x){\mathrm{\mid }}^{2}dx<\mathrm{\infty }\}$. These are the uses of AC inherited in [A6]. For positive integers $n$, $f_n={\mathbf{1}}_{[n,n+1]}$ has norm one, belongs to $D(Q)$, and satisfies $\mathrm{\parallel }Q{f}_n\mathrm{\parallel }\ge n$. Thus this generator is unbounded.

For $t\mathrm{\ne }0$, $\mathrm{\mid }{e}^{itx}-1\mathrm{\mid }\le 2$ gives $\mathrm{\parallel }U(t)-I\mathrm{\parallel }\le 2$. At $x_0=\pi \mathrm{/}t$ the absolute value equals $2$. Given $0<\epsilon <2$, continuity supplies $\delta >0$ such that it exceeds $2-\epsilon$ on the finite interval $J=(x_0-\delta ,x_0+\delta )$. The vector ${\mathbf{1}}_J\mathrm{/}\sqrt{2\delta }$ has norm one and its image under $U(t)-I$ has norm at least $2-\epsilon$. Letting $\epsilon \downarrow 0$ proves $\mathrm{\parallel }U(t)-I\mathrm{\parallel }=2$. At $t=0$ the norm is zero.

To prove the general assertion, let $S$ be a semigroup on a Banach space $X$ as defined in the statement, with $\mathrm{\parallel }S(h)-I\mathrm{\parallel }\to 0$ as $h\downarrow 0$. If $X=\{0\}$ its generator is the zero operator on all of $X$. Otherwise choose $a>0$ so that $\mathrm{\parallel }S(h)\mathrm{\parallel }\le 2$ for $0\le h\le a$. For $0\le s\le T$, write $s=ka+r$ with $0\le r<a$ and $k\le \lfloor T\mathrm{/}a\rfloor$; the semigroup law gives $\mathrm{\parallel }S(s)\mathrm{\parallel }\le M_T:=2^{\lfloor T\mathrm{/}a\rfloor +1}$. Consequently, for $0\le s\le t\le T$, $\mathrm{\parallel }S(t)-S(s)\mathrm{\parallel }\le M_T\mathrm{\parallel }S(t-s)-I\mathrm{\parallel }$. Thus $S$ is uniformly continuous in operator norm on every compact time interval.

For each $f\in H$, $\mathrm{\parallel }U(t)f-f{\mathrm{\parallel }}^2=\int \mathrm{\mid }{e}^{itx}-1{\mathrm{\mid }}^2\mathrm{\mid }f(x){\mathrm{\mid }}^{2}dx\to 0$ by dominated convergence, with majorant $4\mathrm{\mid }f{\mathrm{\mid }}^2$. This applies along every sequence $t\to 0$, hence gives continuity at zero. The group law and isometry give $\mathrm{\parallel }U(t)f-U(t_0)f\mathrm{\parallel }=\mathrm{\parallel }U(t-t_0)f-f\mathrm{\parallel }$, proving continuity at every $t_0$.

On a compact interval define the operator integral of $S$ by tagged Riemann sums. To justify existence, uniform continuity in step 1.4 bounds the difference between a sum and any refinement by the interval length times the modulus of continuity at the original mesh. Comparing two sums through their common refinement proves the Cauchy property as both meshes tend to zero. Completeness in [A5] gives the limit, independent of tags and partitions. The triangle inequality for sums gives $\mathrm{\parallel }{\int }_c^{d}F(s)ds\mathrm{\parallel }\le (d-c){\sup }_{[c,d]}\mathrm{\parallel }F(s)\mathrm{\parallel }$ for the continuous integrands used here. Linearity, subdivision, translation of intervals, and interchange with a fixed bounded operator follow first for sums and then for their limits. Put $B_{\tau }={\int }_0^{\tau }S(s)ds$. Then $\mathrm{\parallel }{\tau }^{-1}{B}_{\tau }-I\mathrm{\parallel }\le {\sup }_{0\le s\le \tau }\mathrm{\parallel }S(s)-I\mathrm{\parallel }<1$ for sufficiently small $\tau >0$.

For such $\tau$, let $C=I-{\tau }^{-1}{B}_{\tau }$ and $q=\mathrm{\parallel }C\mathrm{\parallel }<1$. The series $R={\sum }_{n=0}^{\mathrm{\infty }}{C}^n$ converges in operator norm: its tails are bounded by the geometric tails $\sum q^n$, and [A5] gives completeness. Telescoping finite sums and the product bound in [A4] give $(I-C)R=R(I-C)=I$. Thus $B_{\tau }$ is invertible with bounded inverse ${\tau }^{-1}R$.

For $h>0$, the semigroup law and the integral identities from step 2.2 yield $\frac{S(h)-I}{h}{B}_{\tau }=\frac{1}{h}({\int }_{\tau }^{\tau +h}S(s)ds-{\int }_0^{h}S(s)ds)\to S(\tau )-I$ in operator norm, by continuity at $\tau$ and at zero and the integral bound. Hence $B_{\tau }X\subseteq D(G)$ and $G{B}_{\tau }=S(\tau )-I$. Since $B_{\tau }$ is onto, $D(G)=X$ and $G=(S(\tau )-I)B_{\tau }^{-1}$ is bounded. There is no additional factor $1\mathrm{/}\tau$ in this last formula with the unnormalized integral $B_{\tau }$.

Restrict $U$ to nonnegative times. Its right generator extends the two-sided generator $iQ$: on $D(Q)$ the two-sided limit from step 1.2 in particular gives the right limit. If $U$ were norm continuous at zero, step 4.1 would make this right generator bounded on all of $H$, contradicting the unit vectors $f_n$ in step 1.2. This corroborates the direct computation in step 1.3 and completes all claims.