Functional calculus for elliptic operators on noncommutative tori, I

Abstract

In this paper, we introduce a parametric pseudodifferential calculus on noncommutative n-tori which is a natural nest for resolvents of elliptic pseudodifferential operators. Unlike in some previous approaches to parametric pseudodifferential calculi, our parametric pseudodifferential calculus contains resolvents of elliptic pseudodifferential operators that need not be differential operators. As an application we show that complex powers of positive elliptic pseudodifferential operators on noncommutative n-tori are pseudodifferential operators. This confirms a claim of Fathi-Ghorbanpour-Khalkhali.

Appendices

Appendix A: Improper integrals with values in locally convex spaces

In this appendix we gather a few facts about convergence of improper integrals with values in some locally convex space.

In what follows we let $\mathscr {E}$ be a locally convex space. Given any continuous map $f:[a,b]\rightarrow \mathscr {E}$ its Riemann integral $\int _a^b f(\lambda )d\lambda$ is defined in the same way as the Riemann integral of scalar-valued functions (see, e.g., [12, 26]; see also [23, Appendix B]).

Given a continuous map $f:[a,\infty ) \rightarrow \mathscr {E}$ we say that the integral $\int _a^\infty f(\lambda ) d\lambda$converges in$\mathscr {E}$ when $\lim _{b\rightarrow \infty }\int _a^bf(\lambda )d\lambda$ exists. In this case we set

$$\begin{aligned} \int _a^\infty f(\lambda ) d\lambda = \lim _{b\rightarrow \infty }\int _a^bf(\lambda )d\lambda . \end{aligned}$$

Proposition A.1

Suppose that $T:\mathscr {E}\rightarrow \mathscr {F}$ is a continuous linear map with values in some locally convex space $\mathscr {F}$. Let $f:[a,\infty ) \rightarrow \mathscr {E}$ be a continuous map whose integral $\int _a^\infty f(\lambda )d\lambda$ converges in $\mathscr {E}$. Then the integral $\int _a^\infty T[f(\lambda )]d\lambda$ converges in $\mathscr {F}$, and we have

$$\begin{aligned} \int _a^\infty T\big [f(\lambda )\big ] d\lambda = T \Big ( \int _a^\infty f(\lambda )d\lambda \Big ). \end{aligned}$$

Proof

The composition $T[f(\lambda )]$ is a continuous map from $[a,\infty )$ to $\mathscr {F}$. Moreover, we have $\int _a^b T[f(\lambda )]d\lambda = T(\int _a^b f(\lambda )d\lambda )$ for $0<a<b$ (see, e.g., [23, Proposition B.5]). Thus, in $\mathscr {F}$ we have

$$\begin{aligned} \lim _{b\rightarrow \infty } \int _a^b T\big [f(\lambda )\big ] d\lambda = \lim _{b\rightarrow \infty } T \Big ( \int _a^b f(\lambda )d\lambda \Big ) = T \Big ( \int _a^\infty f(\lambda )d\lambda \Big ). \end{aligned}$$

This proves the result. $\square$

Proposition A.2

In the special case $\mathcal {E}=\mathbb {C}$ the above results shows that, for every $\varphi \in \mathscr {E}'$, the integral $\int _a^\infty \varphi [f(\lambda )] d\lambda$ is convergent, and we have

$$\begin{aligned} \int _a^\infty \varphi \big [f(\lambda )\big ] d\lambda = \varphi \Big ( \int _a^\infty f(\lambda )d\lambda \Big ). \end{aligned}$$

(A.1)

Note that this property uniquely determines the value of $\int _a^\infty f(\lambda )d\lambda$.

Proposition A.3

Let $f:[a,\infty )\rightarrow \mathscr {E}$ be a continuous map such that $\int _a^\infty f(\lambda )d\lambda$ converges in $\mathscr {E}$. Let $\phi :[a',\infty )\rightarrow [a, \infty )$ be a $C^1$-diffeomorphism. Then $\int _{a'}^\infty \phi '(\lambda ) (f\circ \phi )(\lambda ) d\lambda$ converges in $\mathscr {E}$, and we have

$$\begin{aligned} \int _{a'}^\infty \phi '(\lambda ) (f\circ \phi )(\lambda ) d\lambda = \int _a^\infty f(\lambda )d\lambda . \end{aligned}$$

Proof

The assumption that $\phi$ is a diffeomorphism ensures us that $\phi$ is increasing, $\phi (a')= a$, and $\lim _{\lambda \rightarrow \infty }\phi (\lambda )=\infty$. Therefore, by using [23, Proposition B.7] we get

$$\begin{aligned} \int _{a'}^b \phi '(\lambda ) (f\circ \phi )(\lambda ) d\lambda = \int _{a}^{\phi (b)} f(\lambda )d\lambda \longrightarrow \int _{a}^{\infty }f(\lambda )d\lambda \qquad \text {as }b \rightarrow \infty . \end{aligned}$$

This proves the result. $\square$

It is convenient to interpret the above notion of convergence in term of Lebesgue integrations of maps with values in locally convex spaces. A natural setting for this notion of integration is the setting of quasi-complete locally convex Suslin spaces (see [46]; see also [23, Appendix B]). This encompasses a large class of locally convex spaces, including separable Fréchet spaces such as $\mathscr {A}_\theta$. However, in the precise setting of this paper, we can bypass this as follows.

In what follows, given any Borel set $Y\subset \mathbb {C}$, we say that a measurable map $f:Y\rightarrow \mathscr {E}$ is absolutely integrable when

$$\begin{aligned} \int _Y {\mathfrak {p}}\big [f(\lambda )\big ] d\lambda <\infty \qquad \text {for every continuous semi-norm }{\mathfrak {p}}\text { on }\mathscr {E}. \end{aligned}$$

(7.2)

Recall that a locally convex space is quasi-complete when every bounded Cauchy net is convergent. For instance, Fréchet spaces and their weak duals, as well as inductive limits of Fréchet spaces are quasi-complete (see, e.g., [47]).

Proposition A.4

Assume that $\mathscr {E}$ is quasi-complete. Then, for every absolutely integrable continuous map $f:[a,\infty )\rightarrow \mathscr {E}$, the integral $\int _a^\infty f(\lambda ) d\lambda$ converges in $\mathscr {E}$.

Proof

Let $f:[a,\infty )\rightarrow \mathscr {E}$ be an absolutely integrable continuous map. Let $\mathfrak {p}$ be a continuous semi-norm on $\mathscr {E}$. Given any $b\geqslant a$ we have

$$\begin{aligned} \mathfrak {p}\Big [ \int _a^{b} f(\lambda )d\lambda \Big ] \leqslant \int _a^{b}\mathfrak {p}\big [ f(\lambda ) \big ]d\lambda \leqslant \int _a^{\infty }\mathfrak {p}\big [ f(\lambda ) \big ]d\lambda <\infty . \end{aligned}$$

Similarly, given any $c\geqslant 0$, we have

$$\begin{aligned}&\mathfrak {p}\Big [ \int _a^{b+c} f(\lambda )d\lambda - \int _a^{b} f(\lambda )d\lambda \Big ] = \mathfrak {p}\Big [ \int _b^{b+c} f(\lambda )d\lambda \Big ] \\&\quad \leqslant \int _b^{b+c}\mathfrak {p}\big [ f(\lambda ) \big ]d\lambda \leqslant \int _b^{\infty }\mathfrak {p}\big [ f(\lambda ) \big ]d\lambda . \end{aligned}$$

The absolute-integrability condition (7.2) ensures us that $\int _b^{\infty }\mathfrak {p}\big [ f(\lambda ) \big ]d\lambda \rightarrow 0$ as $b\rightarrow \infty$. Therefore, we see that, for every continuous semi-norm $\mathfrak {p}$ on $\mathscr {E}$, we have

$$\begin{aligned} \sup _{c\geqslant 0}\mathfrak {p}\Big [ \int _a^{b+c} f(\lambda )d\lambda - \int _a^{b} f(\lambda )d\lambda \Big ]\longrightarrow 0 \qquad \text {as } b\rightarrow \infty . \end{aligned}$$

This shows that $\{ \int _a^{b} f(\lambda )d\lambda ; b\geqslant a\}$ is a bounded Cauchy net in $\mathscr {E}$. As $\mathcal {E}$ is quasi-complete, it then follows that $\lim _{b\rightarrow \infty } \int _a^{b} f(\lambda )d\lambda$ exists in $\mathscr {E}$. That is, the integral $\int _a^\infty f(\lambda ) d\lambda$ converges in $\mathscr {E}$. The proof is complete. $\square$

As an immediate consequence we have the following dominated convergence result.

Proposition A.5

Suppose that $\mathscr {E}$ is quasi-complete. Let $f:[a,\infty ) \rightarrow \mathscr {E}$ be a continuous map such that, for every continuous semi-norm $\mathfrak {p}$ on $\mathscr {E}$, there is a function $g_\mathfrak {p}(\lambda ) \in L^1[a,\infty )$ such that

$$\begin{aligned} \mathfrak {p}\big [ f(\lambda )\big ] \leqslant g_\mathfrak {p}(\lambda ) \qquad \forall \lambda \geqslant a. \end{aligned}$$

Then $f(\lambda )$ is absolutely integrable, and the integral $\int _a^\infty f(\lambda ) d\lambda$ converges in $\mathscr {E}$.

The above considerations extend verbatim to integrals along rays $L_\phi (a)=e^{i\phi }[a,\infty )$ with $a\geqslant 0$ and $0\leqslant \phi < 2\pi$. If $f: L_\phi (a)\rightarrow \mathscr {E}$ is a continuous map, then the integral $\int _{L_\phi (a)} f(\lambda )d\lambda$ converges in $\mathscr {E}$ when $\lim _{b\rightarrow \infty } \int _a^b f(e^{i\phi }\lambda ) d(e^{i\phi }\lambda )$ exists in $\mathscr {E}$. In this case, and if $L_\phi (a)$ is outward-oriented (i.e., it is directed toward $\infty$), then we set

$$\begin{aligned} \int _{L_\phi (a)} f(\lambda )d\lambda = \lim _{b\rightarrow \infty } \int _a^b f(e^{i\phi }\lambda ) d(e^{i\phi }\lambda ). \end{aligned}$$

There is a change of sign when $L_\phi (a)$ is inward-oriented.

More generally, we can consider keyhole contours, by which we mean contours of the form,

$$\begin{aligned} \Gamma = L_{\phi _1}(a) \cup C_{\phi _1,\phi _2}(a) \cup L_{\phi _2}(a), \qquad a\geqslant 0, \quad \phi _1>\phi _2 \geqslant \phi _1-2\pi , \end{aligned}$$

where $C_{\phi _1,\phi _2}(a)$ is the arc of circle $\{ae^{i\phi }; \ \phi _1\geqslant \phi \geqslant \phi _2\}$. The orientation of $\Gamma$ is chosen so that it agrees with the clockwise-orientation on $C_{\phi _1,\phi _2}(a)$. Thus, $L_{\phi _1}(a)$ is inward-oriented, whereas $L_{\phi _2}(a)$ is outward-oriented. For instance when $\phi _1=\pi$, $\phi _2=0$ and $a=0$ the contour $\Gamma$ is just the real line $(-\infty ,\infty )$.

Fig. 4

Examples of Keyhole Contours

Given a continuous map $f:\Gamma \rightarrow \mathscr {E}$ we say that the contour integral $\int _\Gamma f(\lambda ) d\lambda$ converges in $\mathscr {E}$ when the integrals $\int _{L_{\phi _1}(a)} f(\lambda )d\lambda$ and $\int _{L_{\phi _2}(a)} f(\lambda )d\lambda$ both converge in $\mathscr {E}$. In this case we set

$$\begin{aligned} \int _\Gamma f(\lambda )d\lambda = \int _{L_{\phi _1}(a)} f(\lambda )d\lambda + \int _{C_{\phi _1,\phi _2}(a)} f(\lambda )d\lambda + \int _{L_{\phi _2}(a)} f(\lambda )d\lambda , \end{aligned}$$

where the middle integral in the r.h.s. is defined as a Riemann integral.

At the exception of Proposition A.3 all the previous results of this section hold verbatim for this kind of improper integrals. We also have the following version of Proposition A.3.

Proposition A.6

Suppose that $\Gamma$ is a keyhole contour, and let $f:\Gamma \rightarrow \mathscr {E}$ be a continuous map whose integral $\int _\Gamma f(\lambda )d\lambda$ converges in $\mathcal {E}$. Then, for every $t>0$, the integral $\int _{t\Gamma } f(t^{-1}\lambda )d\lambda$ converges in $\mathscr {E}$, and we have

$$\begin{aligned} \int _{t\Gamma } f(t^{-1}\lambda )d\lambda = t\int _{\Gamma } f(\lambda )d\lambda . \end{aligned}$$

We are mostly interested in the case where $f(\lambda )$ is an ${\text {Hol}}^d(\Lambda )$-map, where $\Lambda$ is some open pseudo-cone containing the contour $\Gamma$. Note that keyhole contours themselves are pseudo-cones.

Proposition A.7

Assume that $\mathscr {E}$ is quasi-complete. Let $f(\lambda )\in {\text {Hol}}^d(\Lambda ;\mathscr {E})$, $d<-1$, where $\Lambda$ is some open pseudo-cone. In addition, let $\Gamma$ be a keyhole contour contained in $\Lambda$.

• (1) The contour integral $\int _\Gamma f(\lambda ) d\lambda$ converges in $\mathscr {E}$.

• (2) Let $T:\mathscr {E}\rightarrow \mathscr {F}$ be a continuous linear map with values in some other locally convex space $\mathscr {F}$. Then, the integral $\int _\Gamma T[f(\lambda )] d\lambda$ converges in $\mathscr {F}$, and we have

  $$\begin{aligned} \int _\Gamma T\big [f(\lambda )\big ] d\lambda = T \Big ( \int _\Gamma f(\lambda )d\lambda \Big ). \end{aligned}$$

• (3) Let $\Gamma '\subset \Lambda$ be either a keyhole contour or a clockwise-oriented Jordan curve, and assume there is a domain $\Omega \subset \Lambda$ such that $\partial \Omega = \Gamma \cup \Gamma '$. Then, we have

  $$\begin{aligned} \int _{\Gamma } f(\lambda )d\lambda = \int _{\Gamma '} f(\lambda )d\lambda . \end{aligned}$$

Proof

As $\Gamma$ is a pseudo-cone $\subset \!\subset \Lambda$, for every continuous semi-norm $\mathfrak {p}$ on $\mathscr {E}$, there is $C_{\Gamma \mathfrak {p}}>0$ such that

$$\begin{aligned} \mathfrak {p}\big [ f(\lambda )\big ] \leqslant \left( 1+|\lambda |\right) ^d \qquad \forall \lambda \in \Gamma . \end{aligned}$$

As $d<-1$ and $\mathscr {E}$ is quasi-complete, it follows from Proposition A.5 that $f(\lambda )$ is absolutely integrable and the integral $\int _\Gamma f(\lambda )d\lambda$ converges in $\mathscr {E}$. This proves the first part. Combining it with Proposition A.1 gives the 2nd part.

It remains to prove the 3rd part. Let $\Gamma '\subset \Lambda$ be either a keyhole contour or a clockwise-oriented Jordan curve, and assume there is a domain $\Omega \subset \Lambda$ such that $\partial \Omega = \Gamma \cup \Gamma '$. In addition, let $\varphi \in \mathscr {E}'$. The composition $\varphi \circ f(\lambda )$ is contained in ${\text {Hol}}^d(\Lambda )$, and so this is an integrable holomorphic function on $\Omega$. Therefore, if we orient $\partial \Omega$ suitably, then we get

$$\begin{aligned} 0 = \int _{\partial \Omega } \varphi \left[ f(\lambda )\right] d\lambda = \int _{\Gamma }\varphi \left[ f(\lambda )\right] d\lambda - \int _{\Gamma '} \varphi \left[ f(\lambda )\right] d\lambda . \end{aligned}$$

Combining this with (A.1) we obtain

$$\begin{aligned} \varphi \Big ( \int _{\Gamma } f(\lambda )d\lambda \Big ) = \varphi \Big ( \int _{\Gamma '} f(\lambda )d\lambda \Big ) \qquad \forall \varphi \in \mathscr {E}'. \end{aligned}$$

As $\mathcal {E}'$ separates the point of $\mathscr {E}$ by Hahn-Banach theorem, we see that $\int _{\Gamma } f(\lambda )d\lambda = \int _{\Gamma '} f(\lambda )d\lambda$. This proves the 3rd part. The proof is complete. $\square$

Appendix B: Proof of Proposition 5.29

In this Appendix, we include a proof of Proposition 5.29. Recall that the topology of $\mathscr {L}(\mathscr {A}_\theta ',\mathscr {A}_\theta )$ is generated by the semi-norms,

$$\begin{aligned} R\longrightarrow \sup _{u\in \mathscr {B}}\Vert \delta ^\alpha (Ru) \Vert , \qquad \mathscr {B}\subset \mathscr {A}_\theta '\text { bounded} , \quad \alpha \in \mathbb {N}_0^n. \end{aligned}$$

We have the following description of bounded sets in $\mathscr {A}_\theta '$.

Lemma B.1

A subset $\mathscr {B}\subset \mathscr {A}_\theta '$ is bounded in $\mathscr {A}_\theta '$ if and only if there $s\in {\mathbb {R}}$ such that $\mathscr {B}$ is contained and bounded in $\mathscr {H}_\theta ^{(s)}$.

Proof

This is a special case of a general result for inductive limits of compact inclusions of Banach spaces, but we shall give a proof for reader’s convenience. By Proposition 3.27 the strong topology of $\mathscr {A}_\theta '$ agrees with the inductive limit of the $\mathscr {H}_\theta ^{(s)}$-topologies. Equivalently, this is the strongest locally convex topology with respect to which the inclusion of $\mathscr {H}_\theta ^{(s)}$ into $\mathscr {A}_\theta '$ is continuous for every $s\in {\mathbb {R}}$. Thus, a basis of neighborhoods of the origin in $\mathscr {A}_\theta '$ consists of all convex balanced sets $\mathscr {U}$ such that $\mathscr {U}\cap \mathscr {H}_\theta ^{(s)}$ is a neighborhood of the origin in $\mathscr {H}_\theta ^{(s)}$ for every $s\in {\mathbb {R}}$.

Bearing this in mind, a subset $\mathscr {B}\subset \mathscr {A}_\theta '$ is bounded when, given any neigborhood $\mathscr {U}$ of the origin in $\mathscr {A}_\theta '$, there is $t>0$ such that $t\mathscr {B}\subset \mathscr {U}$. As the inclusion of $\mathscr {H}_\theta ^{(s)}$, $s\in {\mathbb {R}}$, into $\mathscr {A}_\theta '$ is continuous, it is immediate that any bounded set of $\mathscr {H}^{(s)}_\theta$ is bounded in $\mathscr {A}_\theta '$.

Conversely, let $\mathscr {B}$ be a bounded set of $\mathscr {A}_\theta '$. With a view toward contradiction assume that $\mathscr {B}$ is not a bounded set in any $\mathscr {H}_\theta ^{(s)}$. Given $s\in {\mathbb {R}}$ and $r>0$ let us denote by $B^{(s)}(r)$ the open ball in $\mathscr {H}_\theta ^{(s)}$ of radius r about the origin. The above assumption means that, for every $s\in {\mathbb {R}}$, we cannot find $r>0$ such that $\mathscr {B}\subset B^{(s)}(r)$. In particular, there is $u^{(0)}\in \mathscr {B}$ such that $u^{(0)}\not \in B^{(0)}(1)$, i.e., $\sum |u_k^{(0)}|^2>1$, and so there is an integer $N_0$ such that $\sum _{|k|\leqslant N_0} |u_k^{(0)}|^2>1$. Likewise, there is $u^{(1)}\in \mathscr {B}$ such that $\sum \langle k\rangle ^{-2} |u_k^{(1)}|^2>2 \sum _{|k|\leqslant N_0} |u_k^{(0)}|^2$, and so there is also an integer $N_1$ such that $\sum _{|k|\leqslant N_1} \langle k\rangle ^{-2}|u_k^{(1)}|^2> 2 \sum _{|k|\leqslant N_0} |u_k^{(0)}|^2$. Repeating this argument allows us to construct sequences $(N_\ell )_{\ell \geqslant 0}\subset \mathbb {N}_0$ and $(u^{(\ell )})_{\ell \geqslant 0}\subset \mathscr {B}$ such that

$$\begin{aligned} \sum _{|k|\leqslant N_{\ell +1}} \langle k\rangle ^{-2(\ell +1)}\big |u_k^{(\ell +1)}\big |^2> 2 \sum _{|k|\leqslant N_{\ell }} \langle k\rangle ^{-2\ell }\big |u_k^{(\ell )}\big |^2 \qquad \text {for all }\ell \geqslant 0. \end{aligned}$$

(B.1)

For $\ell \geqslant 0$ set $\mu _\ell = 2^{-\ell } \sum _{|k|\leqslant N_{\ell }} \langle k\rangle ^{-2\ell }|u_k^{(\ell )}|^2$, and define

$$\begin{aligned} \mathscr {U}= \bigcap _{\ell \geqslant 0} \biggl \{u \in \mathscr {A}_\theta '; \ \sum _{|k|\leqslant N_{\ell }} \langle k\rangle ^{-2\ell }|u_{k}|^2<\mu _\ell \biggr \}. \end{aligned}$$

Note that (B.1) ensures us that $(\mu _\ell )_{\ell \geqslant 0}$ is an increasing sequence.

Claim

$\mathscr {U}$ is a neighborhood of the origin in $\mathscr {A}_\theta '$.

Proof of the Claim

For $\ell \geqslant 0$ set $p_\ell (u)= (\sum _{|k|\leqslant N_{\ell }} \langle k\rangle ^{-2\ell }|u_{k}|^2)^{\frac{1}{2}}$, $u\in \mathscr {A}_\theta '$. This defines semi-norms on $\mathscr {A}_\theta '$. In particular, we see that $\mathscr {U}$ is a convex balanced set. Therefore, in order to prove the claim it is enough to show that $\mathscr {U}\cap \mathscr {H}_\theta ^{(s)}$ is a neighborhood of the origin in $\mathscr {H}_\theta ^{(s)}$ for every $s\in {\mathbb {R}}$. In fact, as the inclusion of $\mathscr {H}_\theta ^{(s)}$ into $\mathscr {H}_\theta ^{(s')}$ is continuous for $s>s'$, it is enough to show this for $s=-\ell _0$ with $\ell _0\in \mathbb {N}$.

Given $\ell _0\in \mathbb {N}$, let $\epsilon \in (0,\sqrt{\mu _{\ell _0}})$. We observe that if $u\in B^{(-\ell _0)}(\epsilon )$ and $\ell \geqslant \ell _0$, then

$$\begin{aligned} \sum _{|k|\leqslant N_\ell } \langle k\rangle ^{-2\ell } |u_k|^2 \leqslant \sum _{k\in \mathbb {Z}^n} \langle k\rangle ^{-2\ell _0} |u_k|^2<\epsilon ^2<\mu _{\ell _0}\leqslant \mu _\ell . \end{aligned}$$

This shows that $B^{(-\ell _0)}(\epsilon )\subset \{u\in \mathscr {A}_\theta '; \ p_\ell (u)<\sqrt{\mu _\ell }\}$ for $\ell \geqslant \ell _0$. Thus,

$$\begin{aligned} \mathscr {U}\cap B^{(-\ell _0)}(\epsilon ) = \bigcap _{\ell< \ell _0}\left\{ u \in B^{(-\ell _0)}(\epsilon ); \ p_\ell (u)<\sqrt{\mu _\ell }\right\} . \end{aligned}$$

As the semi-norms $p_\ell$ are continuous on $\mathscr {H}_\theta ^{(-\ell _0)}$, it then follows that $\mathscr {U}\cap B^{(-\ell _0)}(\epsilon )$ is an open set of $\mathscr {H}_\theta ^{(-\ell _0)}$, and so $\mathscr {U}\cap \mathscr {H}_\theta ^{(-\ell _0)}$ is a neighborhood of the origin in $\mathscr {H}_\theta ^{(-\ell _0)}$ for every $\ell _0\in \mathbb {N}$. The proof is complete. $\square$

The above claim leads us to a contradiction as follows. As $\mathscr {B}$ is bounded in $\mathscr {A}_\theta '$ and $\mathscr {U}$ is a neighborhood of the origin in $\mathscr {A}_\theta '$, there is $t>0$ such that $t \mathscr {B}\subset \mathscr {U}$. In particular, $tu^{(\ell )}\in \mathscr {U}$ for all $\ell \geqslant 0$. Now, choose $\ell$ so that $2^{-\ell }\leqslant t^2$. Then we have

$$\begin{aligned} \sum _{|k|\leqslant N_\ell }\langle k\rangle ^{-2\ell } t^2|u_k^{(\ell )}|^2 \geqslant \sum _{|k|\leqslant N_\ell } 2^{-\ell } \langle k\rangle ^{-2\ell } |u_k^{(\ell )}|^2 =\mu _\ell . \end{aligned}$$

Thus, $tu^{(\ell )}$ cannot be contained in $\mathscr {U}$. This is a contradiction. Therefore, if $\mathscr {B}$ is a bounded set of $\mathscr {A}_\theta '$, then it must be a bounded set of $\mathscr {H}_\theta ^{(s)}$ for some $s\in {\mathbb {R}}$. This proves the result. $\square$

We are now in a position to prove Proposition 5.29.

Proof of Proposition 5.29

Given $s,t\in {\mathbb {R}}$, the continuity of the inclusions $\mathscr {H}_\theta ^{(s)}\subset \mathscr {A}_\theta '$ and $\mathscr {A}_\theta \subset \mathscr {H}_\theta ^{(t)}$ implies that the natural embedding of $\mathscr {L}(\mathscr {A}_\theta ',\mathscr {A}_\theta )$ into $\mathscr {L}(\mathscr {H}_\theta ^{(s)},\mathscr {H}_\theta ^{(t)})$ is continuous, and so $\Vert \cdot \Vert _{s,t}$ is a continuous semi-norm on $\mathscr {L}(\mathscr {A}_\theta ',\mathscr {A}_\theta )$.

Moreover, we know by Proposition 3.27 that the topology of $\mathscr {A}_\theta$ is generated by the Sobolev norms $\Vert \cdot \Vert _{t}$, $t\in {\mathbb {R}}$. Therefore, the topology of $\mathscr {L}(\mathscr {A}_\theta ',\mathscr {A}_\theta )$ is generated by the semi-norms,

$$\begin{aligned} R\longrightarrow \sup _{u\in \mathscr {B}}\Vert Ru \Vert _t , \qquad \mathscr {B}\subset \mathscr {A}_\theta '\ \text {bounded} , \quad t\in {\mathbb {R}}. \end{aligned}$$

Bearing this in mind, let $t\in {\mathbb {R}}$ and let $\mathscr {B}$ be a bounded subset of $\mathscr {A}_\theta '$. By Lemma B.1 there is $s\in {\mathbb {R}}$ such that $\mathscr {B}$ is a bounded subset of $\mathscr {H}_\theta ^{(s)}$, i.e., there is $\epsilon >0$ such that $\mathscr {B}\subset \{u\in \mathscr {H}_\theta ^{(s)};\ \Vert u\Vert _s \leqslant \epsilon \}$. Thus, for all $R\in \mathscr {L}(\mathscr {A}_\theta ',\mathscr {A}_\theta )$, we have

$$\begin{aligned} \sup _{u \in \mathscr {B}} \Vert Ru\Vert _t \leqslant \sup _{\begin{array}{c} u \in \mathscr {H}_\theta ^{(s)}\\ \Vert u\Vert _s\leqslant \epsilon \end{array}} \Vert Ru\Vert _{t} \leqslant \epsilon \sup _{\begin{array}{c} u \in \mathscr {H}_\theta ^{(s)}\\ \Vert u\Vert _s\leqslant 1 \end{array}} \Vert Ru\Vert _{t}= \epsilon \Vert R\Vert _{s,t}. \end{aligned}$$

All this ensures us that the norms $\Vert \cdot \Vert _{s,t}$, $s,t\in {\mathbb {R}}$, generate the topology of $\mathscr {L}(\mathscr {A}_\theta ',\mathscr {A}_\theta )$. The proof is complete. $\square$