2 The construction

This section presents the basic construction of finite-dimensional approximations that we will use and modify in later sections. Here we study a separable Hilbert space with a (scaled) unitary operator having a cyclic vector. We use the cyclic vector and polynomials to build finite-dimensional approximations such that the original Hilbert space embeds isometrically into their metric ultraproduct. We also get a vector space homomorphism from the space of polynomials (over a compact set $S\subset \mathbb {C}$ ) into the metric ultraproduct. In Section 3 we will construct a suitable measure to make the homomorphism an isometry.

The construction will be used to find similar approximations for the case where the operator is bounded and self-adjoint in Section 4. In Section 5 we will study other norms for the ultraproduct space that will be essential in studying distributions in Sections 5 and 6. In Section 7 we will have a short look at what one can do if the operator doesn’t have a cyclic vector.

So we let H be a separable complex Hilbert space and A an operator on H of the form $qU$ , where q is a nonzero complex number and U is unitary. Then A is normal and bounded and $A^{*}\circ A=A\circ A^{*}=r_{A}I$ where $r_{A}$ is a positive real, I is the identity operator, and $A^{*}$ is the adjoint of A. This is enough for the constructions we do in this paper (we will use the construction when A is unitary), but we use notation allowing for more general operators, as there are also other operators allowing for this sort of approximations (see the example at the end of this section).

From a polynomial $P(X,Y)\in \mathbb {C} [X,Y]$ over the complex numbers $\mathbb {C}$ we get an operator $P(A,A^{*})$ on H the natural way, e.g., $X^{2}Y(A,A^{*})=A\circ A\circ A^{*}$ , and we say that $\phi \in H$ is cyclic if the set $\{ P(A,A^{*})(\phi )\vert \ P\in \mathbb {C} [X,Y]\}$ is dense in H. We will here study the case where there is a cyclic vector $\phi $ in H and we pick $\phi $ so that in addition its norm is $1$ . Notice that H can always be split into countably many complete subspaces so that they are orthogonal to each other, closed under A and $A^{*}$ and each of them has a cyclic vector, and in Section 7 we will see how to combine the constructions from this decomposition.

For all $N\in {\omega }$ , let $H_{N}$ be the subspace of H generated by

(1) $$ \begin{align} \{A^{i}(A^{*})^{j}(\phi )\vert\ i\le N,\ j\le N\}. \end{align} $$

In particular, note that $H_0$ is the space generated by $\phi $ , and $H_N\subset H_{N+1}$ . Notice that as finite dimensional spaces these are complete spaces. We write $H^{-}_{N}$ for the subspace of H generated by

(2) $$ \begin{align} \{A^{i}(A^{*})^{j}(\phi )\vert\ i< N,\ j\le N\} \end{align} $$

and $H^{+}_{N}$ for the subspace of H generated by

(3) $$ \begin{align} \{A^{i}(A^{*})^{j}(\phi )\vert\ 0<i\le N,\ j\le N\}. \end{align} $$

As $AA^*=A^*A=r_AI$ , the operators A and $A^*$ swap these spaces: A maps $H^{-}_{N}$ onto $H^{+}_{N}$ and $A^{*}$ maps $H^{+}_{N}$ onto $H^{-}_{N}$ . To get approximating operators on $H_N$ we can modify A and $A^*$ by a wraparound trick: Let $W^{-}$ be the orthogonal complement of $H^{-}_{N}$ in $H_{N}$ and $W^{+}$ be the orthogonal complement of $H^{+}_{N}$ in $H_{N}$ . Then $W^{+}$ and $W^{-}$ have the same dimension and we let U be a unitary operator from $W^{-}$ onto $W^{+}$ . Now we define an operator $A_{N}$ on $H_{N}$ as follows. If $u\in H^{-}_{N}$ , then $A_{N}(u)=A(u)$ and if $u\in W^{-}$ , then $A_{N}(u)=\sqrt {r_{A}}U(u)$ (by $\sqrt {r_{A}}$ we mean the one that is positive). We get the adjoint of $A_{N}$ as follows: If $u\in H^{+}_{N}$ , then we let $A^{*}_{N}(u)=A^{*}(u)$ and if $u\in W^{+}$ , then $A^{*}_{N}(u)=\sqrt {r_{A}}U^{-1}(u)$ . It is easy to check that $A^{*}_{N}$ is indeed the adjoint of $A_{N}$ and that $A^{*}_{N}\circ A_{N}=A_{N}\circ A^{*}_{N}=r_{A}I$ and so $A_{N}$ is normal. Notice that because of how A and $A^*$ cancel out (up to multiplication by a constant), $H^{+}_{N}$ is also generated by the set $\{A^{i}(A^{*})^{j}(\phi )\vert \ i\le N,\ j< N\}$ . Thus $H_N^-$ and $H_N^+$ are the subspaces on which $A_N$ and $A_N^*$ respectively behave ‘correctly’ and

(4) $$ \begin{align} A_{N}(A^{i}(A^{*})^{j}(\phi))=A(A^{i}(A^{*})^{j}(\phi ))=A^{i+1}(A^{*})^{j}(\phi ) \textrm{ if } i<N,\,j\leq N \end{align} $$

and

(5) $$ \begin{align} A^{*}_{N}(A^{i}(A^{*})^{j}(\phi ))=A^{i}(A^{*})^{j+1}(\phi )\textrm{ if }j<N,\, i\leq N. \end{align} $$

In $H_{N}$ we can also define operators $P(A_{N},A_{N}^{*})$ for polynomials $P(X,Y)$ in the natural way.

For all $N\in {\omega }$ , we let $D_{N}$ be the dimension of $H_{N}$ and we choose eigenvectors $u_{N}(n)$ , $n<D_{N}$ , of $A_{N}$ with eigenvalues $\lambda _{N}(n)$ . Notice that

(6) $$ \begin{align} u_{N}(n)\ \text{is also an eigenvector of } A^{*}_N \text{ with eigenvalue } \overline{\lambda_{N}(n)}, \end{align} $$

where $\overline {\lambda _{N}(n)}$ is the complex conjugate of $\lambda _{N}(n)$ . Since $A_{N}$ is normal, we can choose these so that they form an orthonormal basis of $H_{N}$ . Denote $\xi _{N}(n)=\langle u_{N}(n) |\phi \rangle $ , i.e.,

(7) $$ \begin{align} \phi =\sum_{n=0}^{D_{N}-1}\xi_{N}(n)u_{N}(n), \end{align} $$

and we choose the vectors $u_{N}(n)$ so that in addition $\xi _{N}(n)$ is a non-negative real number. Then we have

(8) $$ \begin{align} A_N(\phi)=\sum_{n=0}^{D_{N}-1}\lambda_N(n)\xi_{N}(n)u_{N}(n)\ \text{and}\ A^*_N(\phi)=\sum_{n=0}^{D_{N}-1}\overline{\lambda_N(n)}\xi_{N}(n)u_{N}(n). \end{align} $$

One should note that even though the spaces $H_N$ form an increasing sequence, the bases do not extend each other, as the operators $A_N$ (as well as the operators $A_N^*$ ) do not extend each other (because of the wraparound modification).

Next we want to tie a spectral decomposition of A to our finite-dimensional approximation. So we choose a natural number M so that it is strictly greater than both the operator norm of A and $\sqrt {r_{A}}$ , and let

(9) $$ \begin{align} S=\{\lambda\in\mathbb{C}\vert\ -M\le Re(\lambda )\le M, -M\le Im(\lambda )\le M\}. \end{align} $$

Notice that there is a real $\varepsilon>0$ such that for all $N<{\omega }$ and $n<D_{N}$ , $\vert \lambda _{N}(n)\vert <M-\varepsilon $ since the norm of $A_{N}$ is at most the maximum of the norm of A and $\sqrt {r_{A}}$ . For all $X\subseteq \mathbb {C}$ , let $C(X)$ be the vector space of all bounded continuous functions from X to $\mathbb {C}$ . Of these our main interest is in $C(S)$ . Notice that since S is compact, every $f\in C(S)$ is uniformly continuous. We let $D(S)$ be the subspace of $C(S)$ that consists of all functions $f_{P}(\lambda )=P(\lambda ,\overline \lambda )$ , where $P\in \mathbb {C}[X,Y]$ and $\overline \lambda $ is the complex conjugate of $\lambda $ . On $D(S)$ we can again define operators $A_{D}$ and $A^{*}_{D}$ in a natural way

(10) $$ \begin{align} A_D(f_P)=f_{XP}\quad\text{and}\quad A^*_D(f_P)=f_{YP} \end{align} $$

(e.g., $A_{D}(f_{X^{i}Y^{j}})=f_{X^{i+1}Y^{j}}$ ). Notice that $A_{D}(f_{P})(\lambda )=\lambda f_{P}(\lambda )$ and $A^{*}_{D}(f_{P})(\lambda )=\overline \lambda f_{P}(\lambda )$ .

Let’s define some mappings that will be crucial in this paper:

Definition 2.1. We define $G: D(S)\to H$ and $F_N:D(S)\to H_N$ as follows: For $P\in \mathbb {C}[X,Y]$ , let

(11) $$ \begin{align} G(f_{P})=P(A,A^{*})(\phi ) \end{align} $$

and note that it is a homomorphism from $D(S)$ to H (as vector spaces), and that its image is dense in H (by the cyclicity assumption). Further let

(12) $$ \begin{align} F_{N}(f_{P})=\sum_{n<D_{N}}\xi_{N}(n)f_{P}(\lambda_{N}(n))u_{N}(n), \end{align} $$

and note by comparing to Equation (8) that $F_{N}(f_{P})=P(A_{N},A_{N}^{*})(\phi )$ , and also that this is a homomorphism from $D(S)$ to $H_{N}$ . Also note, that as $\lVert {A_N}\rVert $ is always bounded by $\max \{\lVert {A}\rVert ,\sqrt {r_A}\}$ , for each $f_P\in D(S)$ , $\{F_N(f_P):N\geq 1\}$ is bounded.

Now the inner product in $H_{N}$ (inherited from H), gives us one norm $\Vert \cdot \Vert _{2}$ on  $H_{N}$ . Using this, there is a natural way of defining metric ultraproducts of the Hilbert spaces $H_N$ . However, we want to consider also other norms in the ultraproduct, so we take as our starting point a classical ultraproduct of the structures $H_N$ and ‘dig out’ various metric ultraproduct spaces from it. We look at the $\lVert {\cdot }\rVert _2$ -norm here, and consider the other norms in Section 5.

Let ${\mathcal {U}}$ be a nonprincipal ultrafilter on ${\omega }$ . Depending on what one wishes to calculate using the finite dimensional approximations, here there is room to adjust the ultrafilter accordingly. For example, in [Reference Hirvonen and Hyttinen6] our calculations required that for all $m\in {\omega }-\{ 0\}$ , $\{ N\in {\omega } -\{ 0\}\vert \ \sqrt {N}\in {\omega } ,\ m|\sqrt {N}\}\in {\mathcal {U}}$ , and similar properties can be added according to need.

Let $H^{u}$ be the (classical) ultraproduct $\Pi _{N\in {\omega }}H_{N}/{\mathcal {U}}$ and $\mathbb {C}^{u}$ be the ultrapower $\mathbb {C}^{{\omega }}/{\mathcal {U}}$ of the field of complex numbers $\mathbb {C}$ . Then $H^{u}$ is a vector space over $\mathbb {C}^{u}$ and by identifying elements of $\mathbb {C}$ with their images under the canonical embedding of $\mathbb {C}$ into $\mathbb {C}^{u}$ , also a vector space over $\mathbb {C}$ . There is also a pairing $\langle \cdot |\cdot \rangle ^{u}$ from $H^{u}$ to $\mathbb {C}^{u}$ which is obtained as the ultraproduct of the inner products $\langle \cdot |\cdot \rangle _{N}$ of $H_{N}$ . And similarly we get a unary function $\Vert \cdot \Vert ^{u}_{2}$ . Notice that for all $v=(v_{N})_{N\in {\omega }}/{\mathcal {U}}\in H^{u}$ , $\Vert v\Vert ^{u}_{2}= (\sqrt {\langle v_{N}|v_{N}\rangle _{N}})_{N\in {\omega }}/{\mathcal {U}}$ and that $((\sqrt {\langle v_{N}|v_{N}\rangle _{N}})_{N\in {\omega }}/{\mathcal {U}})^{2}=\langle v\vert v\rangle ^{u}$ . Similarly for all $q=(q_{N})_{N\in {\omega }}/{\mathcal {U}}\in \mathbb {C}^{u}$ , we write $\vert q\vert ^{u}$ for $(\vert q_{N}\vert )_{N\in {\omega }}/{\mathcal {U}}$ . Notice that if we let X be the range of this ‘absolute value’, then $\mathbb {R}^{u}=X\cup \{-r\vert \ r\in X\}$ is a real closed field containing the reals. In particular, it is linearly ordered and thus we can compare, e.g., the ‘norm’ $\Vert v\Vert ^{u}$ of $v\in H^{u}$ and a rational number. Thus for all $u,v\in H^{u}$ , if there is $q\in \mathbb {C}$ such that q is infinitely close to $\langle u|v\rangle ^{u}$ , we write $\langle u\vert v\rangle $ for this q. If there is no such q, we write $\langle u|v\rangle =\infty $ .

Now $H^{u}$ contains some structures of interest to us. Let us first look at $H^{\infty }$ : On $H^{u}$ we define an equivalence relation $\sim $ so that $u\sim v$ if $\Vert u-v\Vert ^{u}_{2}$ is infinitesimal (i.e., smaller than $1/n$ for all natural numbers n) and let $H^{\infty }=H^{u}/\sim $ . Now addition of elements of $H^{\infty }$ and for all $q\in \mathbb {C}$ , the multiplication by q are well-defined in $H^{\infty }$ and thus $H^{\infty }$ is a vector space over $\mathbb {C}$ . We define a partial pairing $\langle \cdot |\cdot \rangle $ on $H^{\infty }$ as follows: For all $u,v\in H^{\infty }$ , if there is $q\in \mathbb {C}$ such that for all $u'\in u/\sim $ and $v'\in v/\sim $ , $\langle u'|v'\rangle ^{u}=q$ , then we let $\langle u|v\rangle =q$ and otherwise $\langle u|v\rangle $ is $\infty $ .

The second of these structures is the metric ultraproduct $H^{m}$ : We let $H^{m}$ be the set of all $v/\sim \in H^{\infty }$ such that $\Vert v\Vert ^{u}_{2}$ is finite (i.e., smaller than some natural number). Notice that $H^{m}$ is still a vector space over $\mathbb {C}$ and $\langle \cdot |\cdot \rangle $ is total on $H^{m}$ and, in fact, an inner product. This also gives a norm on $H^{m}$ but it is the same as $\Vert \cdot \Vert _{2}$ defined above. Later, when considering spaces arising from other norms, we may write $H^{m}_{2}$ for $H^{m}$ .

Let $A^{u}$ be the ultraproduct of the operators $A_{N}$ . Since the operators $A_{N}$ are uniformly bounded, $A^{u}$ induces a well-defined operator $A^{\infty }$ on $H^{\infty }$ and the restriction $A^{m}$ of $A^{\infty }$ to $H^{m}$ is a well-defined operator on $H^{m}$ . We can do the same for the $A^{*}_{N}$ ’s and get operators $A^{*u}$ , $A^{*\infty }$ and $A^{*m}$ . Notice that $(A^{m})^{*}=A^{*m}$ since being an adjoint is preserved in metric ultraproducts and that $A^{m}$ and $A^{*m}$ commute since $A_{N}$ and $A^{*}_{N}$ commute for all N and thus $A^{m}$ is normal.

Now the homomorphisms $F_N$ (from Definition 2.1) induce a homomorphism from the vector space $D(S)$ into $H^m$ :

Definition 2.4. Let $F^m: D(S)\to H^m$ be given by

$$ \begin{align*}F^m(f_{P})=(F_{N}(f_{P}))_{N<{\omega}}/{\mathcal{U}}/\sim .\end{align*} $$

This is well-defined, as $\{F_N(f_P):N<{\omega }\}$ is bounded (see Definition 2.1). By $F^u$ we will denote the homomorphism from $D(S)$ to $H^u$ :

$$ \begin{align*}F^u(f_{P})=(F_{N}(f_{P}))_{N<{\omega}}/{\mathcal{U}}.\end{align*} $$

Notice that $F^m(A_{D}(f_{P}))=A^{m}(F^m(f_{P}))$ and similarly for $A^{*}_{D}$ and $A^{*m}$ . We can also embed H into $H^m$ :

Definition 2.5. Let $G^m: H\to H^m$ be given by

$$ \begin{align*}G^{m}(P(A,A^{*})(\phi ))=(P(A,A^{*})(\phi ))_{N<{\omega}}/{\mathcal{U}}/\sim ,\end{align*} $$

and denote $H^{Im}=rng(G^m)$ .

Notice that $G^m$ is well-defined, as

$$ \begin{align*}P(A,A^{*})(\phi )\in H_{N}\end{align*} $$

for all N large enough. Notice also that for all $\psi \in H$ , $G^{m}(A(\psi ))=A^{m}(G^{m}(\psi ))$ and the same for $A^{*}$ and $A^{*m}$ . Also $\Vert G^{m}(\psi )\Vert ^{m}_{2}=\Vert \psi \Vert _{2}$ and so $G^{m}$ is an (isometric) embedding of the Hilbert space H into $H^{m}$ and maps A to $A^{m}\restriction H^{Im}$ and $A^{*}$ to $A^{*m}\restriction H^{Im}$ .

So we have seen the following:

Proof. Now $f=f_{P}$ for some $P\in \mathbb {C} [X,Y]$ . Then

$$ \begin{align*}G^{m}(G(f))=G^{m}(P(A,A^{*})(\phi ))=P(A^{m},A^{*m})(G^{m}(\phi ))=F^m(f),\end{align*} $$

where the last identity follows from (4), (5), and (6) above.

As we are working with several spaces and functions, we have gathered the main ones in Figure 1 to act as a quick reference.

Figure 1

The main spaces and functions.

We now return to the reason for talking about A and $A^*$ instead of A and $A^{-1}$ , namely there are operators that are not obtained just by multiplying a unitary operator by a constant but that still allow for the sort of approximation we developed here.

In Section 4 we will use the method from this section applied to unitary operators, and use them to extend our approximating results to bounded self-adjoint operators. We leave it to future work to either find a way to extend the method of this section to all bounded normal operators or modify the construction for bounded self-adjoint operators from Section 4 to access bounded normal operators via a decomposition into self-adjoint operators.

3 Spectral measure

In this section we show how weighted counting measures in the finite dimensional spaces give rise to a measure on subsets of $\mathbb {C}$ . With this measure the vector space homomorphism $F^m$ from the previous section becomes an isometric embedding of $L_2(S,\mu )$ into the metric ultraproduct.

For all $N<{\omega }$ , we define a measure $\mu _{N}$ for subsets X of $\mathbb {C}$ : Recall that $\phi =\sum _{k=0}^{D_{N}-1}\xi _{N}(k)u_{N}(k)$ and each $\xi _{N}(k)$ is a non-negative real. We let $\mu _{N}(X)$ be the sum of all $\xi _{N}(k)^{2}$ such that $k<D_{N}$ and $\lambda _{N}(k)\in X$ . Note that, as $\lVert {\phi }\rVert =1$ , $\mu _N(X)\in [0,1]$ for all $N<{\omega }$ and $X\subset \mathbb {C}$ . Then we define $\mu ^{n}(X)$ to be the ultralimit $\lim _{{\mathcal {U}}}\mu _{N}(X)$ of the measures $\mu _{N}(X)$ , i.e., $\mu ^{n}(X)$ is the unique real r such that for all $\varepsilon>0$ , the set

$$ \begin{align*}\{ N<{\omega}\vert\ \vert \mu_{N}(X)-r\vert <\varepsilon\}\in {\mathcal{U}}.\end{align*} $$

We think of $\mu ^n$ as a sort of naive measure, as it is our main tool towards building a measure, but it is defined on all subsets of $\mathbb {C}$ and we have no reason to believe it actually would be a measure. However, we can use it to build a Borel measure.

Definition 3.3. Let $Sq^{*}$ be the set of all subsets Y such that there are $r_{i}\in NI$ such that

$$ \begin{align*}Y=\{ \lambda\in S\vert\ r_{0}<Re(\lambda )<r_{1},\ r_{2}<Im(\lambda )<r_{3}\} .\end{align*} $$

Now define $\mu ^{*}(Y)$ for any $Y\subseteq Int(S)$ (interior of S) to be the outer measure based on $\mu ^{n}(X)$ , $X\in Sq^{*}$ , i.e.,

$$ \begin{align*}\mu^{*}(Y)=\inf\{\sum_{i=0}^{\infty}\mu^{n}(X_{i})\vert\ X_{i}\in Sq^{*}, Y\subseteq\cup_{i<{\omega}}X_{i}\}.\end{align*} $$

As usual, we say that $X\subseteq Int(S)$ is measurable if for all $Y\subseteq Int(S)$ , $\mu ^{*}(Y)=\mu ^{*}(Y\cap X)+\mu ^{*}(Y-X)$ . For $X\subseteq S$ (or $X\subseteq \mathbb {C}$ ), we let $\mu ^{*}(X)=\mu ^{*}(X\cap Int(S))$ and X is measurable if $X\cap Int(S)$ is. Since there is $\varepsilon>0$ such that for all $N<{\omega }$ and $k<D_{N}$ , $\vert \lambda _{N}(k)\vert <M-\varepsilon $ , it is enough to look at subsets of $Int(S)$ (everything close to the boundary of S has naive measure zero). We write $\overline X$ for the closure of $X\subseteq \mathbb {C}$ .

Proof. (i): Clearly $\mu ^{*}(Y)\le \mu ^{*}(\overline Y)$ . From the choice of $Sq^{*}$ it follows easily that for all $\varepsilon>0$ , there is $X\in Sq^{*}$ such that $\vert \mu ^{n}(X)-\mu ^{n}(Y)\vert <\varepsilon $ and $\overline Y\subseteq X$ . Thus $\mu ^{*}(\overline Y)\le \mu ^{n}(Y)$ and $\mu ^{n}(\overline Y)=\mu ^{n}(Y)$ . Now for a contradiction, suppose that $\mu ^{*}(Y) +\varepsilon <\mu ^{n}(Y)$ for some $\varepsilon>0$ . Then we can find $Y_{i}\in Sq^*$ , $i<{\omega }$ , such that $Y\subseteq \cup _{i<{\omega }}Y_{i}$ and $\sum _{i<{\omega }}\mu ^{n}(Y_{i}) +\varepsilon <\mu ^{n}(Y)$ . But then we can find $X\in Sq^{*}$ such that $\overline X\subseteq Y$ and $\mu ^{n}(Y-\overline X)<\varepsilon $ . Since $\overline X$ is compact, there is $K<{\omega }$ such that $\overline X\subseteq \cup _{i<K}Y_{i}$ . Now

$$ \begin{align*}\mu^{n}(Y- \overline X)+\sum_{i<K}\mu^{n}(Y_{i})<\mu^{n}(Y),\end{align*} $$

a contradiction by Lemma 3.2.

(ii): By the definition of the outer measure $\mu ^*$ , and by considering coverings, it is enough to show that if $X\in Sq^{*}$ , then $\mu ^{*}(X-Y)+\mu ^{*}(X\cap Y)=\mu ^{*}(X)$ . Clearly $\mu ^{*}(X-Y)+\mu ^{*}(X\cap Y)\ge \mu ^{*}(X)$ . On the other hand there are $k\le 4$ and disjoint $Z_{i}\in Sq^{*}$ , $i<k$ , such that $\cup _{i<k}Z_{i}\subseteq X-Y$ and $X-Y\subseteq \cup _{i<k}\overline Z_{i}$ . Notice that $X\cap Y\in Sq^{*}$ . Now

$$ \begin{gather*}\mu^{*}(X-Y)\le\mu^{*}(\cup_{i<k}\overline Z_{i})\le\sum_{i<k}\mu^{*}(\overline Z_{i})=\\[5pt]=\sum_{i<k}\mu^{*}(Z_{i})=\sum_{i<k}\mu^{n}(Z_{i})=(\sum_{i<k}\mu^{n}(Z_{i}))+\mu^{n}(X\cap Y)-\mu^{n}(X\cap Y)\le\\[5pt]\le\mu^{n}((X\cap Y)\cup\cup_{i<k}Z_{i})-\mu^{n}(X\cap Y)\le\mu^{n}(X)-\mu^{n}(X\cap Y)=\mu^{*}(X)-\mu^{*}(X\cap Y).\end{gather*} $$

Now as usual one can see that the collection of measurable sets is closed under countable unions and complements. In particular every Borel set ( $\subseteq S$ ) is measurable. For measurable X ( $\subseteq S$ ), we write $\mu (X)=\mu ^{*}(X)$ and then $\mu $ is a Borel measure on S. Notice that by Lemma 3.4, $\mu (S)=\mu ^{*}(S)=\mu ^{n}(S)=1$ (recall Equation (7) and that $\phi $ has norm $1$ ).

When working with the measure and later for constructing rigged Hilbert spaces, it will be convenient to have a sequence of refinements in $Sq^*$ .

Remark 3.6. Now by Lemma 3.4, it is easy to see that for all $X\subseteq S$ ,

$$ \begin{align*}\mu^{*}(X)=\inf\{\sum_{i=0}^{\infty}\mu^{n}(X_{i})\vert\ X_{i}\in Sq, X\subseteq\cup_{i<{\omega}}\overline X_{i}\} .\end{align*} $$

We will use this frequently without mentioning it.

Now $L_{2}(S,\mu )$ is a Hilbert space and we write $\Vert \cdot \Vert _{2}$ also for the norm of this space. Elements of $L_{2}(S,\mu )$ are equivalence classes of the equivalence relation $\Vert f-g\Vert _{2}=0$ and we write $[f]_{2}$ for these equivalence classes. Then as usual we write $C(S)$ also for $\{ [f]_{2}\vert \ f\in C(S)\}$ and similarly for $D(S)$ . For each equivalence class $x\in C(S)$ we fix a continuous $f_{x}:S\rightarrow \mathbb {C}$ so that $x=[f_{x}]_{2}$ . We will see later (in Lemma 3.12) that this determines $f_{x}$ on $\sigma (A)$ and what $f_x$ does outside $\sigma (A)$ doesn’t matter. Now later when a function gives a value in $C(S)$ which is an equivalence class, we can think of it also as a continuous function from S to $\mathbb {C}$ . Thus $L_{2}(S,\mu )$ and $L_{2}(\sigma (A),\mu \restriction \sigma (A))$ are naturally isomorphic since every $f\in C(\sigma (A))$ extends to a (bounded uniformly) continuous function $f'\in C(S)$ .

Our next step is to show that the mappings $F^m:D(S)\to H^m$ and $G:D(S)\to H$ are not only vector space homomorphisms, but also isometries under the $L_2$ norm, which will let us define $F^m([f]_{2})=F^m(f)$ and $G([f]_{2})=G(f)$ to get isometric embeddings of $L_2(S,\mu )$ into $H^m$ and H respectively.

Proof. This proof is essentially the same as the proof that shows that continuous functions are Riemann integrable. Let $1>\varepsilon >0$ , $m=1+ \sup \{\vert f_{P}(\lambda )\vert \ \vert \ \lambda \in S\}$ and choose $k>0$ so that if $\vert \lambda -\lambda '\vert <4M/k$ (which happens for all $\lambda ,\lambda '\in \overline Y$ , for $Y\in Sq_k$ ), then $\vert f_{P}(\lambda )-f_{P}(\lambda ')\vert <\varepsilon /8m$ . Then we can choose $W\in {\mathcal {U}}$ such that for all $N\in W$ and $Y\in Sq_{k}$ , both $\vert \mu (Y)-\mu _{N}(Y)\vert $ and $\vert \mu (\overline Y)-\mu _{N}(\overline Y)\vert $ are $ {<\varepsilon /(4m^{2}\vert Sq_{k}\vert )}$ . Notice that if we let X be the union of all $\overline Y-Y$ , $Y\in Sq_{k}$ , then $\mu (X)=0$ , and

$$ \begin{align*}\lVert{\sum_{\lambda_{N}(k)\in X}\xi_{N}(k)f_{P}(\lambda_{N}(k))u_{N}(k)}\rVert_2^2\leq\mu_N(X)m^2<\varepsilon/4.\end{align*} $$

Then

$$ \begin{align*} \vert\ (\Vert F_{N}(f_{P})\Vert_{2})^{2}-(\Vert f_{P}\Vert_{2})^{2}\ \vert =\vert\ \sum_{k<D_N}\xi_N(k)^2\lvert{f_P(\lambda_N(k))}\rvert^2 - \int_S\lvert{f_P(x)}\rvert^2d\mu \ \vert\\ =\vert\ \sum_{\lambda_N(k)\in X}\xi_N(k)^2\lvert{f_P(\lambda_N(k))}\rvert^2 +\sum_{Y\in Sq_k}\sum_{\lambda_N(k)\in Y}\xi_N(k)^2\lvert{f_P(\lambda_N(k))}\rvert^2\\ - \sum_{Y\in Sq_k}\int_Y\lvert{f_P(x)}\rvert^2d\mu \ \vert\qquad\ \quad\qquad\\ \leq \varepsilon/4 +\sum_{Y\in Sq_k}\lvert{\sum_{\lambda_N(k)\in Y}\xi_N(k)^2\lvert{f_P(\lambda_N(k))}\rvert^2 - \int_Y\lvert{f_P(x)}\rvert^2d\mu}\rvert, \end{align*} $$

and for each $Y\in Sq_k$

$$ \begin{align*} \lvert{\sum_{\lambda_N(k)\in Y}\xi_N(k)^2\lvert{f_P(\lambda_N(k))}\rvert^2 - \int_Y\lvert{f_P(x)}\rvert^2d\mu}\rvert\qquad\qquad\qquad\\ \leq \int_Y \left(\frac{\sum_{l_N(k)\in Y}\xi_N(k)^2\lvert{\lvert{f_P(\lambda_N(k))}\rvert^2-\lvert{f_P(x)}\rvert^2}\rvert}{\mu(Y)}\right.\\ \left.\quad +\lvert{\frac{\sum_{\lambda_N(k)\in Y}\xi_N(k)^2\lvert{f_P(x)}\rvert^2}{\mu(Y)}-\lvert{f_P(x)}\rvert^2}\rvert\right)d\mu\\ \leq \int_Y \left(\frac{\mu_N(Y)2m\frac{\varepsilon}{8m}}{\mu(Y)}+\lvert{\frac{\mu_N(Y)}{\mu(Y)}-1}\rvert m^2\right)d\mu\\ =\mu_N(Y)\frac{\varepsilon}{4}+\lvert{\mu_N(Y)-\mu(Y)}\rvert m^2\\ < \mu_N(Y)\frac{\varepsilon}{4}+\frac{\varepsilon}{4\lvert{Sq_k}\rvert}. \end{align*} $$

Putting this together (and recalling $\mu _N(S)=1$ ), we get

$$ \begin{align*}\vert\ (\Vert F_{N}(f_{P})\Vert_{2})^{2}-(\Vert f_{P}\Vert_{2})^{2}\ \vert \le\varepsilon/4 +\varepsilon /4 +\varepsilon/4<\varepsilon \end{align*} $$

from which the claim follows.

We have been looking at $L_2(S,\mu )$ , where S was chosen to be a ‘big enough’ set. However, $\mu $ only puts weight on part of this set, and next we will show that set is exactly the spectrum of A. As pointed out above, there is a rational number ${0<q<M}$ such that for all $N<{\omega }$ , the norm of $A_{N}$ is $<q$ since the norm of $A_{N}$ is at most the norm of A. Thus if we let $S^{-}=\{\lambda \in \mathbb {C}\vert \ -M< Re(\lambda )<M, -M< Im(\lambda )<M\}$ then $\mu (S^{-})=\mu (S)=1$ .

Looking back at the constructions of this and the past section, one notes that the finite dimensional spaces $H_N$ depend on the choice of $\phi $ , and thus it is likely that one gets different $H^u$ and $H^m$ depending on the choice of cyclic vector. We have not looked into this, as the whole of these spaces are too big for our interest. As every $H_N$ has a part, where the operator is replaced by a wraparound-mapping, these erroneous parts will show up in the ultraproduct, and we are really interested only in parts stemming from the original operator. There are subspaces where these errors disappear, and one of them is $H^{Im}$ , which is canonical, as we have showed it to be isometrically isomorphic to H. A natural question arises:

As for the canonicity of our measure $\mu $ we have the following:

However, note that in general $\sigma (A)\neq \sigma (A^m)$ , as the latter can be strictly larger. Thus we are not considering elementary embeddings into the (metric) ultraproduct, but only embeddings. This is enough for our purposes, as we only use the construction for calculating equations.

4 Self-adjoint operators

In this section we show that the constructions from Sections 2 and 3 can be generalized to (bounded) self adjoint operators.

We will need the following facts, some of which seem hard to find in the literature (in particular item (iv) of the fact). Our version is not the best possible.

Proof. (i): The first claim is the spectral radius formula (for a proof see, e.g., Chapter VI of [Reference Reed and Simon9]) the others are proved via polynomial approximations, see Chapter VII of [Reference Reed and Simon9].

(ii): The first claim follows from the fact that B and $r I$ commute. The rest is easy.

(iii): Obviously A must be such that if $v\in H_{0}$ and $u\in H_{1}$ , then $A(v+u)=A_{0}(v)+A_{1}(u)$ and this is a bounded operator. Also clearly $\sigma (A_{0})\cup \sigma (A_{1})\subseteq \sigma (A)$ and since spectra are closed the other direction follows by an easy calculation. If $A_{0}$ and $A_{1}$ are self adjoint, then for all $v_{0},v_{1}\in H_{0}$ and $u_{0},u_{1}\in H_{1}$ ,

$$ \begin{align*}\langle A(v_{0}+v_{1})|u_{0}+u_{1}\rangle=\langle A_{0}(v_{0})|u_{0}\rangle+\langle A_{1}(v_{1})|u_{1}\rangle=\end{align*} $$

$$ \begin{align*}=\langle v_{0}|A_{0}(u_{0})\rangle+\langle v_{1}|A_{1}(u_{1})\rangle=\langle v_{0}+v_{1}|A(u_{0}+u_{1})\rangle,\end{align*} $$

where the first and the third identity follow from the choice of $H_{0}$ and $H_{1}$ . Finally if $A_{0}$ and $A_{1}$ are unitary, easy calculations show that A is norm preserving and onto.

(iv): Let $U'=e^{i\pi }U$ . Then $U'$ is unitary, and if $\lambda \in \sigma (U')$ , then $\lambda =e^{iq}$ for some $q\in [{-}r+\pi ,r+\pi ]$ . Now by [Reference Stone11, Theorem 8.4], there is a unique self-adjoint operator $A'$ such that $U'=e^{i2\pi A'}$ and the spectrum of $A'$ is a subset of $[0,1]$ and 0 is not in the point spectrum of $A'$ . By item (i), the spectrum of $2\pi A'$ is a subset of $[\pi -r,\pi +r]$ . Let $A=2\pi A'-\pi I$ . Then $e^{iA}=U$ , $\sigma (A)\subseteq [{-}r,r]$ and the uniqueness follows from the uniqueness of $A'$ and (ii).

From now on, in this section we assume that $Re(e^{i})\in NI$ . If this is not true, we can replace $1$ by a real q close to $1$ such that $Re(e^{iq})\in NI$ and work with it.

Let B be a bounded self-adjoint operator on a separable Hilbert space H. It will turn out that it does not matter whether we look at B or $rB$ for some positive real r, and thus we may assume that the norm of B is $\leq 1$ . Also we may assume that $B(v)\ne 0$ for all $v\in H-\{ 0\}$ because we can always restrict B to the orthogonal complement of its kernel (which is closed under B since B is self-adjoint). Let $A=e^{iB}$ . Then A is unitary and not the identity by Fact 4.1. Also by the same fact, if $\lambda \in \sigma (A)$ , then $\lambda =e^{iq}$ for some $q\in [-1,1]$ .

In the proof of the following lemma, the assumption that the norm of B is small plays an important role, stemming from Fact 4.1. We will need the uniqueness of the self-adjoint operator in point (iv) of the fact, and for that we need to avoid a ‘wraparound’. Our formulation of the fact allows for B to have norm smaller that $\pi /2$ , and we have chosen 1 as it is convenient in calculations.

By Lemma 4.2, if we may assume that H has a cyclic vector with respect to the operator A as in Section 2, we have spaces $H_{N}$ , operators $A_{N}$ , functions G, $G^{m}$ , $F^m$ and $F_{N}$ , a spectral measure $\mu $ etc. as in sections 2 and 3. Notice that each $A_{N}$ is a unitary operator and thus for all $N<{\omega }$ and $n< D_{N}$ , $\lambda _{N}(n)=e^{iq}$ for some $q\in ]-\pi ,\pi ]$ . It follows that $\mu (S-S_{0})=0$ when $S_{0}$ is the set $\{ e^{iq}\vert \ q\in ]-\pi ,\pi ]\}$ .

We define operators $B_{N}$ as follows: $B_{N}(u_{N}(n))=qu_{N}(n)$ if $A_{N}(u_{N}(n))=e^{iq}u_{N}(n)$ and $q\in ]-\pi ,\pi ]$ . We write $B^{m}$ for the metric ultraproduct (with the filter ${\mathcal {U}}$ as in the previous sections) of the operators $B_{N}$ . Then $A_{N}=e^{iB_{N}}$ for all $N<{\omega }$ and since the operators $B_{N}$ are uniformly bounded (by $\pi $ ) it is easy to see that $A^{m}=e^{iB^{m}}$ . Also each $B_{N}$ is self-adjoint and thus also $B^{m}$ is self-adjoint. We want to show the following:

This follows immediately from Fact 4.1(iv) if we know that $H^{Im}$ is closed under $B^{m}$ . So what we need to show is the following:

Proof. Suppose not. Note that from Corollaries 2.7 and 3.8 it follows that $H^{Im}:=rng(G^m)=rng(F^m)$ . Let $H_{0}$ be the closure of $H^{Im}$ under $B^{m}$ i.e., the complete subspace of $H^{m}$ generated by the set

$$ \begin{align*}\{ (B^{m})^{k}(F^m(f_{P}))\vert\ P\in \mathbb{C}[X,Y],\ k<{\omega}\}\end{align*} $$

(keep in mind that $B^{m}$ is bounded). Then $H^m$ is a complete subspace of $H_0$ , closed under $A^m$ and $A^{m*}$ , so we get a contradiction as in the proof of Lemma 4.2, unless the norm is too large to guarantee uniqueness. But this we can rule out by showing the following:

Claim 4.4.1. The operator norm of $B^{m}\restriction H_{0}$ is $\le 1$ .

Proof. For a contradiction suppose that there is $\varepsilon>0$ , $k<{\omega }$ and $P\in \mathbb {C} [X,Y]$ such that $\Vert B^{m}((B^{m})^{k}(F^m(f_{P})))\Vert>(1+\varepsilon )\Vert (B^{m})^{k}(F^m(f_{P}))\Vert $ . Since A is not the identity, $\mu (\{ 1\} )<1$ . So there are $\delta>0$ and $V\subseteq \sigma (A)$ such that $\mu (V)>\delta $ , for all $\lambda \in V$ , $\vert f_{P}(\lambda )\vert>\delta $ and if $\lambda =e^{iq}\in V$ , then $\vert q\vert>\delta $ .

For all $N<{\omega }$ , let $H^{0}_{N}$ be the subspace of $H_{N}$ generated by those $u_{N}(n)$ for which $\lambda _{N}(n)=e^{iq}$ for some $q\in [-\pi ,\pi ]$ such that $\vert q\vert>1$ and let $H^{1}_{N}$ be the subspace of $H_{N}$ generated by those $u_{N}(n)$ for which $\lambda _{N}(n)=e^{iq}$ for some $q\in [-\pi ,\pi ]$ such that $\vert q\vert \le 1$ . Then $H^{1}_{N}$ is the orthogonal complement of $H^{0}_{N}$ . Further let $H^2_N$ be the subspace of $H_N$ generated by those $u_N(n)$ for which $\lambda _N(n)\in V$ . Let $v_{i}(N)$ be the projection of $F_{N}(f_{P})$ to $H^{i}_{N}$ , $i<3$ . Let $S^{*}$ be the set of all $\lambda \in S$ such that $Re(\lambda )\ge Re(e^{i})$ . Notice that since $Re(e^{i})\in NI$ (see the assumption immediately after the proof of Fact 4.1), $\mu (S^{*})$ and $\mu (S-S^{*})$ are ultralimits of $\mu _{N}(S^{*})$ and $\mu _{N}(S-S^{*})$ . Let s be the supremum of all $\vert f_{P}(\lambda )\vert $ for $\lambda \in S$ .

Now

1. (a)

   $$ \begin{align*}&\lVert {(B_N)^k(F_N(f_P))}\rVert \geq \lVert {(B_N)^k(v_2(N))}\rVert =\\&\quad =\lVert{(B_N)^k\sum_{\lambda_N(n)\in V}\xi_N(n)f_P(\lambda_N(n))u_N(n)}\rVert\geq \delta^{k+3/2}\end{align*} $$
   in a set of N’s that belongs to ${\mathcal {U}}$ ,

2. (b)

   $$ \begin{align*}&\lVert {B_N((B_N)^k(F_N(f_P)))}\rVert =\lVert {(B_N)^{k+1}(v_0(N)+v_1(N))}\rVert \leq\\&\quad\pi^{k+1}\sqrt{\mu_N(S-S^*)}s+\lVert{v_1(N)}\rVert\end{align*} $$
   in a set of N’s that belongs to ${\mathcal {U}}$ ,

3. (c) $\mu _{N}(S-S^{*})\le (\varepsilon \delta ^{k+3/2}/(2\pi ^{k+1}s))^{2}$ in a set of N’s that belong to ${\mathcal {U}}$ (since $\mu (S-S^{*})=0$ by the assumption $\lVert {B}\rVert \leq 1$ and the construction of the measure for A).

By putting these together, we get

$$ \begin{align*}\lVert{B_N((B_N)^k(F_N(f_P)))}\rVert\leq \pi^{k+1}\sqrt{\mu_N(S-S^*)}s+\lVert{v_1(N)}\rVert\leq (\frac{\varepsilon}{2}+1)\lVert{(B_N)^k(F_N(f_P))}\rVert \end{align*} $$

in a set of N’s that belongs to ${\mathcal {U}}$ , which is a contradiction.

Now we can define an operator $B_{D}$ on $L_{2}(S,\mu )$ the natural way: For $f\in C(S)$ , we let $B_{D}(f)(\lambda )=K_B(\lambda )f(\lambda )$ , where $K_B(\lambda )=q$ if $q\in ]-\pi ,\pi ]$ and $\lambda =pe^{iq}$ for some real number p. Now it is easy to see that $e^{iB_{D}}=A_{D}$ and thus by Fact 4.1, $B^{m}\restriction H^{Im}=F^m B_{D}(F^m)^{-1}$ .

If we want for B exactly the same setup as for A we can do the following: Let ${S_{B}=]-\pi ,\pi ]}$ and define a Borel measure $\mu _{B}$ on $S_{B}$ by letting $\mu _{B}(X)=\mu (K_{B}^{-1}(X))$ . Then we can define an operator $B^{\prime }_{D}$ on $L_{2}(S_{B},\mu _{B})$ so that for all $f\in C(S_{B})$ , $B^{\prime }_{D}(f)(q)=qf(q)$ . Finally we notice that if we define a continuous $F_{B}: L_{2}(S,\mu )\rightarrow L_{2}(S_{B},\mu _{B})$ so that for $f\in C(S)$ , $F_{B}(f)(q)=f(e^{iq})$ we get an isometric isomorphism and $B^{\prime }_{D}=F_{B}B_{D}F_{B}^{-1}$ .

5 Rigged Hilbert spaces

Rigged Hilbert spaces are a way of accounting for objects that don’t fit into a Hilbert space, such as Dirac delta functions. A rigged Hilbert space consists of a Hilbert space H and a subspace $\Phi $ (with a finer norm topology than the one given by the norm in H). One then has

$$ \begin{align*}\Phi\subset H\subset \Phi^*, \end{align*} $$

where $\Phi ^*$ is the dual space of $\Phi $ . The elements in $\Phi ^*$ are antilinear functionals, and if $\Phi $ was suitably chosen with respect to a cyclic, essentially self adjoint operator B, then the functionals act as generalized eigenvectors corresponding to values in the spectrum of B (see, e.g., [Reference de la Madrid Modino8]).

In this section we set out to find something similar to $\Phi ^*$ by considering different norms in our ultraproduct space $H^u$ , and finding extensions of $H^{Im}$ . Here, as well as in the next section, it does not matter whether we look at a self-adjoint bounded B or $e^{iB}$ , and thus we return to the assumptions of sections 2 and 3, i.e., that A is bounded, $A\circ A^{*}=A^{*}\circ A=r_{A}I$ and $\phi $ is a cyclic vector. Also we let $H_{N}$ , $A_{N}$ etc. be as in these sections.

We will look at two norms, that end up being essentially an $L_{\infty }$ -norm and an $L_{1}$ -norm. The latter of these will be called $\Vert \cdot \Vert _{0}$ in order to avoid confusion. Strictly speaking both of these may be only seminorms in the finite dimensional spaces $H_{N}$ , but we will call them norms for simplicity. In the ultraproduct we will fraction out all vectors to which the seminorm gives an infinitesimal value, and then the seminorms become norms, and we obtain spaces $H^m_0$ and $H^m_\infty $ somewhat similarly to how we obtained $H^m$ for the $\lVert {\cdot }\rVert _2$ -norm before. Then we will show how generalized distributions embed into $H^m_0$ such that values of the distribution can be calculated via a pairing function resembling an inner product.

Definition 5.1. Consider the following functions from $H_{N}$ to $\mathbb {C}$ : Let $X\subseteq \mathbb {C}$ be a closed set such that $X\subseteq S$ . Then

$$ \begin{align*}\Vert\sum_{n=0}^{D_{N}-1}a_{n}u_N(n)\Vert^{X}_{\infty}=\sup\{\xi_{N}(n)^{-1}\vert a_{n}\vert\ \vert\ \lambda_{N}(n)\in X\} ,\end{align*} $$

where by $0^{-1}$ we mean $0$ , and

$$ \begin{align*}\Vert\sum_{n=0}^{D_{N}-1}a_{n}u_N(n)\Vert_{0}=\sum_{n=0}^{D_{N}-1}\xi_{N}(n)\vert a_{n}\vert .\end{align*} $$

Keep in mind that for all N and n, $\xi _{N}(n)$ is a non-negative real number $\le 1$ .

We define the $\lVert {\cdot }\rVert ^X_\infty $ -norm this way to get something that corresponds to an $L_\infty $ -norm. When the map $F_N$ maps functions over S into vectors in $H_N$ , it carries with it a weight factor that needs to be removed in order to achieve this. And disregarding the values where $\xi _N(n)=0$ corresponds to disregarding values in a set of measure 0.

When constructing $H^m$ from the $\lVert {\cdot }\rVert _2$ -norm we just fractioned out infinitesimals and discarded elements of infinite length. For our other norms we need to be a bit more careful. We start with $\Vert \cdot \Vert _{0}$ . As with the Hilbert space norm, this gives first a function from $H^{u}$ to $\mathbb {C}^{u}$ , denoted by $\Vert \cdot \Vert ^{u}_{0}$ . Note that the function is always defined, with values in a real closed field, but for some vectors the value may be $>n$ for all natural numbers n. Further we get an equivalence relation on $H^u$ by $v\sim _0w$ iff $\lVert {v-w}\rVert ^u_0$ is infinitesimal.

Definition 5.2. Let $n<{\omega }$ , $Y\subseteq Sq_{n}$ , $R^{Y}_{0}=\cup _{R\in Y}R$ , $S^{Y}_{0}=\cup _{R\in Y}\overline R$ , $R^{Y}_{1}=\cup _{R\in Sq_{n}-Y}R$ and $S^{Y}_{1}=\cup _{R\in Sq_{n}-Y}\overline R$ . We say that Y is good if the distance between $\sigma (A)$ and $S^{Y}_{1}$ is $>0$ (then $\sigma (A)\subseteq S^{Y}_{0}$ ). We say that

$$ \begin{align*}(\sum_{k=0}^{D_{N}-1}a^{N}_{k}u_{N}(k))_{N<{\omega}}/{\mathcal{U}}\in H^{u}\end{align*} $$

(and $(\sum _{k=0}^{D_{N}-1}a^{N}_{k}u_{N}(k))_{N<{\omega }}$ ) is $0$ -good if for all good Y

$$ \begin{align*}(\sum_{\lambda_{N}(k)\in S^{Y}_{1}}\xi_{N}(k)\vert a^{N}_{k}\vert )_{N<{\omega}}/{\mathcal{U}}/\sim =0.\end{align*} $$

Then we let $H^{m0}$ be the set of all $0$ -good $f/{\mathcal {U}}\in H^{u}$ such that $\Vert f/{\mathcal {U}}\Vert ^{u}_{0}<n$ for some natural number n. And we let $H^{m}_{0}=H^{m0}/\sim _{0}$ . Finally, we define the norm $\Vert \cdot \Vert _{0}$ on $H^{m}_{0}$ as before: $\Vert f/{\mathcal {U}}/\sim _{0}\Vert _{0}=q$ if $q\in \mathbb {C}$ is the unique element such that for all $\varepsilon>0$ ,

$$ \begin{align*}\{ N<{\omega}\vert\ \vert\Vert f(N)\Vert_{0}-q\vert <\varepsilon\}\in {\mathcal{U}}.\end{align*} $$

With this norm $H^{m}_{0}$ is a Banach space (addition and scalar multiplication are defined as before). In order to simplify the notation, for all $f\in \Pi _{N\in {\omega }}H_{N}$ , we write $[f]_{0}=f/{\mathcal {U}}/\sim _{0}$ . Notice that if $f,g\in H^{u}$ and $\Vert f-g\Vert ^{u}_{0}$ is infinitesimal and f is $0$ -good, then so is g. We also have the following:

Proof. For the first claim, note that $\mu (S)$ is bounded and, for a good Y, $\mu ^n(S^Y_1)=\mu (S^Y_1)=0$ , so as f is bounded, $F^u(f)$ is 0-good and has bounded $\lVert {\cdot }\rVert ^u_0$ -norm.

For the second claim, note that f is uniformly continuous and thus Riemann integrable, so we can for each $l<{\omega }$ and $X\in Sq_l$ pick any element $c_X\in X$ and then

$$ \begin{align*}\lVert{f}\rVert_1=\lim_{l\to\infty}\sum_{X\in Sq_l}\mu(X)\lvert{f(c_X)}\rvert. \end{align*} $$

On the other hand,

$$ \begin{align*}\lVert{F^u(f)}\rVert^u_0=\lim_{N,{\mathcal{U}}}\sum_{k<D_N}\xi_N(k)^2\lvert{f(\lambda_N(k))}\rvert =\lim_{N,{\mathcal{U}}}\sum_{X\in Sq_l}\sum_{k<D_N,\lambda_N(k)\in X}\xi_N(k)^2\lvert{f(\lambda_N(k))}\rvert \end{align*} $$

for any l, and if l is large enough so that values of f differ only slightly within each $X\in Sq_l$ , we can replace all the eigenvalues within X with the $c_X$ from above and have

$$ \begin{align*}\sum_{X\in Sq_l}\sum_{k<D_N,\lambda_N(k)\in X}\xi_N(k)^2\lvert{f(\lambda_N(k))}\rvert\approx \sum_{X\in Sq_l}\mu_N(X)\lvert{f(c_X)}\rvert. \end{align*} $$

For any l, we also have that for $X\in Sq_l$ , $\mu (X)=\mu ^n(X)=\lim _{N,{\mathcal {U}}}\mu _N(X)$ .

So for any $\varepsilon>0$ , we can choose an l large enough to get (for all N).

$$ \begin{align*}\lvert{\sum_{k<D_N}\xi_N(k)^2\lvert{f(\lambda_N(k))}\rvert-\sum_{X\in Sq_l}\mu_N(X)\lvert{f(c_X)}\rvert}\rvert<\varepsilon.\\[-45pt] \end{align*} $$

Let us then look at the norm $\Vert \cdot \Vert _{\infty }$ First we will need a modified version of the $\sup $ -norm on $C(S)$ :

Then $\Vert \cdot \Vert ^\sigma _{\infty }$ is a norm on $L_{\infty }(S,\mu )$ .

Definition 5.5. For all $\varepsilon>0$ , we let $X_{\varepsilon }$ be the set of all $\lambda \in \mathbb {C}$ whose distance to $\sigma (A)$ is $\le \varepsilon $ and we let $\Vert (v_{N})_{N<{\omega }}/{\mathcal {U}}\Vert ^{\varepsilon }_{\infty }$ be the unique $q\in \mathbb {C}$ such that for all $\delta>0$

$$ \begin{align*}\{ N<{\omega}\vert\ \vert\Vert v_{N}\Vert^{X_{\varepsilon}}_{\infty}-q\vert <\delta\}\in {\mathcal{U}}\end{align*} $$

if there is such a q and otherwise the value is $\infty $ (i.e., we do as before).

Now let $v=(\sum _{k=0}^{D_{N}-1}a^{N}_{k}u_{N}(k))_{N<{\omega }}/{\mathcal {U}}\in H^{u}$ . Then we let $\Vert v\Vert _{\infty }=\lim _{\varepsilon \rightarrow 0}\Vert v\Vert ^{\varepsilon }_{\infty }$ if there are $p<{\omega }$ and $W\in {\mathcal {U}}$ for which $\vert a^{N}_{k}(\xi _{N}(k))^{-1}\vert <p$ for all $N\in W$ and $k<D_{N}$ and otherwise we let $\Vert v\Vert _{\infty }=\infty $ . Notice that thus having bounded $\Vert \cdot \Vert _\infty $ -norm in $H^u$ requires the $\Vert \cdot \Vert ^S_\infty $ -norm to be bounded in a ${\mathcal {U}}$ -large set of $H_N$ s although the value is determined by what happens ‘close to $\sigma (A)$ ’. We then let $H^{m\infty }$ be the set of all $v\in H^{u}$ for which $\Vert v\Vert _{\infty }<\infty $ and $H^{m}_{\infty }=H^{m\infty }/\sim _{\infty }$ , where $\sim _{\infty }$ is the relation $\Vert v-u\Vert _{\infty }=0$ . We write $[f]_{\infty }$ for $f/{\mathcal {U}}/\sim _{\infty }$ .

The definition above may seem cumbersome, as we cannot restrict attention to the spectrum directly. This is because, although our measure will find the right spectrum eventually (cf. Lemma 3.12), our construction does not give any direct control of this convergence. Each $H_N$ has a discrete spectrum, but it is not even guaranteed that the eigenvalues in $H_N$ are in $\sigma (A)$ .

Let $H^{m}_{*}=\bigcup _{k\in \{ 0,2,\infty \}}H^{m}_{k}$ . We define a pairing $\langle \cdot |\cdot \rangle $ on $H^{m}_{*}$ the same way as before: For $[f]_{k},[g]_{r}\in H^{m}_{*}$ , $\langle [f]_{k}|[g]_{r}\rangle =q$ if $q\in \mathbb {C}$ is such that for all $f'\in [f]_{k}$ and $g'\in [g]_{r}$ , $\langle f'|g'\rangle ^{u}$ is infinitely close to q. If no such q exists we let $\langle [f]_{k}|[g]_{r}\rangle =\infty $ (meaning that $\langle [f]_{k}|[g]_{r}\rangle $ is not well-defined).

Proof. Let $x=[(x_{N})_{N<{\omega }}]_{\infty }$ , where

$$ \begin{align*}x_{N}=\sum_{n=0}^{D_{N}-1}a^{N}_{n}u_{N}(n)\end{align*} $$

and $y=[(y_{N})_{N<{\omega }}]_{0}$ , where

$$ \begin{align*}y_{N}=\sum_{n=0}^{D_{N}-1}b^{N}_{n}u_{N}(n)\end{align*} $$

and let $q>\Vert x\Vert _{\infty }$ and $r>\Vert y\Vert _{0}$ . Let $p<{\omega }$ , be such that for all $N<{\omega }$ and $n<D_{N}$ , $\vert a^{N}_{n}(\xi _{N}(n))^{-1}\vert <p$ . It is enough to show that

$$ \begin{align*}\vert \langle (x_{N})_{N<{\omega}}|(y_{N})_{N<{\omega}}\rangle\vert <qr.\end{align*} $$

Let $\varepsilon>0$ be such that $\Vert x\Vert ^{\varepsilon }_{\infty }<q$ . Then we choose $k<{\omega }$ so that there is a good $Y\subseteq Sq_{k}$ such that if as before we denote $S^{Y}_{0}=\cup _{R\in Y}\overline R$ and $S^{Y}_{1}=\cup _{R\in Sq_{k}-Y}\overline R$ , then $S-X_{\varepsilon }\subseteq S^{Y}_{1}$ . Then

$$ \begin{align*}\vert \langle x_{N}|y_{N}\rangle\vert\le\sum_{n=0}^{D_{N-1}}\vert a_{n}^{N}\vert\ \vert b^{N}_{n}\vert <\end{align*} $$

$$ \begin{align*}< \sum_{\lambda_{N}(n)\in S^{Y}_{0}}\xi_{N}(n)q\vert b^{N}_{n}\vert +\sum_{\lambda_{N}(n)\in S^{Y}_{1}}\xi_{N}(n)p\vert b^{N}_{n}\vert\le\end{align*} $$

$$ \begin{align*}\le q\sum_{\lambda_{N}(n)\in S^{Y}_{0}}\xi_{N}(n)\vert b^{N}_{n}\vert +p\sum_{\lambda_{N}(n)\in S^{Y}_{1}}\xi_{N}(n)\vert b^{N}_{n}\vert <qr +p\sum_{\lambda_{N}(n)\in S^{Y}_{1}}\xi_{N}(n)\vert b^{N}_{n}\vert\end{align*} $$

in a set of $N<{\omega }$ that belongs to ${\mathcal {U}}$ . Now the claim follows from the assumption that $(y_{N})_{N<{\omega }}/{\mathcal {U}}$ is $0$ -good.

Next we set out to find distributions as vectors in our space $H^u$ .

We now fix a generalized distribution $\theta $ for the remainder of this section.

In general a distribution emerges through its test functions, and we will construct test functions picking out rectangles in the spectrum, similarly to how the measure was built. This will let us ‘copy over’ the values of $\theta $ to our vector. However, we need to be careful to make sure our rectangles are well behaved.

When we talk about $I^{\varepsilon }_{r^{k}_{i}}$ or $J^{\varepsilon }_{r^{k}_{i}}$ we assume that $\varepsilon $ is small enough so that for all $j<2^{k+2}$ , $r^{k}_{j}+\varepsilon <r^{k}_{j+1}-\varepsilon $ .

For all $k<{\omega }$ , $i,j<2^{k+2}$ and $\varepsilon>0$ , we let $B^{k}_{ij}=\{ \lambda \in S\vert \ r^{k}_{i}<Re(\lambda )<r^{k}_{i+1},\ r^{k}_{j}<Im(\lambda )<r^{k}_{j+1}\}$ and $R^{k}_{\varepsilon }=\cup _{i<2^{k+2}+1}(I^{\varepsilon }_{r^k_{i}}\cup J^{\varepsilon }_{r^k_{i}})$ . We say that $\varepsilon>0$ is very nice for $k<{\omega }$ , if it is nice for every $r^k_{i}$ , $i<2^{k+2}$ . As in Lemma 5.12, we can see that $\mu (B^{k}_{ij})=\mu ^{n}(B^{k}_{ij})$ , that if $\varepsilon $ is very nice for k, then $\mu (R^{k}_{\varepsilon })=\mu ^{n}(R^{k}_{\varepsilon })$ , and that excluding countably many $\varepsilon $ , every $\varepsilon $ is very nice for k.

Notice that if $(n,\varepsilon )$ is $\theta $ -good and for all $N<{\omega }$ , $0<\varepsilon ' (N)<\varepsilon (N)$ , and $\varepsilon '$ is very nice for $n(N)$ , then $(n,\varepsilon ')$ is $\theta $ -good. So keeping in mind that the set of $\varepsilon $ very nice for n is dense, we can always assume, e.g., that $\varepsilon (N)$ is such that for all $i<2^{n(N)+2}$ , $\varepsilon (N)<(r^{n(N)}_{i+1}-r^{n(N)}_{i})/3$ .

Proof. Simply let $n(N)$ be the largest natural number $0<n\le N$ for which there is $\varepsilon =\varepsilon (N)$ for which Definition 5.16(ii) and (iii) hold, if such n and $\varepsilon $ exist and otherwise we let $n(N)=0$ and $\varepsilon =1$ . Since for all $\delta>0$ , $n<{\omega }$ and $i,j<2^{n+2}$ , the set

$$ \begin{align*}\{ N<{\omega}\vert\ \vert\mu_{N}(B^{n}_{ij})-\mu (B^{n}_{ij}\vert <\delta\}\end{align*} $$

belongs to ${\mathcal {U}}$ , it is easy to see that now also (i) holds.

Now we fix a $\theta $ -good pair $(n,\varepsilon )$ and in order to simplify the notations, we assume that $W(\theta ,n,\varepsilon )={\omega }$ .

We are finally ready to define the vector in $H^u$ that will correspond to the distribution $\theta $ :

Definition 5.19. We define $u(\theta )=(u_{N}(\theta ))_{N<{\omega }}/{\mathcal {U}}\in H^{u}$ by letting $n=n(N)$ and defining

$$ \begin{align*}u_{N}(\theta )=\sum_{\mu (B^{n}_{ij})\ne 0}\sum_{\lambda_{N}(k)\in B^{n}_{ij}}(\xi_{N}(k)\theta (B^{n}_{ij},1)/\mu_{N}(B^{n}_{ij}))u_{N}(k).\end{align*} $$

Recall that we may assume the relevant $\mu _N(B^n_{ij})\neq 0$ , as this will happen for all N in some $W\in {\mathcal {U}}$ . Also notice that $u(\theta )$ is $0$ -good (immediate by the definition) and that for all $f\in C(S)$ , one can define a generalized distribution $\theta _{f}$ by $\theta _{f}(g)=\langle g|f\rangle $ and then (if $\theta =\theta _{f}$ ) $[F^u(f)]_{0}=[u(\theta _{f})]_{0}$ (easy calculation).

Proof. Let K be such that for all $f\in C(S)$ , $\vert \theta (f)\vert <K\Vert f\Vert ^\sigma _{\infty }$ . It is enough to show that for all $N<{\omega }$ ,

$$ \begin{align*}\sum_{\mu (B^{n}_{ij})\ne 0}\sum_{\lambda_{N}(k)\in B^{n}_{ij}}\vert\xi_{N}(k)^{2}\theta (B^{n}_{ij},1)/\mu_{N}(B^{n}_{ij})\vert =\end{align*} $$

$$ \begin{align*}\sum_{\mu (B^{n}_{ij})\ne 0}\vert\theta (B^{n}_{ij},1)\vert\le 16K,\end{align*} $$

where $n=n(N)$ . Suppose not. Then one can find $X\subseteq \{ B^{n}_{ij}\vert \ i,j<2^{n+2},\ \mu (B^{n}_{ij})>0\}$ such that

$$ \begin{align*}\sum_{B\in X}\vert\theta (B,1)\vert>2K\end{align*} $$

and for all $B\in X$ , $\theta (B,1)$ point roughly to the same direction, e.g., $Re(\theta (B,1))\ge Im(\theta (B,1))\ge 0$ , see the proof of Lemma 5.10. But now for all $B\in X$ , choose a B-function $f_{B}$ so that $\theta (B,1)$ is very close to $\theta (f_{B})$ . Let $f=\sum _{B\in X}f_{B}\in C(S)$ . Then $\Vert f\Vert _{\infty }=1$ but $\vert \theta (f)\vert>K$ , a contradiction.

Proof. Let $f\in C(S)$ . W.l.o.g. we may assume that $\vert f(x)\vert \le 1$ for all $x\in S$ , in particular, $\Vert f\Vert ^\sigma _{\infty }\le 1$ . For all $0<n<{\omega }$ , there is $\varepsilon _{1}(n)$ such that if $x,y\in S$ are such that $\vert x-y\vert <4M/(n(N)+1)$ , then $\vert f(x)-f(y)\vert <\varepsilon _{1}(n(N))$ and $\varepsilon _{1}(n)$ goes to $0$ when n goes to infinity. Then for all $N<{\omega }$ such that $n(N)>0$ , we can find $f_{N}\in C(S)$ such that $\Vert f-f_{N}\Vert _{\infty }<\varepsilon _{1}(n(N))$ , $\vert f_{N}(x)\vert \le 1$ for all $x\in S$ and $f_{N}\restriction (B^{n(N)}_{ij}-R^{n(N)}_{\varepsilon (N)})$ is constant for all $i,j<2^{n+2}$ (here $\varepsilon (N)$ is the one from Definition 5.16, not $\varepsilon _{1}(n(N))$ ). Let $c^{N}_{ij}$ be the constant value. Now we notice that $\theta (f_{N})$ goes to $\theta (f)$ when $n(N)$ goes to infinity and also $\langle [F^u(f_{N})]_{\infty }|[u(\theta )]_{0}\rangle $ goes to $\langle [F^u(f)]_{\infty }|[u(\theta )]_{0}\rangle $ when $n(N)$ goes to infinity by the proof of Lemma 5.7 (and Lemma 5.20). Thus it is enough to show the following: Let $\varepsilon ^{*}>0$ and $N<{\omega }$ be such that $n(N)>0$ , and $K/n(N)<\varepsilon ^{*}$ where $K>0$ is a natural number such that $\vert \theta (f)\vert \le K\Vert f\Vert ^\sigma _{\infty }$ . Then

$$ \begin{align*}\vert \theta (f_{N})-\langle F_{N}(f_{N})|u_{N}(\theta)\rangle\vert \le 5\varepsilon^{*}.\end{align*} $$

We write $n=n(N)$ , $\varepsilon =\varepsilon (N)$ , $\delta =\delta (N)$ etc.

Now for all $i,j<2^{n+2}$ , there is a $B^{n}_{ij}$ -function $f_{ij}$ such that for all $x\in B^{n}_{i,j}-R^{n}_{\varepsilon }$ , $f_{ij}(x)=f_{N}(x)=c^{N}_{ij}$ . Let $f_{2}=\sum _{i,j<2^{n+2}}f_{ij}$ . Then (as $\delta <\varepsilon ^*$ ) it is enough to show that

$$ \begin{align*}\vert \theta (f_{2})-\langle F_{N}(f_{N})|u_{N}(\theta)\rangle\vert \le 4\varepsilon^{*} .\end{align*} $$

For this it is enough to show that for all $i,j <2^{n+2}$ ,

$$ \begin{align*}\vert \theta (f_{ij})-\langle F_{N}(f_{N}\restriction B^{n}_{i,j})|u_{N}(\theta)\rangle\vert \le 3K/(2^{3(n+2)}) ,\end{align*} $$

where by $F_{N}(f_{N}\restriction B^{n}_{i,j})$ we mean $\sum _{\lambda _{N}(k)\in B^{n}_{ij}}\xi _{N}(k)f_{N}(\lambda _{N}(k))u_{N}(k)$ .

Now if $\mu (B^{n}_{ij})=0$ , then $B^{n}_{ij}\cap \sigma (A)=\emptyset $ by the definition of $\tilde {\sigma }(A)$ and Lemma 3.12 and thus $\theta (f_{ij})=0$ since $f_{ij}$ is $\sim _{\infty }$ -equivalent with the constant zero function. Clearly then also $\langle F_{N}(f_{N}\restriction B^{n}_{i,j})|u_{N}(\theta )\rangle =0$ . Thus from now on we may assume that $\mu (B^{n}_{ij})>0$ .

But then it is enough to show that

$$ \begin{align*}\vert \theta (f_{ij})-\langle F_{N}(f_{ij})|u_{N}(\theta)\rangle\vert \le 2K/(2^{3(n+2)}),\end{align*} $$

since

1. (*) if $\mu (B^{n}_{ij})>0$ , then

   $$ \begin{align*}(\sum_{\lambda_{N}(k)\in B^{n}_{ij}\cap R^{n}_{\varepsilon}}\vert\xi_{N}(k)\vert^{2})/\mu_{N}(B^{n}_{ij})\le\delta/\delta'\le 1/(2^{3(n+2)}).\end{align*} $$

But then it is enough to show that

$$ \begin{align*}\vert \theta (B^{n}_{ij},c^{N}_{ij})-\langle F_{N}(f_{ij})|u_{N}(\theta)\rangle\vert \le K/(2^{3(n+2)}) .\end{align*} $$

Since $\theta (B^{n}_{ij}, c^{N}_{ij})=\overline {c^{N}_{ij}}\theta (B^{n}_{ij},1)=\overline {f_{ij}(x)}\theta (B^{n}_{ij},1)$ for any $x\in B^{n}_{ij}-R^{n}_{\varepsilon }$ , using $(*)$ above, this is easy.

We can now compare our construction to a rigged Hilbert space. We have defined our distributions using $C(S)$ as test functions. However, our construction disregards what happens outside $\sigma (A)$ , and in practice we are looking at $C(\sigma (A))$ , which embeds into $H^m_\infty $ . The distributions then embed into $H^m_0$ , so we can dig out the rigged Hilbert space in our ultraproduct. However, the construction embedding distributions into $H^m_0$ can only handle countably many distributions at a time (the way we approximate the test functions over rectangles varies with the distributions), asking for a question to be posed:

Another natural question that arises is how canonical the norms $\lVert {\cdot }\rVert _\infty $ and $\lVert {\cdot }\rVert _0$ are, as they were constructed from the chosen cyclic vector $\phi $ . However, there is no reason to believe that different cyclic vectors would generate the same ultraproduct $H^u$ , so different norms act on different spaces. The crucial thing is that we have shown that the vectors corresponding to distributions behave in the desired way (as distributions). Thus, e.g., computing kernels (the way we consider in the next section) in spaces arising from different cyclic vectors will give the same result.

6 Dirac delta functions and the Feynman propagator

We now turn to the study of the Feynman propagator. In [Reference Hirvonen and Hyttinen6] we studied the kernel vs. Feynman propagator for two simple quantum mechanical systems. To explain the terminology, let us consider a particle in (one-dimensional) space. Its state can be described by a wave function $\varphi $ in position space, where the amplitude $\varphi (x)$ determines the particle’s probability of being in position x. The evolution over a given time interval is described by the propagator $K(x,y)$ which expresses the probability amplitude of the particle traveling form point x to point y in the given time. (Usually the time is given as a third parameter, but with a time independent evolution operator, the question of calculating the kernel can be studied over a fixed time interval.) If the system has eigenvectors $\lvert {x}\rangle $ , $\lvert {y}\rangle $ , corresponding to the positions, one can calculate the propagator as an inner product