An Analytic Operator-Valued Generalized Feynman Integral on Function Space

Abstract

In this paper, we use a generalized Brownian motion process to define an analytic operator-valued Feynman integral. We then establish the existence of the analytic operator-valued generalized Feynman integral. We next investigate a stability theorem for the analytic operator-valued generalized Feynman integral.

1 Introduction

Cameron and Storvick [1] introduced an analytic operator-valued function space integral and showed that the integral satisfied an integral equation related to the Schrödinger equation. The existence of this integral was established as an operator from $L_2(\mathbb R)$ to $L_2(\mathbb R)$. Since then, Johnson and Lapidus [8] established the existence of the operator-valued function space integral as a bounded linear operator on $L_2(\mathbb R^n)$ for certain functionals which only define finite Borel measures on the compact interval [0, T] in $\mathbb R$. These integrals are based on the Wiener integral associated with the Wiener process.

On the other hand, Johnson [7] studied a bounded convergence theorem (stability theorem) for the operator-valued Feynman integral of functionals of the form $F(x)=\exp \{\int _{0}^{T}\theta (x(s))\hbox {d}s\}$. Chang et al. [2] established a stability theorems for the operator-valued Feynman integral of certain functionals involving some Borel measures on an interval (0, T) as a bounded linear operator from $L_1(\mathbb R)$ to $C_0(\mathbb R)$. Moreover, Chang and Lee [5] studied an analytic operator-valued generalized Feynman integral. The integral investigated in [5] is based on the function space integral associated with a generalized Brownian motion process.

The function space $C_{a,b}[0,T]$ induced by a generalized Brownian motion was introduced by Yeh in [10] and was studied extensively in [3, 4, 6]. In this paper, we define an analytic operator-valued generalized Feynman integral on the function space $C_{a,b}[0,T]$. We then establish the existence of the analytic operator-valued generalized Feynman integral and investigate a stability theorem for the analytic operator-valued generalized Feynman integral.

2 Definitions and Preliminaries

Let $D =[0,T]$ and let $(\Omega , {\mathcal {B}},P)$ be a probability measure space. A real-valued stochastic process Y on $(\Omega , {\mathcal {B}}, P)$ and D is called a generalized Brownian motion process if $Y(0,\omega )=0$ almost everywhere and for $0=t_0<t_1< \cdots <t_n \le T$, the n-dimensional random vector $(Y(t_1,\omega ), \ldots ,Y(t_n,\omega ))$ is normally distributed with density function

$$\begin{aligned} \begin{aligned} K(\vec t,\vec \eta )&=\left( (2\pi )^n \prod \limits _{j=1}^n\left( b(t_j)-b(t_{j-1})\right) \right) ^{-1/2} \\&\quad \times \exp \left\{ -\frac{1}{2} \sum \limits _{j=1}^n \frac{\left( \left( \eta _j-a(t_j)\right) -\left( \eta _{j-1}-a(t_{j-1})\right) \right) ^2}{b(t_j)-b(t_{j-1})} \right\} \end{aligned} \end{aligned}$$

where $\vec \eta = (\eta _1,\ldots ,\eta _n)$, $\eta _0=0$, $\vec t = (t_1,\ldots , t_n)$, a(t) is an absolutely continuous real-valued function on [0, T] with $a(0)=0$, $a'(t) \in L^2[0,T]$, and b(t) is a strictly increasing, continuously differentiable real-valued function with $b(0)=0$ and $b'(t) >0$ for each $t \in [0,T]$.

As explained in [11, pp. 18–20], Y induces a probability measure $\mu$ on the measurable space $(\mathbb {R}^D,\mathcal {B}^D )$ where $\mathbb {R}^D$ is the space of all real-valued functions x(t), $t \in D$, and $\mathcal {B}^D$ is the smallest $\sigma$-algebra of subsets of $\mathbb {R}^D$ with respect to which all the coordinate evaluation maps $e_t (x)=x(t)$ defined on ${\mathbb R}^D$ are measurable. The triple $({\mathbb R}^D,\mathcal {B}^D,\mu )$ is a probability measure space. This measure space is called the function space induced by the generalized Brownian motion process Y determined by $a(\cdot )$ and $b(\cdot )$.

We note that the generalized Brownian motion process Y determined by $a(\cdot )$ and $b(\cdot )$ is a Gaussian process with mean function a(t) and covariance function $r(s,t)= \min \{b(s),b(t)\}$. By [11, Theorem 14.2], the probability measure $\mu$ induced by Y, taking a separable version, is supported by $C_{a,b}[0,T]$ (which is equivalent to the Banach space of continuous functions x on [0, T] with $x(0)=0$ under the sup norm). Hence, $(C_{a,b}[0,T],{\mathcal {B}}(C_{a,b}[0,T]),\mu )$ is the function space induced by Y where ${\mathcal {B}}(C_{a,b}[0,T])$ is the Borel $\sigma$-algebra of $C_{a,b}[0,T]$. We then complete this function space to obtain $(C_{a,b}[0,T],{\mathcal {W}}(C_{a,b}[0,T]),\mu )$ where ${\mathcal {W}}(C_{a,b}[0,T])$ is the set of all $\mu$-Carathéodory measurable subsets of $C_{a,b}[0,T]$.

We note that the coordinate process defined by $e_t(x)=x(t)$ on $C_{a,b}[0,T]\times [0,T]$ is also the generalized Brownian motion process determined by a(t) and b(t). For more detailed studies about this function space $C_{a,b}[0,T]$, see [3, 6, 10].

Next, we state the definition of the analytic operator-valued generalized Feynman integral.

Definition 2.1

Let $\mathbb {C}$ be the set of complex numbers, let ${\mathbb C}_{+}=\{\lambda \in \mathbb {C}{:}{} \textit{Re}(\lambda )>0\}$ and let $\tilde{\mathbb C}_{+}=\{\lambda \in \mathbb {C}{:}{} \textit{Re}(\lambda )\ge 0,\lambda \not =0\}$. Also, let C[0, T] denote the space of real-valued continuous functions x on [0, T], and given a real number $\alpha$, let $\nu _{\alpha }$ be the measure on $\mathcal {B}(\mathbb {R})$ such that $\hbox {d}\nu _{\alpha } =\exp \{\alpha \eta ^2\}\hbox {d}\eta$. Next let F be a $\mathbb {C}$-valued functional on C[0, T]. For each $\lambda >0$, $\psi \in L^2(\mathbb {R},\nu _{\alpha })$ and $\xi \in \mathbb {R}$, assume that the functional $F(\lambda ^{-1/2}x+\xi )\psi (\lambda ^{-1/2}x(T)+\xi )$ is $\mu$-integrable with respect to x on $C_{a,b}[0,T]$, and let

$$\begin{aligned} \left( I_{\lambda }(F)\psi \right) (\xi ) =\int _{ C_{a,b}[0,T]}F\left( \lambda ^{-1/2}x+\xi \right) \psi \left( \lambda ^{-1/2}x(T)+\xi \right) \hbox {d}\mu (x). \end{aligned}$$

If $I_{\lambda }(F)\psi$ is in $L^{2}(\mathbb {R},\nu _{-\alpha })$ as a function of $\xi$ and if the correspondence $\psi \rightarrow I_{\lambda }(F)\psi$ gives an element of $\mathcal {L}\equiv \mathcal {L}(L^{2}(\mathbb {R},\nu _{\alpha }),L^{2}(\mathbb {R},\nu _{-\alpha }))$, the space of continuous linear operators from $L^2(\mathbb {R},\nu _{\alpha })$ to $L^2(\mathbb {R},\nu _{-\alpha })$, we say that the operator-valued function space integral $I_{\lambda }(F)$ exists. Next, suppose that there exists an $\mathcal {L}$-valued function which is analytic in $\mathbb {C}_{+}$ and agrees with $I_{\lambda }(F)$ on $(0,\infty )$, then this $\mathcal {L}$-valued function is denoted by $I_{\lambda }^{\mathrm{an}}(F)$ and is called the analytic operator-valued function space integral of F associated with $\lambda$. Finally, suppose that there exists an operator $J_{q}^{\mathrm{an}}$ in $\mathcal {L}(L^2(\mathbb {R},\nu _{\alpha }),L^2(\mathbb {R},\nu _{-\alpha }))$ for some $\alpha >0$ such that

$$\begin{aligned} \left\| I_{ \lambda }^{\mathrm{an}}(F)\psi - J_{q}^{\mathrm{an}}(F)\psi \right\| _{L^{2}(\mathbb {R},\nu _{-\alpha } )} \rightarrow 0 \end{aligned}$$

as $\lambda \rightarrow -iq$ through ${\mathbb C}_{+}$, then $J_{q}^{\mathrm{an}}(F)$ is called the analytic operator-valued generalized Feynman integral of F with parameter q.

3 An Analytic Operator-Valued Function Space Integral

Throughout the rest of this paper, we consider functionals of the form

$$\begin{aligned} F(x) = f\bigg (\int _{0}^{T} \theta (s,x(s))\hbox {d}\eta (s)\bigg ), \end{aligned}$$

(3.1)

where f is an analytic function on $\mathbb C$ and $\theta$ is an appropriate $\mathbb C$-valued function on $[0,T]\times \mathbb {R}$. F(x) is a very important functional in quantum mechanics. We then establish the existence of the analytic operator-valued function space integral for functionals F of the form (3.1).

Let $\mathcal {M}(0,T)$ denote the space of complex Borel measures $\eta$ on the open interval (0, T). Then $\eta \in \mathcal {M} (0,T)$ has a unique decomposition $\eta = \beta + \beta _d$ into its continuous part $\beta$ and its discrete part $\beta _d$ [9]. Let $\delta _{\tau }$ denote the Dirac measure at $\tau \in (0,T)$. For convenience, we let

$$\begin{aligned} \eta = \beta + \omega \delta _{\tau } , \quad \omega \in \mathbb {C} . \end{aligned}$$

(3.2)

Throughout the rest of this paper, we use the following notations:

(1) For $\lambda \in \tilde{\mathbb C}_+$ and $\psi \in L^2(\mathbb {R},\nu _{\alpha })$, let

$$\begin{aligned} \left( C_{( \lambda ,K,L)} \psi \right) (\xi )\equiv \bigg (\frac{ \lambda }{2\pi K}\bigg )^{1/2}\int _{\mathbb {R}} \psi (u) \exp \bigg \{- \frac{1}{2K}\Big ( \sqrt{\lambda }(u -\xi )- L \Big )^2 \bigg \}\hbox {d}u \end{aligned}$$

(3.3)

where K and L are real numbers with $K>0$. Then $C_{(\lambda ,K,L)}$ is in $\mathcal {L}(L^2(\mathbb {R},\nu _{\alpha }), L^2(\mathbb {R},\nu _{-\alpha }) )$.

(2) For each $s\in (0,T)$, let $\theta (s)$ denote the operator of multiplication from $L^2(\mathbb {R},\nu _{-\alpha })$ to $L^2(\mathbb {R},\nu _{\alpha })$ given by

$$\begin{aligned} \left( \theta (s)\psi \right) (\xi )=\theta (s,\xi )\psi (\xi ),\quad \xi \in \mathbb {R}. \end{aligned}$$

(3.4)

(3) Given a positive integer $l_1$, let

$$\begin{aligned} \Delta _{l_1;j}(T) \equiv \left\{ \left( s_1,\ldots ,s_{l_1}\right) |0<s_1<\cdots<s_j<\tau< s_{j+1}<\cdots<s_{l_1}<T\right\} \end{aligned}$$

and let

$$\begin{aligned} \Delta _{l_1}(T)\equiv \left\{ \left( s_1,\ldots ,s_{l_1}\right) |0<s_1<\cdots<s_{l_1}<T \right\} . \end{aligned}$$

Also, for $(s_1,\ldots , s_{l_1})\in \Delta _{l_1 ; j}(T)$ and a positive integer $l_2$, let

$$\begin{aligned} \begin{aligned} \mathcal {L}^{ \lambda }_{l_1;j}&\equiv C_{\left( \lambda ,b(s_1),a(s_1)\right) }\circ \theta (s_1)\circ \cdots \circ \theta (s_j) \circ C_{\left( \lambda ,b(\tau )-b(s_j),a(\tau )-a(s_j)\right) } \circ [\theta (\tau )]^{l_2} \\&\quad \circ C_{\left( \lambda ,b(s_{j+1})-b(\tau ),a(s_{j+1})-a(\tau )\right) }\circ \theta (s_{j+1})\circ \cdots \circ \theta (s_{l_1-1}) \\&\quad \circ C_{\left( \lambda ,b(s_{l_1})-b(s_{l_1-1}),a(s_{l_1})-a(s_{l_1-1})\right) } \circ \theta (s_{l_1}) \circ C_{\left( \lambda ,b(T)-b(s_{l_1}),a(T)-a(s_{l_1})\right) }. \end{aligned} \end{aligned}$$

(3.5)

Finally, for $(s_1,\ldots ,s_{l_1})\in \Delta _{l_1}(T)$, let

$$\begin{aligned} \mathcal {L}^{ \lambda }_{l_1}\equiv C_{\left( \lambda ,b(s_1),a(s_1)\right) }\circ \theta (s_1)\circ \cdots \circ \theta (s_{l_1})\circ C_{\left( \lambda , b(T)-b(s_{l_1}), a(T)-a(s_{l_1})\right) }. \end{aligned}$$

(3.6)

For example, we see that for $s_1\in \Delta _{1;1}(T)=\{s_1| 0< s_1<\tau <T\}$,

$$\begin{aligned} \begin{aligned} \mathcal {L}^{ \lambda }_{1;1}=&\,C_{\left( \lambda ,b(s_1),a(s_1)\right) }\circ \theta (s_1) \circ C_{\left( \lambda ,b(\tau )-b(s_1),a(\tau )-a(s_1)\right) }\circ [\theta (\tau )]^{l_2}\\&\circ C_{\left( \lambda ,b(T)-b(\tau ),a(T)-a(\tau )\right) } \end{aligned} \end{aligned}$$

and for $(s_1,s_2)\in \Delta _2(T)$,

$$\begin{aligned} \begin{aligned} \mathcal {L}^{ \lambda }_{2} =&\,C_{\left( \lambda ,b(s_1),a(s_1)\right) }\circ \theta (s_1) \circ C_{\left( \lambda ,b(s_2)-b(s_1),a(s_2)-a(s_1)\right) }\\&\circ \theta (s_2) \circ C_{\left( \lambda ,b(T)-b(s_2),a(T)-a(s_2)\right) }. \end{aligned} \end{aligned}$$

Hence using Eqs. (3.3)–(3.6), we observe that for $\psi \in L^2(\mathbb {R},\nu _{\alpha })$,

$$\begin{aligned} \begin{aligned} \left( \mathcal {L}^{ \lambda }_{1;1}\circ \psi \right) (\xi ) =&\left( \prod _{j=1}^{3}\frac{ \lambda }{2\pi (b(s_j)- b(s_{j-1}))}\right) ^{1/2} \int _{\mathbb {R}^3} \theta (s_1,u_1) \left[ \theta (\tau ,u_2)\right] ^{l_2} \psi (u_3)\\&\times \exp \left\{ -\sum _{j=1}^{3} \frac{\left[ \left( \sqrt{ \lambda }u_j-a(s_j)\right) -\left( \sqrt{ \lambda }u_{j-1} -a(s_{j-1}) \right) \right] ^2}{2\left( b(s_j)-b(s_{j-1})\right) }\right\} \hbox {d}u_1\hbox {d}u_2\hbox {d}u_3, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \left( \mathcal {L}^{ \lambda }_{2} \circ \psi \right) (\xi ) =&\left( \prod _{j=1}^{3}\frac{ \lambda }{2\pi \left( b(s_j)- b(s_{j-1})\right) }\right) ^{1/2} \int _{\mathbb {R}^3}\prod _{j=1}^{2}\theta (s_j,u_j) \psi (u_3)\\&\times \exp \left\{ -\sum _{j=1}^{3} \frac{\left[ \left( \sqrt{ \lambda }u_j - a(s_j)\right) -\left( \sqrt{ \lambda }u_{j-1}-a(s_{j-1})\right) \right] ^2}{2\left( b(s_j)-b(s_{j-1})\right) }\right\} \hbox {d}u_1\hbox {d}u_2\hbox {d}u_3 \end{aligned} \end{aligned}$$

where $s_0=0$, $a(s_0)=0$, $u_0 = \xi$, $s_2 = \tau$ and $s_3 = T$.

Also we will use the following conventions: for all positive integer l and $\lambda \in \tilde{\mathbb C}_{+}$, let

$$\begin{aligned} B_j^l \left( s_j;|\lambda |\right) \equiv \left( \frac{M_j |\lambda |}{ 2\pi }\right) ^{1/2} \int _{\mathbb {R}} \left| \left[ \theta (s_j,u_j)\right] ^l\right| \exp \left\{ M_j \left| \lambda \right| ^{1/2}\left| u_j\right| \right\} \hbox {d}u_j \end{aligned}$$

(3.7)

for some $M_j>0$, $j=1,\ldots ,{l_1}$. Furthermore, in order to ensure that analytic operator-valued generalized Feynman integral exists, we will assume that $B_j^l (s_j;|\lambda |)$, $a(\cdot )$ and $b(\cdot )$ satisfy the following conditions: for $j=1,\ldots ,l_1$ and $s_{l_1+1}=T$,

• (1) $\int _{0}^{T} B_j^l \left( s_j;\left| \lambda \right| \right) \hbox {d}|\eta |(s) < \infty$

• (2) $\frac{1}{b(s_j) - b(s_{j-1})} \le L_{jn}$

• (3) $\left| a'(s_j^*)\right| \le \left| b'(s_j^*)\right| M_{jn}$

for $s_j^* \in (s_{j-1},s_j)$ and some positive real numbers $L_{jn}$ and $M_{jn}$.

The next lemma plays a key role in the proof of Theorem 3.2.

Lemma 3.1

Let $\mathcal {L}^{ \lambda }_{l_1;j}$ be given by Eq. (3.5). Then for all $l_2\in \mathbb {N}$, $\xi \in \mathbb {R}$, $\lambda \in \tilde{\mathbb {C}}_+$ and $\psi \in L^2(\mathbb {R},\nu _{\alpha })$,

$$\begin{aligned} \left| \left( \mathcal {L}^{ \lambda }_{l_1;j}\circ \psi \right) (\xi )\right|\le & {} \left( \frac{ L_{Tn}^2 |\lambda |^2}{\pi \alpha }\right) ^{\frac{1}{4}} \exp \left\{ \frac{M_{Tn}^2}{2\alpha }|\lambda | + M_{1n} |\lambda |^{1/2} |\xi | \right\} \Vert \psi \Vert _{L^2\left( \mathbb {R},\nu _{\alpha }\right) } \nonumber \\&\times B_1(s_1;|\lambda |)\cdots B_{\tau }^{l_2} (\tau ;|\lambda |) \cdots B_{s_{l_1}}(s_{l_1};|\lambda |) \end{aligned}$$

(3.8)

for some $\alpha >0$.

Proof

Using Eq. (3.3)–(3.5), we have that for all $l\in \mathbb {N}$, $\xi \in \mathbb {R}$ and $\lambda \in \tilde{\mathbb {C}}_+$

$$\begin{aligned}&\left| (\mathcal {L}^{ \lambda }_{l_1;j}\circ \psi )(\xi )\right| \\&\quad =\left| \left( \frac{\lambda }{2\pi b(s_1)}\right) ^{1/2} \times \cdots \times \left( \frac{\lambda }{2\pi \left( b(\tau ) - b(s_j)\right) }\right) ^{1/2} \times \cdots \times \left( \frac{\lambda }{2\pi \left( b(T) - b(s_{l_1})\right) }\right) ^{1/2} \right. \\&\qquad \times \, \int _{\mathbb {R}^{l_1 +2}}\theta (s_1,u_1)\cdots \left[ \theta (\tau ,u_{\tau })\right] ^{l_2} \cdots \theta (s_{l_1},u_{l_1}) \psi (u_{l_1 +1}) \\&\qquad \times \exp \left\{ -\frac{1}{b(s_1)} \left( \sqrt{\lambda }(u_1 - \xi ) - a(s_1)\right) ^2 - \cdots \right. \\&\qquad - \frac{1}{2 \left( b(\tau ) - b(s_j)\right) } \left( \sqrt{\lambda }(u_{\tau } - u_j) - \left( a(\tau ) - a(s_j)\right) \right) ^2 -\cdots \\&\left. \left. \qquad -\, \frac{1}{2 \left( b(T) - b(s_{l_1}) \right) } \left( \sqrt{\lambda }(u_{l_1 +1} - u_{l_1}) - \left( a(T) - a(s_{l_1})\right) \right) ^2 \right\} \hbox {d}u_1\cdots \hbox {d}u_{\tau }\cdots \hbox {d}u_{l_1 +1}\right| \\&\quad \le \left( \frac{L_{1n} |\lambda |}{2\pi }\right) ^{1/2} \times \cdots \times \left( \frac{L_{ \tau n} |\lambda |}{2\pi }\right) ^{1/2} \times \cdots \times \left( \frac{L_{Tn} |\lambda |}{2\pi }\right) ^{1/2}\\&\qquad \times \, \int _{\mathbb {R}^{l_1 +2}}\left| \theta (s_1,u_1)\right| \cdots \left| \left[ \theta (\tau ,u_{\tau })\right] ^{l_{2}}\right| \cdots \left| \theta (s_{l_1},u_{l_1})\right| \left| \psi (u_{l_1 +1}) \right| \\&\qquad \times \, \exp \bigg \{M_{1n} |\lambda |^{1/2}(|u_1| + |\xi |) + M_{2n} |\lambda |^{1/2}(|u_2| + |u_1|) + \cdots \\&\qquad +\,M_{\tau n}|\lambda |^{1/2}(|u_\tau | + |u_j|) + \cdots \\&\qquad +M_{Tn}|\lambda |^{1/2}(|u_{l_1 +1}| + |u_{l_1}|)\bigg \} \hbox {d}u_1\cdots \hbox {d}u_{\tau }\cdots \hbox {d}u_{l_1 +1}\\&\quad \le \exp \{M_{1n} |\lambda |^{\frac{1}{2}} |\xi |\} \bigg (\frac{L_{Tn} |\lambda |}{2\pi }\bigg )^{1/2} \int _{\mathbb {R}} |\psi (u_{l_1 +1})| \exp \{M_{Tn}|\lambda |^{1/2} |u_{l_1 +1}|\} \hbox {d}u_{l_1 +1} \\&\qquad \times \, \bigg (\frac{L_{1n} |\lambda |}{2\pi }\bigg )^{1/2} \int _{\mathbb {R}} |\theta (s_1,u_1)| \exp \{2M_{1n}|\lambda |^{1/2} |u_1|\} \hbox {d}u_1\\&\qquad \qquad \qquad \qquad \quad \qquad \quad \qquad \quad \qquad \vdots \\&\qquad \times \, \bigg (\frac{L_{\tau n} |\lambda |}{2\pi }\bigg )^{1/2} \int _{\mathbb {R}} |[\theta (\tau ,u_{\tau })]^{l_2}| \exp \{2M_{\tau n}|\lambda |^{1/2} |u_{\tau }|\} \hbox {d}u_{\tau }\\&\qquad \qquad \qquad \qquad \quad \qquad \quad \qquad \quad \qquad \vdots \\&\qquad \times \, \bigg (\frac{L_{s_{l_1 n}} |\lambda |}{2\pi }\bigg )^{1/2} \int _{\mathbb {R}} | \theta (s_{l_1},u_{l_1}) | \exp \{2M_{s_{l_1 n}}|\lambda |^{1/2} |u_{l_1}|\} \hbox {d}u_{l_1}\\&\quad \le \bigg (\frac{L^2_{Tn} |\lambda |^2}{ \pi \alpha }\bigg )^{\frac{1}{4}} \exp \bigg \{ M_{1n} |\lambda |^{1/2} |\xi | + \frac{M^2_{Tn}|\lambda | }{2\alpha } \bigg \} \Vert \psi \Vert _{L^2(\mathbb {R},\nu _{\alpha })} \\&\qquad \times \, B_1(s_1;|\lambda |)\cdots B_{\tau }^{l_2} (\tau ;|\lambda |) \cdots B_{s_{l_1}}(s_{l_1};|\lambda |), \end{aligned}$$

which completes the proof of Lemma 3.1. $\square$

In our next theorem, we establish the existence of the analytic operator-valued function space integral for the functional F given by (3.1) with $f(z)=z^n$.

Theorem 3.2

Let $\theta$ be a Borel measurable function on $[0,T]\times \mathbb {R}$. For $n=1,2,\ldots$, let

$$\begin{aligned} F_n (x) = \bigg (\int _{0}^{T}\theta (s,x(s)) \hbox {d}\eta (s)\bigg )^n. \end{aligned}$$

(3.9)

Let $\eta$ be given by (3.2). Then for all $\lambda \in \mathbb {C}_+$ and $\psi \in L^2(\mathbb {R},\nu _{\alpha })$, the analytic operator-valued function space integral of $F_n$, $I_{ \lambda }^{\mathrm{an}}(F_n)$, exists and is given by the formula

$$\begin{aligned} \begin{aligned}&\left( I_{ \lambda }^{\mathrm{an}}(F_n)\psi \right) (\xi ) = \sum _{ \begin{array}{c} l_1 + l_2 = n \\ l_2\not =0 \end{array}}\frac{n! \omega ^{l_2} }{l_2 !} \sum _{j=0}^{l_1} \int _{\Delta _{l_1;j}(T)} \left( \mathcal {L}^{ \lambda }_{l_1;j}\circ \psi \right) (\xi ) \mathrm{d}\prod _{l=1}^{l_1}\beta (s_l)\\ \end{aligned} \end{aligned}$$

(3.10)

where $\beta (s_0)=0$, $s_{l_1 +1} = T$ and $\Delta _{0;j}(T)$ is an empty set.

Proof

Using Eq. (3.1) with $f(z)=z^n$, (3.3), (3.4) and the Fubini theorem, we first obtain that for all $\lambda >0$

$$\begin{aligned} \left( I_{ \lambda }(F)\psi \right) (\xi )=&\,\int _{ C_{a,b}[0,T]} F\left( \lambda ^{-1/2}x + \xi \right) \psi \left( \lambda ^{-1/2}x(T)+\xi \right) \hbox {d}\mu (x)\\ =&\,\int _{ C_{a,b}[0,T]} \bigg (\int _{0}^{T}\theta \left( s, \lambda ^{-1/2}x(s)+\xi \right) \hbox {d}\eta (s)\bigg )^n\psi \left( \lambda ^{-1/2}x(T)+\xi \right) \hbox {d}\mu (x)\\ =&\,\int _{ C_{a,b}[0,T]} \bigg (\int _{0}^{T}\theta \left( s, \lambda ^{-1/2}x(s)+\xi \right) \hbox {d}\beta (s) + \omega \cdot \theta \left( \tau , \lambda ^{-1/2}x(\tau )+\xi \right) \bigg )^n \\&\times \, \psi \left( \lambda ^{-1/2}x(T)+\xi \right) \hbox {d}\mu (x) \\ =&\,\int _{ C_{a,b}[0,T]} \sum _{\begin{array}{c} l_1 + l_2 =n \\ l_2\not =0 \end{array}} \frac{n!}{l_1 ! l_2!} \bigg ( \int _{0}^{T}\theta \left( s, \lambda ^{-1/2}x(s)+\xi \right) \hbox {d}\beta (s) \bigg )^{l_1} \\&\times \, \left( \omega \cdot \theta \left( \tau , \lambda ^{-1/2}x(\tau )+\xi \right) \right) ^{l_2} \psi \left( \lambda ^{-1/2}x(T)+\xi \right) \hbox {d}\mu (x) \\ =&\, \sum _{\begin{array}{c} l_1 + l_2 =n \\ l_2\not =0 \end{array}} \frac{n! \omega ^{l_2}}{l_2 !} \sum _{j=0}^{l_1} \int _{\Delta _{l_1;j}(T)}\bigg [ \int _{ C_{a,b}[0,T]} \theta \left( s_1, \lambda ^{-1/2}x(s_1)+\xi \right) \times \cdots \\&\times \, \theta \left( s_j, \lambda ^{-1/2}x(s_j)+\xi \right) \left[ \theta \left( \tau , \lambda ^{-1/2}x(\tau )+\xi \right) \right] ^{l_2}\\&\times \,\theta \left( s_{j+1}, \lambda ^{-1/2}x(s_{j+1})+\xi \right) \times \cdots \\&\left. \times \, \theta \left( s_{l_1}, \lambda ^{-1/2}x(s_{l_1})+\xi \right) \psi \left( \lambda ^{-1/2}x(T)+\xi \right) \hbox {d}\mu (x)\right] \hbox {d}\beta (s_1)\cdots \hbox {d}\beta (s_{l_1})\\ =&\, \sum _{ \begin{array}{c} l_1 + l_2 = n \\ l_2\not =0 \end{array}}\frac{n! \omega ^{l_2} }{l_2 !} \sum _{j=0}^{l_1} \int _{\Delta _{l_1;j}(T)} \left( \mathcal {L}^{ \lambda }_{l_1;j}\circ \psi \right) (\xi ) \hbox {d}\prod _{l=1}^{l_1}\beta (s_l). \end{aligned}$$

Next we will show that the existence of analytic operator-valued function space integral $I_{\lambda }^{\mathrm{an}}(F_n)$ exists. Using Eq. (3.8), we obtain that for all $\lambda \in \mathbb {C}_+$

$$\begin{aligned}&\sum _{ \begin{array}{c} l_1 + l_2 = n \\ l_2\not =0 \end{array}}\frac{n! \omega ^{l_2} }{l_2 !} \sum _{j=0}^{l_1} \int _{\Delta _{l_1;j}(T)} | (\mathcal {L}^{ \lambda }_{l_1;j}\circ \psi )(\xi )| \hbox {d}\prod _{l=1}^{l_1}|\beta |(s_l) \nonumber \\&\quad =\bigg (\frac{L^2_{Tn} |\lambda |^2}{ \pi \alpha }\bigg )^{\frac{1}{4}} \exp \bigg \{ M_{1n} |\lambda |^{1/2} |\xi | + \frac{M^2_{Tn}|\lambda | }{2\alpha } \bigg \} \Vert \psi \Vert _{L^2(\mathbb {R},\nu _{\alpha })} \nonumber \\&\qquad \times \, \sum _{\begin{array}{c} l_1 + l_2 = n\\ l_2\not =0 \end{array}}\frac{n! \omega ^{l_2} }{l_2!} \sum _{j=0}^{l_1}\int _{\Delta _{l_1;j}(T)} B_1(s_1;|\lambda |)\times \cdots \times B_{\tau }^{l_2} (\tau ;|\lambda |) \times \cdots \nonumber \\&\qquad \times \, B_{s_{l_1}}(s_{l_1};|\lambda |) \hbox {d}|\beta |(s_1)\cdots \hbox {d}|\beta |(s_{l_1}) \nonumber \\&\quad =\bigg (\frac{L^2_{Tn} |\lambda |^2}{ \pi \alpha }\bigg )^{\frac{1}{4}} \exp \bigg \{ M_{1n} |\lambda |^{1/2} |\xi | + \frac{M^2_{Tn}|\lambda | }{2\alpha } \bigg \} \Vert \psi \Vert _{L^2(\mathbb {R},\nu _{\alpha })} \nonumber \\&\qquad \times \, n! \sum _{\begin{array}{c} l_1 + l_2 = n\\ l_2\not =0 \end{array}}\frac{1 }{l_1!l_2!} \bigg (\int _{0}^{T}B(s;|\lambda |)\hbox {d}|\beta |(s) \bigg )^{l_1} (\omega B(\tau ;|\lambda |))^{l_2} \bigg ]\nonumber \\&=\bigg (\frac{L^2_{Tn} |\lambda |^2}{ \pi \alpha }\bigg )^{\frac{1}{4}} \exp \bigg \{ M_{1n} |\lambda |^{1/2} |\xi | + \frac{M^2_{Tn}|\lambda | }{2\alpha } \bigg \} \Vert \psi \Vert _{L^2(\mathbb {R},\nu _{\alpha })}\nonumber \\&\qquad \times \, \bigg (\int _{0}^{T}B(s;|\lambda |)\hbox {d}|\beta |(s) + \omega B(\tau ;|\lambda |) \bigg )^n\nonumber \\&=\bigg (\frac{L^2_{Tn} |\lambda |^2}{ \pi \alpha }\bigg )^{\frac{1}{4}} \exp \bigg \{ M_{1n} |\lambda |^{1/2} |\xi | + \frac{M^2_{Tn}|\lambda | }{2\alpha } \bigg \} \Vert \psi \Vert _{L^2(\mathbb {R},\nu _{\alpha })}\nonumber \\&\qquad \times \, \bigg (\int _{0}^{T}B(s;|\lambda |) \hbox {d}|\eta |(s)\bigg )^{n} < \infty . \end{aligned}$$

(3.11)

Therefore, the analytic operator-valued function space integral $I_{\lambda }^{\mathrm{an}}(F_n)$ exists and is given by Eq. (3.10).

Now we will show that $I_{\lambda }^{\mathrm{an}}(F_n)$ is an element of $\mathcal {L}( L^2(\mathbb {R}, \nu _{\alpha }), L^2(\mathbb {R}, \nu _{-\alpha }))$. Using Eqs. (3.10) and (3.11), it follows that

$$\begin{aligned}&\left\| I_{\lambda }^{\mathrm{an}}(F_n)\psi \right\| ^2_{L^2(\mathbb {R},\nu _{-\alpha })} =\int _{\mathbb {R}} \left| \left( I_{\lambda }^{\mathrm{an}}(F_n)\psi \right) (\xi )\right| ^2 \hbox {d}\nu _{-\alpha }(\xi )\nonumber \\&\quad =\left( \frac{L^2_{Tn} |\lambda |^2}{\pi \alpha }\right) ^{1/2} \left\| \psi \right\| ^2_{L^2(\mathbb {R},\nu _{\alpha })} \bigg (\int _{0}^{T}B(s;|\lambda |) \hbox {d}|\eta |(s)\bigg )^{2n} \exp \left\{ \frac{M^2_{Tn}|\lambda | }{ \alpha } \right\} \nonumber \\&\qquad \times \int _{\mathbb {R}} \exp \left\{ M_{1n} |\lambda |^{1/2}|\xi |\right\} \hbox {d}\nu _{-\alpha }(\xi )\nonumber \\&\quad \le \bigg (\frac{4L^2_{Tn} |\lambda |^2}{\alpha ^2}\bigg )^{1/2} \Vert \psi \Vert ^2_{L^2(\mathbb {R},\nu _{\alpha })} \left( \int _{0}^{T}B(s;|\lambda |) \hbox {d}|\eta |(s)\right) ^{2n} \exp \left\{ \frac{|\lambda |\left( M^2_{1n} + M^2_{Tn}\right) }{\alpha } \right\} . \end{aligned}$$

(3.12)

Hence, we obtain that for all $\lambda \in \mathbb {C}_+$,

$$\begin{aligned} \left\| I_{\lambda }^{\mathrm{an}}(F_n) \right\| \le \left( \frac{4L^2_{Tn} |\lambda |^2}{\alpha ^2}\right) ^{\frac{1}{4}} \left( \int _{0}^{T}B(s;|\lambda |) \hbox {d}|\eta |(s)\right) ^{n} \exp \left\{ \frac{|\lambda |(M^2_{1n} + M^2_{Tn})}{\alpha } \right\} . \end{aligned}$$

Thus, the theorem is proved. $\square$

Let $f(z) = \sum ^{\infty }_{n=1} \beta _n z^n$ be an analytic function on $\mathbb C$ such that

$$\begin{aligned} \sum ^{\infty }_{n=1}| \beta _n | \Psi ^k_n(|\lambda |) <\infty \end{aligned}$$

(3.13)

for all $\lambda \in \tilde{\mathbb C}_{+}$, where

$$\begin{aligned} \Psi ^k_n(|\lambda |)\equiv \left( \frac{4L^2_{Tn} |\lambda |^2}{\alpha ^2}\right) ^{\frac{1}{4}} \left( \int _{0}^{T}B^k(s;|\lambda |) \hbox {d}|\eta |(s)\right) ^{ n} \exp \bigg \{\frac{|\lambda |\left( M^2_{1n} + M^2_{Tn}\right) }{\alpha } \bigg \} \end{aligned}$$

(3.14)

for all positive integers n and k. Let

$$\begin{aligned} F(x) = f\left( \int _{0}^{T}\theta (s,x(s)) \hbox {d}\eta (s)\right) \end{aligned}$$

(3.15)

for $x\in C_{a,b}[0,T]$.

Our aim in this section is to establish the existence of the analytic operator-valued function space integral for the functionals F given by (3.15).

Theorem 3.3

Let F be given by Eq. (3.15). Then for all $\lambda \in \mathbb {C}_+$ and $\psi \in L^2(\mathbb {R},\nu _{\alpha })$, the analytic operator-valued function space integral of F, $I_{\lambda }^{\mathrm{an}}(F)$, exists and is given by the formula

$$\begin{aligned} I_{\lambda }^{\mathrm{an}}(F) \psi = \sum _{n=1}^{\infty } \beta _n I_{\lambda }^{\mathrm{an}}(F_n) \psi \end{aligned}$$

where $I_{\lambda }^{\mathrm{an}}(F_n)$ is given by Eq. (3.10). Furthermore, $I_{\lambda }^{\mathrm{an}}(F)$ is an element of $\mathcal {L} ( L^2(\mathbb {R},\nu _{\alpha }), L^2(\mathbb {R},\nu _{-\alpha }) )$.

Proof

Since $F(x) = \sum _{n=1}^{\infty }\beta _n F_n(x)$, using (3.11) and (3.12) we have

$$\begin{aligned} I_{\lambda }^{\mathrm{an}}(F)\psi = \sum _{n=1}^{\infty }\beta _n I_{\lambda }^{\mathrm{an}}(F_n)\psi \end{aligned}$$

and

$$\begin{aligned} \left\| I_{\lambda }^{\mathrm{an}}(F) \psi \right\| _{L^2(\mathbb {R},\nu _{-\alpha })} \le \sum _{n=1}^{\infty } |\beta _n| \Psi ^1_{n}(|\lambda |) \Vert \psi \Vert _{L^2(\mathbb {R},\nu _{\alpha })} \end{aligned}$$

where $\Psi ^1_{n}(|\lambda |)$ is given by Eq. (3.14) with $k=1$. Next using the condition (3.13), the analytic operator-valued function space integral $I_{\lambda }^{\mathrm{an}}(F)$ exists and $I_{\lambda }^{\mathrm{an}}(F)$ is an element of $\mathcal {L}(L^2(\mathbb {R},\nu _{\alpha }),L^2(\mathbb {R},\nu _{-\alpha }))$. $\square$

4 An Analytic Operator-Valued Generalized Feynman Integral

In Sect. 3, we established the existence of the analytic operator-valued function space integral for the functionals F given by Eq. (3.15). In this section, we establish the existence of the analytic operator-valued generalized Feynman integral for the functionals F. To do this, in Theorem 4.1, we first obtain the analytic operator-valued generalized Feynman integral for the functionals $F_n$ given by (3.9).

Theorem 4.1

Let $F_n$ be given by Eq. (3.9). Then for all $q\in \mathbb {R} {\setminus }\{0\}$, the analytic operator-valued generalized Feynman integral of $F_n$, $J_{q}^{\mathrm{an}}(F_n)$, exists and is given by the formula

$$\begin{aligned} \begin{aligned}&(J_{q}^{\mathrm{an}}(F_n)\psi )(\xi ) = \sum _{ \begin{array}{c} l_1 + l_2 = n \\ l_2\not =0 \end{array}}\frac{n! \omega ^{l_2} }{l_2 !} \sum _{j=0}^{l_1} \int _{\Delta _{l_1;j}(T)} \left( \mathcal {L}^{-iq}_{l_1;j}\circ \psi \right) (\xi ) \hbox {d}\prod _{l=1}^{l_1}\beta (s_l) \end{aligned} \end{aligned}$$

(4.1)

where $\beta (s_0)=0$, $s_{l_1 +1} = T$ and $\Delta _{0;j}(T)$ is an empty set.

Proof

In order to establish Eq. (4.1), it suffices to show that

$$\begin{aligned} \lim _{\lambda \rightarrow -iq} \int _{\mathbb {R}} \left| \left( I_{\lambda }^{\mathrm{an}}(F_n)\right) (\psi ) - \left( J_{q}^{\mathrm{an}}(F_n)\right) (\psi )\right| ^2 \hbox {d}\nu _{-\alpha }(\xi ) = 0. \end{aligned}$$

But, for all $\lambda \in \mathbb {C}_+$, we have

$$\begin{aligned} \left| \left( I_{\lambda }^{\mathrm{an}}(F_n)\right) (\psi ) - \left( J_{q}^{\mathrm{an}}(F_n)\right) (\psi )\right| ^2 \le 2 \left| \left( I_{\lambda }^{\mathrm{an}}(F_n)\right) (\psi )\right| ^2 + 2 \left| \left( J_{q}^{\mathrm{an}}(F_n)\right) (\psi )\right| ^2. \end{aligned}$$

(4.2)

Using a similar method as those used in (3.12), we also see that $|(I_{\lambda }^{\mathrm{an}}(F_n))(\psi )|^2$ and $|(J_{q}^{\mathrm{an}}(F_n))(\psi )|^2$ are in $L^1(\mathbb {R},\nu _{-\alpha })$. Hence, the second expression in Eq. (4.2) is in $L^1(\mathbb {R},\nu _{-\alpha })$. Thus, using the dominated convergence theorem, we obtain the desired result. $\square$

The next theorem is one of the main results in this paper.

Theorem 4.2

Let F be given by Eq. (3.15). Then for all $q\in \mathbb {R}{\setminus }\{0\}$, the analytic operator-valued generalized Feynman integral of F, $J_{q}^{\mathrm{an}}(F)$, exists and is given by the formula

$$\begin{aligned} J_{q}^{\mathrm{an}}(F)\psi = \sum _{n=1}^{\infty } \beta _n J_{q}^{\mathrm{an}}(F_n)\psi \end{aligned}$$

(4.3)

where $J_{q}^{\mathrm{an}}(F_n)$ is given by Eq. (4.1). Furthermore, $J_{q}^{\mathrm{an}}(F)$ is an element of $\mathcal {L}(L^2(\mathbb {R},\nu _{\alpha }), L^2(\mathbb {R},\nu _{-\alpha }) )$.

Proof

Using (3.11) and (3.12) with $\lambda$ replaced with $-iq$, we obtain

$$\begin{aligned} J_{q}^{\mathrm{an}}(F)\psi = \sum _{n=1}^{\infty }\beta _n J_{q}^{\mathrm{an}}(F_n)\psi \end{aligned}$$

and

$$\begin{aligned} \left\| J_{q}^{\mathrm{an}}(F)\psi \right\| _{L^2(\mathbb {R},\nu _{-\alpha })} \le \sum _{n=1}^{\infty } |\beta _n| \Psi ^1_{n}\left( |-iq|\right) \Vert \psi \Vert _{L^2(\mathbb {R},\nu _{\alpha })} \end{aligned}$$

where $\Psi ^1_n(|-iq|)$ is given by Eq. (3.14) with $k=1$. Next using the condition (3.13), we conclude that the analytic operator-valued generalized Feynman integral $J_{q}^{\mathrm{an}}(F)$ exists and $J_{q}^{\mathrm{an}}(F)$ is an element of $\mathcal {L}(L^2(\mathbb {R},\nu _{\alpha }),L^2(\mathbb {R},\nu _{-\alpha }))$. $\square$

The next two lemmas play key roles in the proof of Theorem 4.5.

Lemma 4.3

For each $k=1,2,\ldots$, let $F_n^{(k)}$ be given by (3.9) with $\theta$ replaced with $\theta ^{(k)}$. Then for all $q\in \mathbb {R}{\setminus }\{0\}$, the analytic operator-valued generalized Feynman integral of $F_n^{(k)}$, $J_{q}^{\mathrm{an}}(F_n^{(k)})$, exists and is given by the formula

$$\begin{aligned} \left( J_{q}^{\mathrm{an}}(F_n^{(k)})(\psi )\right) =\sum _{ \begin{array}{c} l_1 + l_2 = n \\ l_2\not =0 \end{array}}\frac{n! \omega ^{l_2} }{l_2 !} \sum _{j=0}^{l_1} \int _{\Delta _{l_1;j}(T)} \left( \mathcal {L}^{-iq}_{l_1;j;k}\circ \psi \right) (\xi ) \mathrm{d}\prod _{l=1}^{l_1}\beta (s_l) \end{aligned}$$

where $\mathcal {L}^{-iq}_{l_1;j;k}$ is given by the right-hand side of Eq. (3.5) with $\theta$ replaced by $\theta ^{(k)}$. Furthermore, we have

$$\begin{aligned} J_{q}^{\mathrm{an}}\left( F^{(k)}\right) \psi = \sum _{n=1}^{\infty } \beta _n J_{q}^{\mathrm{an}}\left( F_n^{(k)}\right) \psi \end{aligned}$$

(4.4)

where $F^{(k)}:C_{a,b}[0,T]\rightarrow \mathbb C$ is given by

$$\begin{aligned} F^{(k)} (x) = f\left( \int _{0}^{T}\theta ^{(k)} (s,x(s))\mathrm{d}\eta (s)\right) \end{aligned}$$

(4.5)

for each $k=1,2,\ldots .$

Proof

The proof is straightforward by replacing $\theta$ with $\theta ^{(k)}$ in Theorem 4.1. $\square$

Lemma 4.4

Let $F_n^{(k)}$ be as in Lemma 4.3. Then for all $q\in \mathbb {R}{\setminus }\{0\}$ and $\psi \in L^2(\mathbb {R},\nu _{\alpha })$,

$$\begin{aligned} \left\| J_{q}^{\mathrm{an}}(F_n^{(k)}) \psi - J_{q}^{\mathrm{an}}(F_n) \psi \right\| _{L^2(\mathbb {R},\nu _{-\alpha })}\rightarrow 0 \quad \text {as}\quad k\rightarrow \infty . \end{aligned}$$

(4.6)

Proof

To establish Eq. (4.6) it will suffice to show that

$$\begin{aligned} \lim _{k\rightarrow \infty } \int _{\mathbb {R}} \left| \left( J_{q}^{\mathrm{an}}(F_n^{(k)}) \psi \right) (\xi ) -\left( J_{q}^{\mathrm{an}}(F_n) \psi \right) (\xi ) \right| ^2 \hbox {d}\nu _{-\alpha }(\xi ) = 0 \end{aligned}$$

for all $\psi \in L^2(\mathbb {R},\nu _{\alpha })$. But using similar methods as those used in (3.11), it follows that for each $n\in \mathbb N$,

$$\begin{aligned}&\left| \left( J_{q}^{\mathrm{an}}(F_n^{(k)}) \psi \right) (\xi ) -\left( J_{q}^{\mathrm{an}}(F_n) \psi \right) (\xi ) \right| ^2\nonumber \\&\quad \le 2 \left| \left( J_{q}^{\mathrm{an}}(F_n^{(k)}) \psi \right) (\xi )\right| ^2 + 2\left| \left( J_{q}^{\mathrm{an}}(F_n) \psi \right) (\xi )\right| ^2 \nonumber \\&\quad \le 2 \left( \frac{L^2_{Tn} q^2}{\pi \alpha }\right) ^{1/2} \Vert \psi \Vert _{L^2\left( \mathbb {R},\nu _{\alpha }\right) }^2 \exp \left\{ 2M_{1n} \sqrt{ |q|}|\xi |+\frac{M^2_{Tn}}{\alpha }|q|\right\} \nonumber \\&\qquad \times \left( \int _{0}^{T}B^{(k)}(s;|-iq|)\hbox {d}|\eta |(s)\right) ^{2n}\nonumber \\&\qquad +\, 2 \left( \frac{L^2_{Tn} q^2}{\pi \alpha }\right) ^{1/2} \Vert \psi \Vert _{L^2\left( \mathbb {R},\nu _{\alpha }\right) }^2 \exp \left\{ 2M_{1n} \sqrt{ |q|}|\xi |+\frac{M^2_{Tn}}{\alpha }|q|\right\} \nonumber \\&\qquad \times \left( \int _{0}^{T}B(s;|-iq|)\hbox {d}|\eta |(s)\right) ^{2n} \end{aligned}$$

(4.7)

where $B^{(k)}$ is given by Eq. (3.7) with $\theta$ replaced with $\theta ^{(k)}$. Also, the last expression of (4.7) is in $L^2(\mathbb {R},\nu _{-\alpha })$ and it dominates the sequence of functions $|(J_{q}^{\mathrm{an}}(F_n^{(k)}) \psi )(\xi )-(J_{q}^{\mathrm{an}}(F_n) \psi )(\xi ) |^2$. Hence using the dominated convergence theorem, we obtain the desired result. Furthermore, using similar methods as those used in (3.12) we have

$$\begin{aligned} \begin{aligned} \left\| J_{q}^{\mathrm{an}}(F_n^{(k)}) \psi \right\| _{L^2\left( \mathbb {R},\nu _{-\alpha }\right) } \le \Psi ^k_{n}(|-iq|) \Vert \psi \Vert _{L^2\left( \mathbb {R},\nu _{\alpha }\right) } \end{aligned} \end{aligned}$$

(4.8)

and

$$\begin{aligned} \begin{aligned} \left\| J_{q}^{\mathrm{an}}(F_n ) \psi \right\| _{L^2(\mathbb {R},\nu _{-\alpha })} \le \Psi ^1_{n}(|-iq|) \Vert \psi \Vert _{L^2(\mathbb {R},\nu _{\alpha })} \end{aligned} \end{aligned}$$

where $\Psi ^k_n(|-iq|)$ is given by Eq. (3.14). $\square$

We are now ready to establish our main result, namely the stability theorem for the analytic operator-valued generalized Feynman integral.

Theorem 4.5

Let $\{\theta ^{(k)}\}$ be a sequence of complex-valued functions such that $\theta ^{(k)}(s,u)\rightarrow \theta (s,u)$, as $k\rightarrow \infty$, for $\eta \times m_{L}$-a.e. (s, u). For $k=1,2,\ldots$, let the functional $F^{(k)}$ on $C_{a,b}[0,T]$ be given by Eq. (4.5). Then for all $q\in \mathbb {R}{\setminus }\{0\}$ and $\psi \in L^2(\mathbb {R},\nu _{\alpha })$,

$$\begin{aligned} \left\| J_{q}^{\mathrm{an}}(F^{(k)})\psi - J_{q}^{\mathrm{an}}(F)\psi \right\| _{L^2(\mathbb {R},\nu _{-\alpha })}\rightarrow 0 \quad \text {as}\quad k\rightarrow \infty \end{aligned}$$

where $J_{q}^{\mathrm{an}}(F^{(k)})$ is given by Eq. (4.4).

Proof

Using Eqs. (4.3), (4.4) and (4.6) we have that

$$\begin{aligned} \begin{aligned} \lim _{k\rightarrow \infty } J_{q}^{\mathrm{an}} (F^{(k)}) \psi&\mathop {=}\limits ^\mathrm{(I)}\lim _{k\rightarrow \infty } \sum _{n=1}^{\infty }\beta _n J_{q}^{\mathrm{an}} \left( F_n^{(k)}\right) \psi \\&\mathop {=}\limits ^\mathrm{(II)}\sum _{n=1}^{\infty } \lim _{k\rightarrow \infty } \beta _n J_{q}^{\mathrm{an}} \left( F_n^{(k)}\right) \psi \\&\mathop {=}\limits ^\mathrm{(III)}\sum _{n=1}^{\infty } \ \beta _n J_{q}^{\mathrm{an}} (F_n) \psi \\&\mathop {=}\limits ^\mathrm{(IV)}J_{q}^{\mathrm{an}} (F) \psi \end{aligned} \end{aligned}$$

is in $L^2(\mathbb {R},\nu _{-\alpha })$. Step $\mathrm (I)$ follows from Lemma 4.3. From Eqs. (3.13) and (4.8), we have

$$\begin{aligned} \begin{aligned}&\left\| \sum _{n=1}^{\infty }\beta _n J_{q}^{\mathrm{an}} (F_n^{(k)}) \psi \right\| _{L^2(\mathbb {R},\nu _{-\alpha })}\\&\quad \le \sum _{n=1}^{\infty }|\beta _n| \left\| J_{q}^{\mathrm{an}} (F_n^{(k)}) \psi \right\| _{L^2(\mathbb {R},\nu _{-\alpha })}\\&\quad \le \sum _{n=1}^{\infty }|\beta _n| \Psi _{n}^{(k)} (|-iq|) \Vert \psi \Vert _{L^2(\mathbb {R},\nu _{\alpha })}<\infty . \end{aligned} \end{aligned}$$

Also, by using Eqs. (4.6) and (4.8), we can show that $J_{q}^{\mathrm{an}} (F_n^{(k)}) \psi \rightarrow J_{q}^{\mathrm{an}} (F_n ) \psi$ in $L^2(\mathbb {R},\nu _{-\alpha })$ as $k\rightarrow \infty$, and hence, $J_{q}^{\mathrm{an}} (F_n ) \psi$ exists. Hence, Step $\mathrm (II)$ now follows. From Lemma 4.4, we obtain Step $\mathrm (III)$. Step $\mathrm (IV)$ then follows from Theorem 4.2. $\square$