Lebesgue Spaces and Operators with Complex Gaussian Kernels

Abstract

This paper examines the boundedness properties and Parseval-type relations for operators with complex Gaussian kernels over Lebesgue spaces. Furthermore, this study includes an exploration of the Gauss–Weierstrass semigroup, serving as a particular example within the framework of our analysis.

1. Introduction and Preliminaries

The study of operators with complex Gaussian kernels is rooted in functional analysis, where their smoothness and positive definiteness play crucial roles [1]. These properties ensure that the associated Hilbert spaces are well structured, allowing for effective approximation of complex functions. The theoretical foundations, particularly in terms of integral operators and reproducing kernel Hilbert spaces, have facilitated significant advancements in approximation theory.

Recent research has focused on the spectral properties of these operators and their implications for stability and convergence in various contexts [2,3]. Understanding the interplay between complex Gaussian kernels and functional spaces has opened new avenues for solving integral equations and enhancing computational methods in applied mathematics. These operators leverage the mathematical properties of complex-valued functions, enhancing the capacity to analyze high-dimensional spaces.

In this paper, we investigate the integral operator that involves a complex Gaussian kernel acting on a suitable complex-valued function f defined on $\mathbb{R}$ by

$$\begin{array}{c}\hfill \left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)={\int }_{-\infty }^{\infty }f\left(x\right)exp\left\{-\beta y^2-ϵx^2+2\delta xy+\zeta y+\gamma x\right\}dx,\end{array}$$

(1)

where $y\in \mathbb{R},\beta ,ϵ,\delta ,\zeta ,\gamma \in \mathbb{C}$.

This topic initially gained attention in Quantum Field Theory, as noted in [4]. The importance of operators with complex Gaussian kernels (1) lies in their connection to the extended oscillator semigroup, a concept introduced by Howe [5] and further developed by Folland ([6], Chapter 5). Lieb [7] expanded operator (1) into n dimensions and examined its properties in the context of Lebesgue spaces. Negrín [8,9] explored (1) in both degenerate ${(\Re \delta )}^2=(\Re \beta )(\Re ϵ)$ and nondegenerate ${(\Re \delta )}^2<(\Re \beta )(\Re ϵ)$ cases. Recently, Maan and Negrín [10] conducted an in-depth study of operators with complex Gaussian kernels (1) within certain Lebesgue spaces. Additional relevant contributions can be found in [11,12].

For $\eta \in \mathbb{R}$, consider the vector space ${\mathcal{E}}_{\eta }$ consisting of all complex-valued measurable functions f on $\mathbb{R}$ such that $e^{\eta x}f\left(x\right)\in L^{\infty }\left(\mathbb{R}\right)$.

A norm ${\parallel ·\parallel }_{\eta }$ is given by

$$\begin{array}{c}\hfill {\parallel f\parallel }_{\eta }={\parallel e^{\eta x}f\left(x\right)\parallel }_{L^{\infty }\left(\mathbb{R}\right)}.\end{array}$$

With this norm, the map

$$\begin{array}{c}\hfill T_{\eta }:{\mathcal{E}}_{\eta }\to L^{\infty }\left(\mathbb{R}\right)\end{array}$$

where for any $f\in {\mathcal{E}}_{\eta }$

$$\begin{array}{c}\hfill \left(T_{\eta }f\right)\left(x\right)=e^{\eta x}f\left(x\right),x\in \mathbb{R},\end{array}$$

is an isometric isomorphism from ${\mathcal{E}}_{\eta }$ to $L^{\infty }\left(\mathbb{R}\right)$. Thus, since $L^{\infty }\left(\mathbb{R}\right)$ is complete, then the space ${\mathcal{E}}_{\eta }$ becomes a Banach space. Observe that ${\mathcal{E}}_0=L^{\infty }\left(\mathbb{R}\right)$.

For $\eta \in \mathbb{R}$, we also consider the vector space ${\mathcal{G}}_{\eta }$ consisting of all complex-valued measurable functions f on $\mathbb{R}$ such that $e^{\eta x^2}f\left(x\right)\in L^{\infty }\left(\mathbb{R}\right)$.

A norm ${\parallel ·\parallel }_{\eta ,G}$ is given by

$$\begin{array}{c}\hfill {\parallel f\parallel }_{\eta ,G}={\parallel e^{\eta x^2}f\left(x\right)\parallel }_{L^{\infty }\left(\mathbb{R}\right)}.\end{array}$$

With this norm, the map

$$\begin{array}{c}\hfill T_{\eta ,G}:{\mathcal{G}}_{\eta }\to L^{\infty }\left(\mathbb{R}\right)\end{array}$$

where for any $f\in {\mathcal{G}}_{\eta }$

$$\begin{array}{c}\hfill \left(T_{\eta ,G}f\right)\left(x\right)=e^{\eta x^2}f\left(x\right),x\in \mathbb{R},\end{array}$$

is an isometric isomorphism from ${\mathcal{G}}_{\eta }$ to $L^{\infty }\left(\mathbb{R}\right)$. Thus, since $L^{\infty }\left(\mathbb{R}\right)$ is complete, then the space ${\mathcal{G}}_{\eta }$ becomes a Banach space. Observe that ${\mathcal{G}}_0=L^{\infty }\left(\mathbb{R}\right)$.

This paper is organized into five sections. Section 1 acts as an introduction, offering fundamental definitions and notations that are used throughout the paper. Section 2 analyzes the boundedness properties of operator (1) in both degenerate and nondegenerate cases across the spaces ${\mathcal{E}}_{\eta }$, resulting in a Parseval-type relation. Section 3 analyzes the boundedness properties of operator (1) in both degenerate and nondegenerate cases across the space ${\mathcal{G}}_{\eta }$, yielding a Parseval-type relation. Section 4 analyzes the Gauss–Weierstrass semigroup as a particular case of operator (1) concerning the results obtained in Section 2 and Section 3. Finally, Section 5 provides concluding remarks.

2. The Operators with Complex Gaussian Kernels over ${\mathcal{E}}_{\mathbf{\eta }}$

For any $f\in {\mathcal{E}}_{\eta }$ and $\Re ϵ>0$, one has

$$\begin{array}{ccc}& & \left|\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)\right|\hfill \\ & & \le {\parallel f\parallel }_{\eta }{\int }_{-\infty }^{\infty }exp\left\{-(\Re \beta )y^2-(\Re ϵ)x^2+2(\Re \delta )xy+(\Re \zeta )y+(\Re \gamma -\eta )x\right\}dx,\hfill \end{array}$$

for all $y\in \mathbb{R}$.

Using the well-known formula

$$\begin{array}{c}\hfill {\left(2\pi c\right)}^{{\displaystyle \frac{-1}{2}}}{\int }_{-\infty }^{\infty }exp\left\{vx-\left({\displaystyle \frac{x^2}{2c}}\right)\right\}dx=exp\left({\displaystyle \frac{c{v}^2}{2}}\right),v\in \mathbb{C},c>0,\end{array}$$

(2)

one has for $\Re ϵ>0$, $\eta \in \mathbb{R}$, $f\in {\mathcal{E}}_{\eta }$

$$\begin{array}{ccc}& & \left|\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)\right|\hfill \\ & & \le {\parallel f\parallel }_{\eta }·\sqrt{{\displaystyle \frac{\pi }{\Re ϵ}}}exp\left\{\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ}}\right)y^2+\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)y+{\displaystyle \frac{{(\Re \gamma -\eta )}^2}{4\Re ϵ}}\right\}.\hfill \end{array}$$

Therefore, for $\eta \in \mathbb{R}$, $f\in {\mathcal{E}}_{\eta }$, $\Re ϵ>0$ and ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ)$, one obtains

$$\begin{array}{ccc}& & \left|\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)\right|\hfill \\ & & \le {\parallel f\parallel }_{\eta }·\sqrt{{\displaystyle \frac{\pi }{\Re ϵ}}}exp\left\{{\displaystyle \frac{{(\Re \gamma -\eta )}^2}{4\Re ϵ}}\right\}·exp\left\{\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)y\right\},\hfill \end{array}$$

(3)

for all $y\in \mathbb{R}$.

Thus, for $0<q<\infty$, $f\in {\mathcal{E}}_{\eta }$, $\eta \in \mathbb{R}$, $\Re ϵ>0$, ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ)$, with w being a measurable function on $\mathbb{R}$ such that $w>0$ a.e. on $\mathbb{R}$, one has

$$\begin{array}{ccc}\hfill {\int }_{-\infty }^{\infty }{\left|\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)\right|}^{q}w\left(y\right)dy& \le & {\parallel f\parallel }_{\eta }^q·{\left(\sqrt{{\displaystyle \frac{\pi }{\Re ϵ}}}\right)}^q{\left(exp\left\{{\displaystyle \frac{{(\Re \gamma -\eta )}^2}{4\Re ϵ}}\right\}\right)}^q\hfill \\ & ·& {\int }_{-\infty }^{\infty }exp\left\{q\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)y\right\}w\left(y\right)dy.\hfill \end{array}$$

Therefore, we have

Proposition 1.

Assume $\Re ϵ>0$, ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ)$ and let w be a measurable function on $\mathbb{R}$ such that $w>0$ a.e. on $\mathbb{R}$ and $0<q<\infty$. If

$${\int }_{-\infty }^{\infty }exp\left\{q\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)y\right\}w\left(y\right)dy<\infty ,$$

then for any $\eta \in \mathbb{R}$

$$\begin{array}{c}\hfill {\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }:{\mathcal{E}}_{\eta }\to L^q(\mathbb{R},w\left(x\right)dx)\end{array}$$

is a bounded linear operator.

Example 1.

Examples of weights for Proposition 1 are the Gaussian ones:

$$\begin{array}{c}\hfill w\left(x\right)=exp(-a{x}^2),x\in \mathbb{R},a>0.\end{array}$$

Remark 1.

For the particular case when $\eta =0$, $w\left(x\right)={(1+x^2)}^{\alpha }$, $\alpha <{\displaystyle \frac{-1}{2}}$, and ${(\Re \delta )}^2=(\Re \beta )(\Re ϵ)$, and $(\Re ϵ)(\Re \zeta )+(\Re \delta )(\Re \gamma -\eta )=0$, the result of Proposition 1 is contemplated in Theorem 4.1 of [13].

For the case when $f\in {\mathcal{E}}_{\eta }$, $\eta \in \mathbb{R}$, $\Re ϵ>0$, ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ)$, and $(\Re ϵ)(\Re \zeta )+(\Re \delta )(\Re \gamma -\eta )=0$, estimate (3) becomes

$$\begin{array}{c}\hfill \left|\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)\right|\le {\parallel f\parallel }_{\eta }·\sqrt{{\displaystyle \frac{\pi }{\Re ϵ}}}exp\left\{{\displaystyle \frac{{(\Re \gamma -\eta )}^2}{4\Re ϵ}}\right\},forally\in \mathbb{R}.\end{array}$$

Then, one has

Proposition 2.

For $\Re ϵ>0$, ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ)$, $\eta \in \mathbb{R}$, and $(\Re ϵ)(\Re \zeta )+(\Re \delta )(\Re \gamma -\eta )=0$; then,

$$\begin{array}{c}\hfill {\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }:{\mathcal{E}}_{\eta }\to L^{\infty }\left(\mathbb{R}\right)\end{array}$$

is a bounded linear operator.

Remark 2.

For the case when $\eta =0$, the result is contempleted in ([14], Theorem 2.1, (iii)).

Theorem 1.

If $f\in {\mathcal{E}}_{\eta }$, $\eta \in \mathbb{R}$, $\Re ϵ>0$, ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ)$, $\Re \beta >0$, and $g\in L^1\left(\mathbb{R},exp\left\{\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)x\right\}dx\right)$, then the following Parseval-type relation holds:

$$\begin{array}{c}\hfill {\int }_{-\infty }^{\infty }\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(x\right)g\left(x\right)dx={\int }_{-\infty }^{\infty }f\left(x\right)\left({\mathcal{F}}_{ϵ,\beta ,\delta ,\gamma ,\zeta }g\right)\left(x\right)dx.\end{array}$$

Proof. 

Applying Fubini’s theorem in the following, we obtain

$$\begin{array}{ccc}& & {\int }_{-\infty }^{\infty }\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)g\left(y\right)dy\hfill \\ & =& {\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f\left(x\right)exp\left\{-\beta y^2-ϵx^2+2\delta xy+\zeta y+\gamma x\right\}dxg\left(y\right)dy\hfill \\ & =& {\int }_{-\infty }^{\infty }f\left(x\right){\int }_{-\infty }^{\infty }g\left(y\right)exp\left\{-ϵx^2-\beta y^2+2\delta xy+\gamma x+\zeta y\right\}dydx\hfill \\ & =& {\int }_{-\infty }^{\infty }f\left(x\right)\left({\mathcal{F}}_{ϵ,\beta ,\delta ,\gamma ,\zeta }g\right)\left(x\right)dx.\hfill \end{array}$$

The Fubini theorem holds here because from (3) one has

$$\begin{array}{ccc}& & {\int }_{-\infty }^{\infty }\left|\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)\right|\left|g\left(y\right)\right|dy\hfill \\ & & \le {\parallel f\parallel }_{\eta }·\sqrt{{\displaystyle \frac{\pi }{\Re ϵ}}}exp\left\{{\displaystyle \frac{{(\Re \gamma -\eta )}^2}{4\Re ϵ}}\right\}\hfill \\ & & ·{\int }_{-\infty }^{\infty }exp\left\{\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)y\right\}\left|g\left(y\right)\right|dy<\infty .\hfill \end{array}$$

Observe that ${\mathcal{F}}_{ϵ,\beta ,\delta ,\gamma ,\zeta }g$ exists for $g\in L^1\left(\mathbb{R},exp\left\{\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)x\right\}dx\right)$.

In fact,

$$\begin{array}{ccc}& & {\int }_{-\infty }^{\infty }\left|g\left(x\right)\right|\left|exp\left\{-ϵy^2-\beta x^2+2\delta xy+\gamma y+\zeta x\right\}\right|dx\hfill \\ & & ={\int }_{-\infty }^{\infty }\left|g\left(x\right)\right|exp\left\{\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)x\right\}\hfill \\ & & ·exp\left\{-(\Re ϵ)y^2-(\Re \beta )x^2+2(\Re \delta )xy+(\Re \gamma )y+(\Re \zeta )x\right\}\hfill \\ & & ·exp\left\{-\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)x\right\}dx,\hfill \end{array}$$

where $\Re \beta >0$ and for each $y\in \mathbb{R}$

$$\begin{array}{ccc}& & exp\left\{-(\Re ϵ)y^2-(\Re \beta )x^2+2(\Re \delta )xy+(\Re \gamma )y+(\Re \zeta )x\right\}\hfill \\ & & ·exp\left\{-\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)x\right\}\hfill \\ & & \le M_{ϵ,\beta ,\delta ,\gamma ,\zeta ,y},\hfill \end{array}$$

for all $x\in \mathbb{R}$ and for some $M_{ϵ,\beta ,\delta ,\gamma ,\zeta ,y}>0$, one obtains for each $y\in \mathbb{R}$

$$\begin{array}{ccc}& & {\int }_{-\infty }^{\infty }\left|g\left(x\right)\right|\left|exp\left\{-ϵy^2-\beta x^2+2\delta xy+\gamma y+\zeta x\right\}\right|dx\hfill \\ & & \le M_{ϵ,\beta ,\delta ,\gamma ,\zeta ,y}{\int }_{-\infty }^{\infty }\left|g\left(x\right)\right|exp\left\{\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)x\right\}dx<\infty .\hfill \end{array}$$

□

Now observe that

$$\begin{array}{ccc}& & D_x\left(exp\left\{-\beta y^2-ϵx^2+2\delta xy+\zeta y+\gamma x\right\}\right)\hfill \\ & & =\left(-2ϵx+2\delta y+\gamma \right)exp\left\{-\beta y^2-ϵx^2+2\delta xy+\zeta y+\gamma x\right\}.\hfill \end{array}$$

Then,

$$\begin{array}{ccc}& & \left(D_x+2ϵx\right)\left(exp\left\{-\beta y^2-ϵx^2+2\delta xy+\zeta y+\gamma x\right\}\right)\hfill \\ & & =\left(2\delta y+\gamma \right)exp\left\{-\beta y^2-ϵx^2+2\delta xy+\zeta y+\gamma x\right\}.\hfill \end{array}$$

So, for $k\in \mathbb{N}$ and denoting $A_x\equiv D_x+2ϵx$, one obtains

$$\begin{array}{ccc}& & A_x^k\left(exp\left\{-\beta y^2-ϵx^2+2\delta xy+\zeta y+\gamma x\right\}\right)\hfill \\ & & ={\left(2\delta y+\gamma \right)}^{k}exp\left\{-\beta y^2-ϵx^2+2\delta xy+\zeta y+\gamma x\right\},\hfill \end{array}$$

(4)

where $A_x^k$ denotes the k-th power of the operator $A_x$.

Denote by $C_c^k\left(\mathbb{R}\right)$ the space of compactly supported functions on $\mathbb{R}$ which are k-times differentiable functions with continuity.

Theorem 2.

If $f\in C_c^k\left(\mathbb{R}\right),k\in \mathbb{N}$, $\Re ϵ>0$, ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ)$, $\Re \beta >0$, and $g\in L^1\left(\mathbb{R},exp\left\{\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)x\right\}dx\right)$, then the following Parseval-type relation holds:

$$\begin{array}{c}\hfill {\int }_{-\infty }^{\infty }\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(x\right)g\left(x\right){\left(2\delta x+\gamma \right)}^{k}dx={\int }_{-\infty }^{\infty }{A^{\prime }}_x^k\left(f\left(x\right)\right)\left({\mathcal{F}}_{ϵ,\beta ,\delta ,\gamma ,\zeta }g\right)\left(x\right)dx,\end{array}$$

(5)

where $A_x^{\prime }\equiv -D_x+2ϵx$.

Proof. 

The scheme of proof follows from the fact that for $f\in C_c^k\left(\mathbb{R}\right)$, and denoting $A_x^{\prime }\equiv -D_x+2ϵx$ and using (4), it follows that

$$\begin{array}{c}\hfill \left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }\left({A^{\prime }}_x^{k}f\right)\right)\left(y\right)={\left(2\delta y+\gamma \right)}^k\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right),y\in \mathbb{R}.\end{array}$$

(6)

Since $C_c^k\left(\mathbb{R}\right)\subseteq {\mathcal{E}}_{\eta }$, for any $\eta \in \mathbb{R}$, and using Theorem 1 and relation (6), one obtains (5).

Thus, the result holds. □

From estimate (3), one obtains for $f\in {\mathcal{E}}_{\eta }$, $\eta \in \mathbb{R}$, $\Re ϵ>0$, and ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ)$ that

$$\begin{array}{c}\hfill exp\left\{-\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)y\right\}\left|\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)\right|\le {\parallel f\parallel }_{\eta }·\sqrt{{\displaystyle \frac{\pi }{\Re ϵ}}}exp\left\{{\displaystyle \frac{{(\Re \gamma -\eta )}^2}{4\Re ϵ}}\right\},\end{array}$$

for all $y\in \mathbb{R}$, and so

$$\begin{array}{c}\hfill \parallel {\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }{f\parallel }_{-\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)}\le {\parallel f\parallel }_{\eta }·\sqrt{{\displaystyle \frac{\pi }{\Re ϵ}}}exp\left\{{\displaystyle \frac{{(\Re \gamma -\eta )}^2}{4\Re ϵ}}\right\}.\end{array}$$

Therefore, the following proposition follows.

Proposition 3.

For $\eta \in \mathbb{R}$, $\Re ϵ>0$, ${(\Re \delta )}^2\le (\Re ϵ)(\Re \beta )$, one has

$$\begin{array}{c}\hfill {\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }:{\mathcal{E}}_{\eta }\to {\mathcal{E}}_{-\left(\Re \zeta +{\displaystyle \frac{(\Re \delta )(\Re \gamma -\eta )}{\Re ϵ}}\right)}\end{array}$$

which is a bounded linear operator.

Remark 3.

For the particular case when $\Re ϵ>0$, ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ)$, $\eta \in \mathbb{R}$ and $(\Re ϵ)(\Re \zeta )+(\Re \delta )(\Re \gamma -\eta )=0$, this result agrees with Proposition 2.

3. The Operators with Complex Gaussian Kernels over ${\mathcal{G}}_{\mathbf{\eta }}$

Also, using the well-known formula (2) given in Section 2, one has for $\eta \in \mathbb{R}$, $\Re ϵ>\eta$, and $(\Re ϵ-\eta )(\Re \zeta )+(\Re \delta )(\Re \gamma )=0,f\in {\mathcal{G}}_{\eta }$ that

$$\begin{array}{ccc}& & \left|\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)\right|\hfill \\ & & \le {\parallel f\parallel }_{\eta ,G}·\sqrt{{\displaystyle \frac{\pi }{\Re ϵ-\eta }}}exp\left\{\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)y^2\right\}·exp\left\{{\displaystyle \frac{{(\Re \gamma )}^2}{4(\Re ϵ-\eta )}}\right\},\hfill \end{array}$$

(7)

for all $y\in \mathbb{R}$.

Thus, for $0<q<\infty$, $\Re ϵ>\eta$, $(\Re ϵ-\eta )(\Re \zeta )+(\Re \delta )(\Re \gamma )=0,f\in {\mathcal{G}}_{\eta }$, with w being a measurable function on $\mathbb{R}$ such that $w>0$ a.e. on $\mathbb{R}$, one has

$$\begin{array}{ccc}& & {\int }_{-\infty }^{\infty }{\left|\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)\right|}^{q}w\left(y\right)dy\hfill \\ & & \le {\parallel f\parallel }_{\eta ,G}^q·{\left(\sqrt{{\displaystyle \frac{\pi }{\Re ϵ-\eta }}}\right)}^q{\int }_{-\infty }^{\infty }exp\left\{q\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)y^2\right\}w\left(y\right)dy\hfill \\ & & ·exp\left\{{\displaystyle \frac{q{(\Re \gamma )}^2}{4(\Re ϵ-\eta )}}\right\}.\hfill \end{array}$$

Therefore, one has

Proposition 4.

Assume $\eta \in \mathbb{R}$, $\Re ϵ>\eta$ and $(\Re ϵ-\eta )(\Re \zeta )+(\Re \delta )(\Re \gamma )=0$ and let w be a measurable function on $\mathbb{R}$ such that $w>0$ a.e. on $\mathbb{R}$ and $0<q<\infty$. If ${\int }_{-\infty }^{\infty }exp\left\{q\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)y^2\right\}w\left(y\right)dy<\infty$, then

$$\begin{array}{c}\hfill {\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }:{\mathcal{G}}_{\eta }\to L^q(\mathbb{R},w\left(x\right)dx)\end{array}$$

is a bounded linear operator.

Example 2.

Examples of weights for Proposition 4 are

$$\begin{array}{c}\hfill w\left(x\right)=exp(-a{x}^r),x\in \mathbb{R},a>0,r>2.\end{array}$$

Proposition 5.

For $\eta \in \mathbb{R}$, $\Re ϵ>\eta$, ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ-\eta )$, and $(\Re ϵ-\eta )(\Re \zeta )+(\Re \delta )(\Re \gamma )=0$,

$$\begin{array}{c}\hfill {\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }:{\mathcal{G}}_{\eta }\to L^{\infty }\left(\mathbb{R}\right)\end{array}$$

is a bounded linear operator.

Proof. 

The scheme of proof follows from the fact that for the case when $\eta \in \mathbb{R}$, $f\in {\mathcal{G}}_{\eta }$, ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ-\eta )$, $\Re ϵ>\eta$, $(\Re ϵ-\eta )(\Re \zeta )+(\Re \delta )(\Re \gamma )=0$, estimate (7) becomes:

$$\begin{array}{c}\hfill \left|\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)\right|\le {\parallel f\parallel }_{\eta ,G}·\sqrt{{\displaystyle \frac{\pi }{\Re ϵ-\eta }}}exp\left\{{\displaystyle \frac{{(\Re \gamma )}^2}{4(\Re ϵ-\eta )}}\right\},\mathrm{for}\mathrm{all}y\in \mathbb{R}.\end{array}$$

Thus, ${\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }:{\mathcal{G}}_{\eta }\to L^{\infty }\left(\mathbb{R}\right)$ is a bounded linear operator. □

Theorem 3.

If $f\in {\mathcal{G}}_{\eta }$, $\eta \in \mathbb{R}$, $\Re ϵ>\eta$, $\Re \delta \ne 0$, ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ-\eta )$, $(\Re ϵ-\eta )(\Re \zeta )+(\Re \delta )(\Re \gamma )=0$, $g\in L^1\left(\mathbb{R},exp\left\{\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)x^2\right\}dx\right)$, then the following Parseval-type relation holds:

$$\begin{array}{c}\hfill {\int }_{-\infty }^{\infty }\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(x\right)g\left(x\right)dx={\int }_{-\infty }^{\infty }f\left(x\right)\left({\mathcal{F}}_{ϵ,\beta ,\delta ,\gamma ,\zeta }g\right)\left(x\right)dx.\end{array}$$

Proof. 

Applying Fubini’s theorem in the following, we obtain

$$\begin{array}{ccc}& & {\int }_{-\infty }^{\infty }\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)g\left(y\right)dy\hfill \\ & & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f\left(x\right)exp\left\{-\beta y^2-ϵx^2+2\delta xy+\zeta y+\gamma x\right\}dxg\left(y\right)dy\hfill \\ & & ={\int }_{-\infty }^{\infty }f\left(x\right){\int }_{-\infty }^{\infty }g\left(y\right)exp\left\{-ϵx^2-\beta y^2+2\delta xy+\gamma x+\zeta y\right\}dydx\hfill \\ & & ={\int }_{-\infty }^{\infty }f\left(x\right)\left({\mathcal{F}}_{ϵ,\beta ,\delta ,\gamma ,\zeta }g\right)\left(x\right)dx.\hfill \end{array}$$

The Fubini theorem holds here because from (7) one has

$$\begin{array}{ccc}\hfill {\int }_{-\infty }^{\infty }\left|\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)\right|\left|g\left(y\right)\right|dy& \le & {\parallel f\parallel }_{\eta ,G}·\sqrt{{\displaystyle \frac{\pi }{\Re ϵ-\eta }}}exp\left\{{\displaystyle \frac{{(\Re \gamma )}^2}{4(\Re ϵ-\eta )}}\right\}\hfill \\ & ·& {\int }_{-\infty }^{\infty }exp\left\{\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)y^2\right\}\left|g\left(y\right)\right|dy<\infty .\hfill \end{array}$$

Observe that ${\mathcal{F}}_{ϵ,\beta ,\delta ,\gamma ,\zeta }g$ exists for $g\in L^1\left(\mathbb{R},exp\left\{\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)x^2\right\}dx\right)$.

In fact,

$$\begin{array}{ccc}& & {\int }_{-\infty }^{\infty }\left|g\left(x\right)\right|\left|exp\left\{-ϵy^2-\beta x^2+2\delta xy+\gamma y+\zeta x\right\}\right|dx\hfill \\ & & ={\int }_{-\infty }^{\infty }\left|g\left(x\right)\right|exp\left\{\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)x^2\right\}\hfill \\ & & ·exp\left\{-(\Re ϵ)y^2-(\Re \beta )x^2+2(\Re \delta )xy+(\Re \gamma )y+(\Re \zeta )x\right\}\hfill \\ & & ·exp\left\{-\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)x^2\right\}dx\hfill \\ & & <\infty ,\hfill \end{array}$$

where $\Re \delta \ne 0$ and for each $y\in \mathbb{R}$

$$\begin{array}{ccc}& & exp\left\{-(\Re ϵ)y^2-(\Re \beta )x^2+2(\Re \delta )xy+(\Re \gamma )y+(\Re \zeta )x\right\}\hfill \\ & & ·exp\left\{-\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)x^2\right\}\hfill \\ & & \le M_{ϵ,\beta ,\delta ,\gamma ,\zeta ,y},\hfill \end{array}$$

for all $x\in \mathbb{R}$ and for some $M_{ϵ,\beta ,\delta ,\gamma ,\zeta ,y}>0$.

Thus, for $\Re \delta \ne 0$ and for each $y\in \mathbb{R}$,

$$\begin{array}{ccc}& & {\int }_{-\infty }^{\infty }\left|g\left(x\right)\right|\left|exp\left\{-ϵy^2-\beta x^2+2\delta xy+\gamma y+\zeta x\right\}\right|dx\hfill \\ & & \le M_{ϵ,\beta ,\delta ,\gamma ,\zeta ,y}{\int }_{-\infty }^{\infty }\left|g\left(x\right)\right|exp\left\{\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)x^2\right\}dx<\infty .\hfill \end{array}$$

□

Theorem 4.

If $f\in C_c^k\left(\mathbb{R}\right),k\in \mathbb{N}$, $\eta \in \mathbb{R}$, $\Re ϵ>\eta$, $\Re \delta \ne 0$, ${(\Re \delta )}^2\le (\Re \beta )(\Re ϵ-\eta )$, $(\Re ϵ-\eta )(\Re \zeta )+(\Re \delta )(\Re \gamma )=0$, and $g\in L^1\left(\mathbb{R},exp\left\{\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)x^2\right\}dx\right)$, then the following Parseval-type relation holds:

$$\begin{array}{c}\hfill {\int }_{-\infty }^{\infty }\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(x\right)g\left(x\right){\left(2\delta x+\gamma \right)}^{k}dx={\int }_{-\infty }^{\infty }{A^{\prime }}_x^k\left(f\left(x\right)\right)\left({\mathcal{F}}_{ϵ,\beta ,\delta ,\gamma ,\zeta }g\right)\left(x\right)dx,\end{array}$$

(8)

where $A_x^{\prime }\equiv -D_x+2ϵx$.

Proof. 

The scheme of proof follows from the fact that $C_c^k\left(\mathbb{R}\right)\subseteq {\mathcal{G}}_{\eta }$ for any $\eta \in \mathbb{R}$, and using Theorem 3 and relation (6) one obtains (8).

Thus, the result holds. □

From estimate (7), one obtains for $f\in {\mathcal{G}}_{\eta }$, $\eta \in \mathbb{R}$

$$\begin{array}{c}\hfill exp\left\{-\left(-\Re \beta +{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)y^2\right\}\left|\left({\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }f\right)\left(y\right)\right|\le {\parallel f\parallel }_{\eta }·\sqrt{{\displaystyle \frac{\pi }{\Re ϵ-\eta }}}exp\left\{{\displaystyle \frac{{(\Re \gamma )}^2}{4(\Re ϵ-\eta )}}\right\},\end{array}$$

for all $y\in \mathbb{R}$, and so

$$\begin{array}{c}\hfill \parallel {\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }{f\parallel }_{\left(\Re \beta -{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)}\le {\parallel f\parallel }_{\eta ,G}·\sqrt{{\displaystyle \frac{\pi }{\Re ϵ-\eta }}}exp\left\{{\displaystyle \frac{{(\Re \gamma )}^2}{4(\Re ϵ-\eta )}}\right\}.\end{array}$$

Therefore, it follows that

Proposition 6.

For $\eta \in \mathbb{R}$, $\Re ϵ>\eta$, $(\Re ϵ-\eta )(\Re \zeta )+(\Re \delta )(\Re \gamma )=0$, one has

$$\begin{array}{c}\hfill {\mathcal{F}}_{\beta ,ϵ,\delta ,\zeta ,\gamma }:{\mathcal{G}}_{\eta }\to {\mathcal{G}}_{\left(\Re \beta -{\displaystyle \frac{{(\Re \delta )}^2}{\Re ϵ-\eta }}\right)}\end{array}$$

which is a bounded linear operator.

4. The Gauss–Weierstrass Semigroup as a Particular Case

The Gauss–Weierstrass semigroup on $\mathbb{R}$ (see ([15], p. 521) and [16]) is given by

$$\begin{array}{c}\hfill \left(e^{z\Delta }f\right)\left(y\right)={\left(4\pi z\right)}^{-{\displaystyle \frac{1}{2}}}{\int }_{-\infty }^{\infty }f\left(x\right)exp\left\{-{\displaystyle \frac{{(y-x)}^2}{4z}}\right\}dx,\end{array}$$

where $\Re z\ge 0$ and $z\ne 0$.

Except for the factor ${\left(4\pi z\right)}^{-{\displaystyle \frac{1}{2}}}$, this integral operator corresponds to the particular case of (1) when the parameters are given by

$$\begin{array}{c}\hfill \beta =ϵ=\delta ={\displaystyle \frac{1}{4z}}and\zeta =\gamma =0,\Re z\ge 0,z\ne 0.\end{array}$$

Observe that for this case

$$\begin{array}{c}\hfill \Re \beta =\Re ϵ=\Re \delta ={\displaystyle \frac{\Re z}{{4\left|z\right|}^2}},\Re z\ge 0,z\ne 0.\end{array}$$

Now, as a consequence of Proposition 1, the next proposition follows.

Proposition 7.

Let $\eta \in \mathbb{R}$, $\Re z>0$ and let w be a measurable function on $\mathbb{R}$ such that $w>0$ a.e. on $\mathbb{R}$ and $0<q<\infty$. If ${\int }_{-\infty }^{\infty }exp\left\{-q\eta y\right\}w\left(y\right)dy<\infty$, then

$$\begin{array}{c}\hfill e^{z\Delta }:{\mathcal{E}}_{\eta }\to L^q(\mathbb{R},w\left(x\right)dx)\end{array}$$

is a bounded linear operator.

Also, from Proposition 2, one has

Proposition 8.

For $\Re z>0$, then

$$\begin{array}{c}\hfill e^{z\Delta }:L^{\infty }\left(\mathbb{R}\right)\to L^{\infty }\left(\mathbb{R}\right)\end{array}$$

is a bounded linear operator.

Remark 4.

This result is contempleted in ([14], Theorem 3.1, (iii)).

From Theorem 1, one obtains

Theorem 5.

If $f\in {\mathcal{E}}_{\eta }$, $\eta \in \mathbb{R}$, $g\in L^1\left(\mathbb{R},exp\left\{-\eta x\right\}dx\right)$, $\Re z>0$, then the following Parseval-type relation holds:

$$\begin{array}{c}\hfill {\int }_{-\infty }^{\infty }\left(e^{z\Delta }f\right)\left(x\right)g\left(x\right)dx={\int }_{-\infty }^{\infty }f\left(x\right)\left(e^{z\Delta }g\right)\left(x\right)dx.\end{array}$$

Also, from Theorem 2 one has

Theorem 6.

If $f\in C_c^k\left(\mathbb{R}\right),k\in \mathbb{N}$, $\eta \in \mathbb{R}$, $g\in L^1\left(\mathbb{R},exp\left\{-\eta x\right\}dx\right)$, $\Re z>0$, then the following Parseval-type relation holds:

$$\begin{array}{c}\hfill {\int }_{-\infty }^{\infty }\left(e^{z\Delta }f\right)\left(x\right)g\left(x\right){\left({\displaystyle \frac{x}{2z}}\right)}^{k}dx={\int }_{-\infty }^{\infty }{A_x^{\prime }}^k\left(f\left(x\right)\right)\left(e^{z\Delta }g\right)\left(x\right)dx,\end{array}$$

where $A_x^{\prime }\equiv -D_x+{\displaystyle \frac{1}{2z}}x$.

Proposition 9.

For $\eta \in \mathbb{R}$, $\Re z>0$, one has

$$\begin{array}{c}\hfill e^{z\Delta }:{\mathcal{E}}_{\eta }\to {\mathcal{E}}_{{\displaystyle \frac{{4\left|z\right|}^2\eta }{\Re z}}}\end{array}$$

is a bounded linear operator.

Proof. 

The proof is an immediate consequence of Proposition 3. □

Remark 5.

For $\Re z>0$ and $\eta =0$, this result agrees with Proposition 2.

As a consequence of Proposition 4, the next proposition follows.

Proposition 10.

Let $\eta \in \mathbb{R}$, $z\ne 0$, $\Re z\ge 0$, ${\Re z>4\left|z\right|}^2\eta$ and let w be a measurable function on $\mathbb{R}$ such that $w>0$ a.e. on $\mathbb{R}$ and $0<q<\infty$. If ${\int }_{-\infty }^{\infty }exp\left\{{\displaystyle \frac{q\eta \Re z}{(\Re z-4|z{|}^2\eta )}}{y}^2\right\}w\left(y\right)dy<\infty$, then

$$\begin{array}{c}\hfill e^{z\Delta }:{\mathcal{G}}_{\eta }\to L^q(\mathbb{R},w\left(x\right)dx)\end{array}$$

is a bounded linear operator.

Also, from Proposition 5, one has

Proposition 11.

For $\eta \in \mathbb{R}$, $z\ne 0$, $\Re z\ge 0$, $\eta \le 0$, then

$$\begin{array}{c}\hfill e^{z\Delta }:{\mathcal{G}}_{\eta }\to L^{\infty }\left(\mathbb{R}\right)\end{array}$$

is a bounded linear operator.

From Theorem 3, one obtains

Theorem 7.

If $f\in {\mathcal{G}}_{\eta }$, $\eta \in \mathbb{R}$, $z\ne 0$, $\Re z\ge 0$, ${\Re z>4\left|z\right|}^2\eta$, and $g\in L^1\left(\mathbb{R},exp\left\{{\displaystyle \frac{\eta \Re z}{(\Re z-4|z{|}^2\eta )}}{y}^2\right\}dx\right)$, then the following Parseval-type relation holds:

$$\begin{array}{c}\hfill {\int }_{-\infty }^{\infty }\left(e^{z\Delta }f\right)\left(x\right)g\left(x\right)dx={\int }_{-\infty }^{\infty }f\left(x\right)\left(e^{z\Delta }g\right)\left(x\right)dx.\end{array}$$

Also, from Theorem 4, one has

Theorem 8.

If $f\in C_c^k\left(\mathbb{R}\right),k\in \mathbb{N}$, $\eta \in \mathbb{R}$, $z\ne 0$, $\Re z\ge 0$, ${\Re z>4\left|z\right|}^2\eta$, and $g\in L^1\left(\mathbb{R},exp\left\{{\displaystyle \frac{\eta \Re z}{(\Re z-4|z{|}^2\eta )}}{y}^2\right\}dx\right)$, then the following Parseval-type relation holds:

$$\begin{array}{c}\hfill {\int }_{-\infty }^{\infty }\left(e^{z\Delta }f\right)\left(x\right)g\left(x\right){\left({\displaystyle \frac{x}{2z}}\right)}^{k}dx={\int }_{-\infty }^{\infty }{A_x^{\prime }}^k\left(f\left(x\right)\right)\left(e^{z\Delta }g\right)\left(x\right)dx,\end{array}$$

where $A_x^{\prime }\equiv -D_x+{\displaystyle \frac{1}{2z}}x$.

Also, the following proposition is obtained from Proposition 6.

Proposition 12.

For $\eta \in \mathbb{R}$, $z\ne 0$, $\Re z\ge 0$, ${\Re z>4\left|z\right|}^2\eta$, one has

$$\begin{array}{c}\hfill e^{z\Delta }:{\mathcal{G}}_{\eta }\to {\mathcal{E}}_{{\displaystyle \frac{-\eta \Re z}{(\Re z-4|z{|}^2\eta )}}}\end{array}$$

which is a bounded linear operator.

5. Conclusions

This paper investigates the boundedness properties of operators with complex Gaussian kernels across both degenerate and nondegenerate cases within the Lebesgue spaces ${\mathcal{E}}_{\eta }$ and ${\mathcal{G}}_{\eta }$. This analysis results in the derivation of Parseval-type relations. Additionally, this study examines the Gauss–Weierstrass semigroup as a specific instance arising from these operators with complex Gaussian kernels.