1 Introduction

In 1938, Delsarte introduced the notion of generalized translation; see [13]. His starting point was the modification of Taylor’s formula by the eigenfunctions of an ordinary differential operator: \(D_xf(\lambda , \cdot )=\lambda f(\lambda ,x)\). The Taylor series of \(f=f(\lambda , z)\) around \(z_0\) at y, with the notation \(x=y-z_0\), can be expressed as

$$\begin{aligned} f(\lambda ,x)=\sum _{k=0}^\infty \lambda ^k \varphi _k(x), \end{aligned}$$

where \(D_x\varphi _0(x)=0\) and \(D_x\varphi _k(x)=\varphi _{k-1}(x)\) (\(\varphi _{-1}=0\)). Moreover, \(\varphi _k(0)=0\) if \(k>0\), and \(\varphi _0(0)=1\).

He illustrated his idea with the following two examples:

\(D_x(f)=f'\), \(f(\lambda ,x)=e^{\lambda x}\), \(\varphi _k(x)=\frac{x^k}{k!}\) and

\(D_x(f)=f''+\frac{2\alpha +1}{x}f'\), \(f(\lambda ,x)=j_\alpha (i\sqrt{\lambda } x)\), \(\varphi _k(x)=\frac{\Gamma (\alpha +1)}{\Gamma (k+1)\Gamma (k+\alpha +1)}\left( \frac{x}{2}\right) ^{2k}\); and introduced the translation operator below:

$$\begin{aligned} T^t_x:=\sum _{k=0}^\infty \varphi _k(t)D_x^k. \end{aligned}$$

The convergence of the series above implies that \(T^t_xf(\lambda ,\cdot )=f(\lambda ,x)f(\lambda ,t)\). Examination of the second example leads to the next chain of ideas. Denoting by \(u(x,t):=T^t_xf\) the construction ensures that the translated functions are solutions to the next Cauchy problem

$$\begin{aligned} D_tu=D_xu, \hspace{4pt}\hspace{4pt}u(x,0)= f(x), \hspace{4pt}\hspace{4pt}\frac{\partial }{\partial t}u(x,0)=0. \end{aligned}$$

The initial study of the properties of generalized translation is due to Delsarte [13] and Levitan; see, e.g., [34]. Braaksma and Snoo in a series of papers dealt with the problem of introducing general translations via certain hyperbolic Cauchy problems; see, e.g., [10, 11]. Examination of product formulae with respect to classical orthogonal polynomials led to definition of Laguerre and Jacobi translations; see, e.g., [18, 20, 26].

General translation has widespread applications. Estimation of the operator norm leads to a maximum principle with respect to the hyperbolic problem in question. It gives certain convolution structures; see, e.g., [6, 32]. It also leads to estimation of p-Christoffel function; see, e.g., [4, 5]. As in the standard case approximation theoretic problems, as smoothness, best approximation or approximation by Cesaro means can be examined by general translation; see, e.g., [27, 37]. By general translation, compactness criteria can be derived too; see [31].

Here, we extend the notion of translation as solutions to Cauchy or Dirichlet problems in parabolic and elliptic cases as well. This approach allows to handle, e.g., the heat-diffusion semigroup as a translation. Although several different examples are listed and investigated below, this one is highlighted, because unlike the old and the other new examples, this one inherits the comfortable semigroup property of the standard translation, i.e., \(T^{t_1}T^{t_2}=T^{t_1+t_2}\). Moreover parabolicity allows to draw conclusions in d-dimension, while the variable of the translation is one-dimensional. We show that the general translations introduced below possess the properties of the standard one.

The paper is organized as follows. In the next section, different types of examples of translation are given. In Sect. 3, the notion of “regularity with respect to compactness” is introduced and is applied to derive Kolmogorov–Riesz-type characterizations in certain \(L^p\)-spaces. The rest of this section is devoted to show that all the listed translations are regular. In Sect. 4, Pego-type characterizations of compact sets of \(L^2\) are derived by convolution method. Section 5 deals with the approximation theoretic aspect of translation that is with modulus of smoothness and K-functional.

2 The translation operator

We define the translation operator as follows.

Definition 2.1

Let \(\Omega =I\times J\subset {\mathbb {R}}^d\times {\mathbb {R}}_{+}\). Let L be a linear partial differential operator of order at most two and take a function \(f(x)\in C(I)\) or \(f\in L^p_\mu (I)\), where \(\mu \) is a Radon measure on I. Define translation as the next linear operator: the translation of f

$$\begin{aligned} T_x^tf:=T^tf(x)=u(x,t), \end{aligned}$$

where u(x, t) is the solution to

$$\begin{aligned}{} & {} Lu=0, \hspace{4pt}(x,t)\in \Omega , \end{aligned}$$
(2.1)
$$\begin{aligned}{} & {} u(x,0)=f(x), \hspace{4pt}x\in I, \end{aligned}$$
(2.2)

where the last equality is meant in \(\sup \)-norm or in p-norm.

Remark 2.2

All of our examples can be given by appropriate integral transformations as well. The corresponding kernel functions are denoted by \(W_t(x,y)\), \(K_t(x,y)\), etc. according to the standard notation. Here, \(t>0\), and \(x,y\in {\mathbb {R}}^d\), \(d\ge 1\). That is

$$\begin{aligned} T^tf(x):=\int _IW_t(x,y)f(y)\textrm{d}y. \end{aligned}$$
(2.3)

It also makes the solution well defined.

Below different types of examples are introduced. Of course, by the same chain of ideas, several further examples can be constructed. Here are the ones we study subsequently.

2.1 Heat-diffusion semigroup associated with Hermite functions

This semigroup, from different point of views, is investigated by several authors; see, e.g., [1, 28, 41] and the references therein.

In \({\mathbb {R}}^d\), the eigenfunctions of the d-dimensional harmonic oscillator (Hermite operator)

$$\begin{aligned} D_x=\Delta _x-|x|^2, \end{aligned}$$
(2.4)

are the d-dimensional Hermite functions

$$\begin{aligned} \tilde{h}_\nu =\tilde{h}_{n_1}\dots \tilde{h}_{n_d}, \end{aligned}$$

where \(\nu =(n_1, \dots ,n_d)\) \(n_i\in {\mathbb {N}}\) and

$$\begin{aligned} \tilde{h}_{k}(x)=h_k(x)e^{-\frac{x^2}{2}}=\frac{1}{\sqrt{2^k k! \sqrt{\pi }}}H_k(x)e^{-\frac{x^2}{2}} \end{aligned}$$

(\(H_k\) are the Hermite polynomials; cf. [42].) The associated heat-diffusion semigroup is given by its kernel function defined on \({\mathbb {R}}^d\times {\mathbb {R}}^d \times {\mathbb {R}}_+\)

$$\begin{aligned} W_t(x,y)=\sum _{n=0}^\infty e^{-(2n+d)t}\sum _{|\nu |=n}\tilde{h}_\nu (x)\tilde{h}_\nu (y). \end{aligned}$$
(2.5)

Thus, for and appropriate f, denoting by \(\tilde{f}(y):=f(y)e^{-\frac{y^2}{2}}\)

Example 2.3

$$\begin{aligned} T^tf(x):=\int _{{\mathbb {R}}^d}W_t(x,y)\tilde{f}(y)\textrm{d}y. \end{aligned}$$

Denoting by \(u(x,t):=T^tf(x)\), if \(\tilde{f}\in L^p({\mathbb {R}}^d)\), \(1\le p \le \infty \), then

$$\begin{aligned} \left( \frac{\partial }{\partial t}-D_x\right) u(x,t)=0 \end{aligned}$$
(2.6)

and

$$\begin{aligned} \lim _{t\rightarrow 0+}\Vert T^tf-\tilde{f}\Vert _p=0; \end{aligned}$$

cf. [41, Proposition 2.5 and Theorem 2.6].

2.2 Further Poisson integrals: elliptic and parabolic equations

Following the previous track of thoughts translation of a function in \(f\in L^p({\mathbb {R}})\) can be defined by Poisson integrals. First, we take the simplest elliptic and parabolic equations on the upper half-plane and define translations by standard convolution.

In the next example, \(T_x^tf\) is a harmonic function on the (open) upper half-plane with limit f on the real line and with limit zero at infinity, cf. e.g., [30], that is the solution to this Dirichlet problem

$$\begin{aligned} \Delta u(x,t)=0, \hspace{4pt}\hspace{4pt}(x,t) \in {\mathbb {R}}\times {\mathbb {R}}_+, \hspace{4pt}\hspace{4pt}u(x,0)=f(x). \end{aligned}$$

It is given by the Poisson integral below.

Example 2.4

Let \(f\in L^p({\mathbb {R}})\), \(1\le p <\infty \), or \(f\in C_b({\mathbb {R}})\)

$$\begin{aligned} T^tf(x):=\frac{1}{\pi }\int _{\mathbb {R}}f(\xi )\frac{t}{(\xi -x)^2+t^2}\textrm{d}t=f*K_{t,e}(x), \end{aligned}$$
(2.7)

where \(C_b({\mathbb {R}})\) stands for continuous and bounded functions on the real line.

Considering the Cauchy problem

$$\begin{aligned} u_t(x,t)=b^2\Delta u(x,t) \hspace{4pt}\hspace{4pt}(x,t)\in {\mathbb {R}}\times (0,\infty ), \hspace{4pt}u(x,0)=f(x), \end{aligned}$$

the next translation can be defined by the corresponding Poisson integral.

Example 2.5

Let \(f\in L^p({\mathbb {R}})\) (\(1\le p \le \infty \))

$$\begin{aligned} T^tf(x):=\int _{\mathbb {R}}f(\xi )\frac{e^{-\frac{(x-\xi )^2}{4b^2t}}}{2b\sqrt{\pi }\sqrt{t}}\textrm{d}t=f*K_{t,p}(x). \end{aligned}$$
(2.8)

We continue with an elliptic example again, where the Poisson integral is not of convolution type.

Let \(P_n^{(\lambda )}(\cos \vartheta )\) be the nth ultraspherical polynomial (\(\lambda >-\frac{1}{2}\), \(\vartheta \in [0,\pi ]\)) orthogonal with respect to \(\textrm{d}\mu (\vartheta )=\sin ^{2\lambda }\vartheta \textrm{d}\vartheta \); see, e.g., [42]. An \(f \in L^p_\mu \) with some \(1 \le p \le \infty ([0,\pi ])\) has an expansion \(f\sim \sum _{k=0}^\infty a_k P_k^{(\lambda )}(\cos \vartheta )\) with \(a_k=\gamma _k\int _0^\pi f(\vartheta )P_k^{(\lambda )}(\cos \vartheta )\textrm{d}\mu (\vartheta )\). In [35], the authors examined the next Poisson integral of f

$$\begin{aligned} u(x,t)=f(r,\vartheta )=\sum _{k=0}^\infty a_k r^kP_k^{(\lambda )}(\cos \vartheta )=\int _0^\pi P(r,\vartheta ,\varphi ) f(\varphi )\textrm{d}\mu (\varphi ), \end{aligned}$$

where \(0<r<1\), \(x=r\cos \vartheta \), \(t=r\sin \vartheta \). The Poisson kernel is

$$\begin{aligned} P(r,\vartheta ,\varphi )=\sum _{k=0}^\infty \gamma _k r^k P_k^{(\lambda )}(\cos \vartheta )P_k^{(\lambda )}(\cos \varphi ). \end{aligned}$$

Then, u(x, t) satisfies the differential equation

$$\begin{aligned} u_{xx}+u_{tt}+\frac{2\lambda }{t}u_t=0. \end{aligned}$$

Since \(\Vert f(r,\vartheta )-f(\vartheta )\Vert _{\mu ,p}\rightarrow 0\) as \(r \rightarrow 1\) if \(1\le p<\infty \) and also for \(p=\infty \) if f is continuous, in view of Definition 2.1, we have the next example. As above, the closed form of the Poisson kernel implies the definition below.

Example 2.6

Let \(D=[0,\pi ]\times [0,1)\), \(f \in L^p_\mu ([0,\pi ])\)

$$\begin{aligned} T^tf(x)=u(x,t)=T^rf(\vartheta )=f(r,\vartheta )=\int _0^\pi P(r,\vartheta ,\varphi ) f(\varphi )\textrm{d}\mu (\varphi ), \end{aligned}$$

where

$$\begin{aligned}{} & {} P(r,\vartheta ,\varphi )\nonumber \\{} & {} \quad =\frac{\lambda (1-r^2)}{\pi }\int _0^\pi \frac{\sin ^{2\lambda -1}\xi }{\left( 1-2r(\cos \vartheta \cos \varphi +\sin \vartheta \sin \varphi \cos \xi )+r^2\right) ^{\lambda +1}}\textrm{d}\xi . \end{aligned}$$
(2.9)

Considering the next differential operator

$$\begin{aligned} D_x=L^{\alpha }:=\frac{1}{2}\left( -\frac{\textrm{d}^2}{\textrm{d}x^2}+x^2+\frac{\alpha ^2-\frac{1}{4}}{x^2}\right) \end{aligned}$$

with eigenfunctions

$$\begin{aligned} L^{\alpha }\varphi _n^{(\alpha )}=(2n+\alpha +1)\varphi _n^{(\alpha )}, \end{aligned}$$

where

$$\begin{aligned} \varphi _n^{(\alpha )}(x)=\sqrt{\frac{\Gamma (n+1)}{\Gamma (n+\alpha +1)}} e^{-\frac{x^2}{2}}L_n^{(\alpha )}(x^2)\sqrt{2x} \end{aligned}$$
(2.10)

(\(L_n^{(\alpha )}\) is the standard Laguerre polynomial; see, e.g., [42]), we define a translation by the solution to the parabolic Cauchy problem

$$\begin{aligned} u_t+L^{\alpha }u=u_t-\frac{1}{2}u_{xx}+\frac{1}{2} \left( x^2+\frac{\alpha ^2-\frac{1}{4}}{x^2}\right) u=0, \hspace{4pt}\hspace{4pt}u(x,0)=g(x), \end{aligned}$$

\(x,t\in (0,\infty )\), \(\alpha \ge -\frac{1}{2}\), cf. [12].

Example 2.7

$$\begin{aligned} T^tg(x):=\int _0^\infty W_t^\alpha (x,y)g(y)\textrm{d}y, \end{aligned}$$

where

$$\begin{aligned} W_t^\alpha (x,y)=2\sqrt{xy}\frac{e^{-t}}{1-e^{-2t}}I_\alpha \left( \frac{2xye^{-t}}{1-e^{-2t}}\right) e^{-\frac{1}{2}(x^2+y^2)\frac{1+e^{-2t}}{1-e^{-2t}}}, \end{aligned}$$

where \(I_\alpha (z)=\sum _{k=0}^\infty \frac{\left( \frac{z}{2}\right) ^{2k+\alpha }}{\Gamma (k+1)\Gamma (k+\alpha +1)}\) is the modified Bessel function.

2.3 Fourier method: hyperbolic equation

The first—well-known—examples are given by hyperbolic equations generated by Sturm–Liouville-type operators; see, e.g., [13, 34]. Consider the operator

$$\begin{aligned} D_x=\frac{\partial ^2}{\partial x^2}+q(x)\frac{\partial }{\partial x}-r(x). \end{aligned}$$

If

$$\begin{aligned} D_x\varphi (x)=a\varphi (x), \end{aligned}$$

denoting by \(u(x,t)=\varphi (x)\varphi (t)\), u(x, t) fulfils the equation below

$$\begin{aligned} u_{xx}-u_{tt}+q(x)u_x-q(t)u_t-(r(x)-r(t))u=0. \end{aligned}$$
(2.11)

In Laguerre and Bessel cases, \(q(x)=\frac{2\alpha +1}{x}\), where \(\alpha \ge -\frac{1}{2}\) and \(r(x)=x^2\) or \(r(x)=0\), respectively. The eigenfunctions of \(D_x\) are the Laguerre functions \({\mathcal {L}}_n^{(\alpha )}(x)=\frac{n!\Gamma (\alpha +1)}{\Gamma (n+\alpha +1)}L_n^{(\alpha )}(x^2)e^{-\frac{x^2}{2}}\), and the Bessel functions \(j_\alpha (\lambda x)\), where \(L_n^{(\alpha )}(x)\) and \(j_\alpha (\lambda x)\) stand for the classical Laguerre polynomials and the entire Bessel functions, respectively. These examples are studied by several authors; see, e.g., [4,5,6, 26, 37].

In Jacobi case with \(\alpha \ge \beta \ge -\frac{1}{2}\), \(\alpha > -\frac{1}{2}\) by the same argument, \(q(x)=\frac{(\beta +1)\cos x-\beta }{\sin x}\) and \(r(x)=\frac{\alpha (\alpha -2\beta +1)\cos x}{4(1-\cos x)}\). The eigenfunctions are the Jacobi functions, \({\mathcal {P}}^{(\alpha ,\beta )}_n(x)=c_nP^{(\alpha ,\beta )}_n(\cos x)\sin ^\alpha \frac{x}{2}\), where \(P_n^{(\alpha , \beta )}(x)\)-s are the classical Jacobi polynomials. For Jacobi translation, see, e.g., [3, 20].

In the listed cases, the eigenfunctions of the Sturm–Liouville operators \(\{u_k\}_{k=0}^\infty \) form Schauder bases in the corresponding (weighted) spaces, that is, if the initial condition is given by \(f(x)\sim \sum _{k=0}^\infty a_ku_k(x)\), then the solution to the Cauchy problem can be given as \(T^tf(x)=u(x,t)\sim \sum _{k=0}^\infty a_ku_k(x)u_k(t).\)

Thus, translation generated by a hyperbolic equation is as follows:

$$\begin{aligned} T^tf(x)=u(x,t) \end{aligned}$$

if

$$\begin{aligned} L_hu=0, \hspace{4pt}\hspace{4pt}(x,t)\in I\times I; \hspace{4pt}\hspace{4pt}u(x,0)=f(x), \hspace{4pt}\hspace{4pt}u_t(x,0)=0,\hspace{4pt}x\in I. \end{aligned}$$
(2.12)

Here, \(L_h\) is given by (2.11), where \(I={\mathbb {R}}_+\) in Laguerre and Bessel cases and in Jacobi case \(I=(0,\pi )\).

It is pointed out in a series of papers that Bessel, Laguerre, and Jacobi translations are bounded operators; see the cited papers above and the references therein. The method described above implies the symmetry of the translation derived by a Sturm–Liouville equation; see [10, 11].

Next, we give an example by Jacobi functions. Let \(\alpha \ge \beta \ge -\frac{1}{2}\), \(\alpha > -\frac{1}{2}\) again and let us denote by \(\varrho =\alpha +\beta +1\), and now, let \(\tau \in {\mathbb {R}}\). Let

$$\begin{aligned} w(x):=2^{2\varrho }\sinh ^{2\alpha +1}x\cosh ^{2\beta +1}x. \end{aligned}$$
(2.13)

With \(q(x)=\frac{w'}{w}(x)\), \(\varphi _\tau ^{(\alpha ,\beta )}(x)\) fulfils the differential equation

$$\begin{aligned} y''+qy'+(\varrho ^2+\tau ^2)y=0, \end{aligned}$$

where

$$\begin{aligned} \varphi _\tau ^{(\alpha ,\beta )}(x):=\varphi _\tau (x)=_2F_1\left( \frac{\varrho +i\tau }{2}, \frac{\varrho -i\tau }{2};\alpha +1;-\sinh ^2x\right) \end{aligned}$$

are the Jacobi functions. Thus, the translation is defined as it follows; cf., e.g., [21, 38, 39] and the references therein.

Example 2.8

Let f be a suitable even function on \({\mathbb {R}}\). \(T^tf(x)=u(x,t)\) if

$$\begin{aligned} u_{xx}-u_{tt}+q(x)u_x-q(t)u_t=0, \hspace{4pt}\hspace{4pt}u(x,0)=f(x), \hspace{4pt}\hspace{4pt}u_t(x,0)=0,\hspace{4pt}x\in {\mathbb {R}}, \end{aligned}$$

where

$$\begin{aligned} q(x)=(2\alpha +1)\coth x+(2\beta +1)\tanh x. \end{aligned}$$

3 Regularity with respect to compactness

In this section, we derive Kolmogorov–Riesz-type compactness criteria in certain \(L^p\)-spaces. For the original theorem, see, e.g., [2]. This theorem has several extensions to different function spaces with standard translation; see, e.g., [7, 15, 24, 25, 29, 40] and with Bessel and Laguerre translations, see [31]. In all but one of our examples, the translation is not symmetric. This implies that one of the criteria of compactness has to be splitted into two different assumptions; see Definition 3.4 below.

We start with some notation. Let \(\mu \) and \(\nu \) be Radon measures on I and J, respectively, and \(1\le p<\infty \). \(\Vert f\Vert _{\mu (x),p}=\left( \int _I|f|^p\textrm{d}\mu \right) ^{\frac{1}{p}}\). \(T^0f(x)=f(x)\).

We define the “norm” of the translation operator \(M_T\) as

$$\begin{aligned} M_T:=M_{T,p}=\sup _{t}\sup _{\Vert f\Vert _{\mu ,p}\le 1}\Vert T^tf(x)\Vert _{\mu (x),p}, \hspace{4pt}\hspace{4pt}1\le p \le \infty . \end{aligned}$$
(3.1)

We introduce the next notation: Let \(a>0\), \(M_0>0\) fixed, \(B_a:=\{\Vert x\Vert \le a\}\cap I\), or \(B_a:=(-a,a)\cap J\); \(A:=\int _{B_a}1\textrm{d}\nu (t)\)

$$\begin{aligned} M_{a,R}f(x):=\left\{ \begin{array}{ll}\frac{1}{A}\int _{B_a}T^tf(x)\textrm{d}\nu (t) \hspace{4pt}|x|<R,\\ 0 \hspace{4pt}|x|\ge R.\end{array}\right. \end{aligned}$$

Definition 3.1

A translation, \(T^t_x\), is regular with respect to p, \(\mu \), \(\nu \) if it fulfils the properties below.

  1. (T1)

    There is a dense set \({\mathcal {E}}_1\subset L^p_\mu (I)\), such that for all \(g\in {\mathcal {E}}_1\), \(0<t<M_0\), and \(\varepsilon >0\), there is a \(\delta =\delta (\varepsilon , M_0, g)>0\), such that for all \(0\le |h| \le \delta \)

    $$\begin{aligned} \left( \int _I \left| \left( T^{t+h} g(x)-T^{t} g(x)\right) \right| ^p\textrm{d}\mu (x)\right) ^{\frac{1}{p}} <\varepsilon . \end{aligned}$$
    (3.2)
  2. (T2)

    There is a dense set \({\mathcal {E}}_2\subset L^p_\mu (I)\), such that for all \(g\in {\mathcal {E}}_2\), and positive numbers \(\varepsilon \), a, R, there is a \(\delta =\delta (g,\varepsilon ,a,R)>0\), such that if \(0\le |h| \le \delta \), then

    $$\begin{aligned} \left| M_{a,R}g(x+h)-M_{a,R}g(x)\right| <\varepsilon , \hspace{4pt}\hspace{4pt}x, x+h \in B_R. \end{aligned}$$
    (3.3)
  3. (T3)

    There is a finite constant, c(a, R), so that for all \(f \subset L^p_\mu (I)\)

    $$\begin{aligned} |M_{a,R}f(x)| \le c(a,R)\Vert f\Vert _{\mu ,p}, \hspace{4pt}\hspace{4pt}\forall \hspace{4pt}x\in I. \end{aligned}$$
    (3.4)

Remark 3.2

If \(T_x^y=T_y^x\) and \(\mu =\nu \), then (3.2) implies (3.3), and if \(M_T<\infty \), it also implies (3.4). Indeed

$$\begin{aligned}{} & {} \left| M_ag(x+h)-M_ag(x)\right| \\{} & {} \quad \le \frac{1}{A} \left( \int _{B_a}1\textrm{d}\mu (y)\right) ^{\frac{1}{p'}}\left( \int _{B_a}|T^{x+h}g(y)-T^{x} g(y)|^p\textrm{d}\mu (y)\right) ^{\frac{1}{p}}, \end{aligned}$$

which ensures (3.3). (3.4) can be obtained in the same way.

Definition 3.3

A set \(K\subset L^p_\mu (I)\) is equivanishing, or we say it fulfils property \(\mathbf{P_a}\) if, for all \(\varepsilon >0\), there is an \(R>0\) such that for all \(f\in K\)

$$\begin{aligned} \left( \int _{I\setminus B_R} |f(x)|^p\textrm{d}\mu (x)\right) ^{\frac{1}{p}}<\varepsilon . \end{aligned}$$

Definition 3.4

A set \(K\subset L^p_\mu (I)\) is equicontinuous in mean if it fulfils the next properties.

\(\mathbf{P_{b_1}}\):

: For all \(\varepsilon \) and \(M_0\) positive numbers, there is a \(\delta >0\) (independent of f), such that for all \(t\in B_{M_0}\), \(0\le |h| \le \delta \) and \(f\in K\) (3.2) is satisfied.

\(\mathbf{P_{b_2}}\):

: For all \(\varepsilon >0\), there is a \(\delta =\delta (a,R)>0\) (independent of f), such that for all \(0\le |h| \le \delta \) and \(f\in K\), (3.3) is satisfied.

Theorem 3.5

With the notation above, let \(T=T^t_x\) be a bounded translation in (3.1)-sense and suppose that it is regular with respect to p, \(\mu \), \(\nu \). Let \(K\subset L^p_\mu (I)\) be a bounded set. Then, K is precompact if and only if it is equivanishing and equicontinuous in mean.

Proof

First, we prove that if K is bounded, equivanishing, and equicontinuous in mean and the translation has the property T3, then K is precompact.

For an arbitrary \(\varepsilon >0\), let R be chosen according to \(\mathbf{P_a}\). Let \(a>0\) be fixed and will be chosen later, and define

$$\begin{aligned} F_{a,R}:=\{M_{a, R}f(x): f\in K, \hspace{4pt}x\in B_R\}. \end{aligned}$$

Then, T3 and the boundedness of K imply that \(F_{a,R}\) is (uniformly) bounded. According to \(\mathbf{P_{b_2}}\), \(F_{a,R}\) is equicontinuous; thus, it is precompact.

Let \(M_af_1, \dots , M_af_n\) be an \(\varepsilon _1\)-net in \(F_{a,R}\), where \(\varepsilon _1=\frac{\varepsilon }{(\mu (B_R))^{\frac{1}{p}}}\). We show that \(f_1, \dots ,f_n\) is a \(5\varepsilon \)-net in K. Recalling the definition of R and A

$$\begin{aligned}{} & {} \Vert M_{a,R}f-f\Vert _{\mu ,p}\le \left\| M_{a,R}f-f|_{B_R}\right\| _{\mu ,p} +\left( \int _{I\setminus B_R}\left| f\right| ^p\textrm{d}\mu \right) ^{\frac{1}{p}}\\{} & {} \quad \le \left( \int _{B_R}\left| \frac{1}{A}\int _{B_a}\left( T^{t}f(x) -f(x)\right) \textrm{d}\nu (t)\right| ^p\textrm{d}\mu (x)\right) ^{\frac{1}{p}}+ \varepsilon \\{} & {} \quad \le \frac{1}{A}\int _{B_a}\left( \int _I\left| T^{t}f(x)-f(x) \right| ^p\textrm{d}\mu (x)\right) ^{\frac{1}{p}}\textrm{d}\nu (t)+ \varepsilon \\{} & {} \quad \le \sup _{0\le |t| \le a}\left\| T^{t}f(x)-f(x)\right\| _{\mu ,p}+ \varepsilon . \end{aligned}$$

Considering \(\mathbf{P_{b_1}}\), if a is small enough, \(\Vert M_{a,R}f-f\Vert _{\mu ,p}\le 2\varepsilon \) for all \(f\in K\). Let \(f\in K\) be arbitrary. Selecting \(f_i\), such that \(|M_{a,R}f(x)-M_{a,R}f_i(x)|\le \varepsilon _1\)

$$\begin{aligned} \Vert M_{a,R}f-M_{a,R}f_i\Vert _{\mu ,p}\le \varepsilon _1(\mu (B_R))^{\frac{1}{p}}=\varepsilon . \end{aligned}$$

Thus, by triangle inequality

$$\begin{aligned} \Vert f-f_i\Vert _{\mu ,p}\le 5\varepsilon . \end{aligned}$$

On the other hand, suppose that K is precompact. Let \(\varepsilon >0\) be arbitrary and \(\varphi _1, \dots ,\varphi _n\) be an \(\frac{\varepsilon }{2}\)-net in K. As \(C_0\), that is the set of compactly supported continuous functions, is dense in \(L^p_\mu (I)\), there are \(g_1, \dots , g_n \in C_0\), so that for all \(f \in K\), there is an i such that \(\Vert g_i-f\Vert _{\mu ,p}<\varepsilon \). Then, R is appropriate with respect to \(\mathbf{P_a}\) if \(B_R\) contains the supports of \(g_i\), \(i=1,\dots n\).

Since T is bounded, selecting \(g_1, \dots , g_n \in {\mathcal {E}}_1\) as above, triangle inequality together with T1 ensures \(\mathbf{P_{b_1}}\).

Similarly, selecting \(g_1, \dots , g_n \in {\mathcal {E}}_2\) as above, T2, T3 and the triangle inequality imply \(\mathbf{P_{b_2}}\).\(\square \)

Remark 3.6

For compactness, it is enough to assume that the initial condition, (2.2) is fulfilled uniformly. The importance of the extension above will be shown in the next section.

In the rest of this section, we show that all the translations listed above are regular in sense of Definition 3.1. At first, we make some observations which will be useful in parabolic cases below and in the last section.

Remark 3.7

For the next observation, let us write the differential equations of the listed examples in the next form

$$\begin{aligned} D_{(1),t}u(x,t)=D_{(2),x}u(x,t), \hspace{4pt}\hspace{4pt}u(x,0)=f(x). \end{aligned}$$
(3.5)

\(T^tf(x)=u(x,t)\) can be defined by an integral transformation, that is

$$\begin{aligned} T^tf(x)=\int _If(\xi )K(x,t,\xi )w(\xi )\textrm{d}\xi . \end{aligned}$$

Thus, the kernel function fulfils the differential equation above. As the kernel function is symmetric at least in x and \(\xi \), that is

$$\begin{aligned} K(x,t,\xi )=K(\xi ,t,x), \end{aligned}$$

we have

$$\begin{aligned} D_{(1),t}K(x,t,\xi )=D_{(2),x}K(x,t,\xi )=D_{(2),\xi }K(x,t,\xi ). \end{aligned}$$

These observations imply the next lemma.

Lemma 3.8

Let g be a smooth function on I and assumes that its derivatives disappears quickly enough at the boundary of I. Then, in the listed cases

$$\begin{aligned} D_{(2)}T^tg=T^tD_{(2)}g. \end{aligned}$$

Proof

Recall, that in our one-dimensional examples

$$\begin{aligned} D_{(2)}h=h''+\frac{w'}{w}h'+rh. \end{aligned}$$

Integrating by parts and considering the boundary condition, we have

$$\begin{aligned} D_{(2),x}T^tg(x)= & {} \int _Ig(\xi )D_{(2),x}K(x,t,\xi )w(\xi )\textrm{d}\xi =\int _Ig(\xi )D_{(2),\xi }K(x,t,\xi )w(\xi )\textrm{d}\xi \\{} & {} =\int _I-\frac{\partial }{\partial \xi }K(x,t,\xi )(gw)'(\xi )\\{} & {} \quad +(gw)(\xi )\left( \frac{w'(\xi )}{w(\xi )} \frac{\partial }{\partial \xi }K(x,t,\xi )+r(\xi )\right) w(\xi )\textrm{d}\xi \\{} & {} =\int _IK(x,t,\xi )D_{(2),\xi }g(\xi )w(\xi )\textrm{d}\xi =T^tD_{(2)}g(x). \end{aligned}$$

If \(w\equiv 1\), it is more direct. Let us see, for instance, the heat-diffusion semigroup. Recalling that \(I={\mathbb {R}}^d\)

$$\begin{aligned}{} & {} \int _Ig(\xi )\left( \Delta _\xi W_t(x,\xi )-|\xi |^2W_t(x,\xi )\right) \textrm{d}\xi \\{} & {} \quad =\int _I\left( \Delta _\xi g(\xi )-g(\xi )|\xi |^2\right) W_t(x,\xi )\textrm{d}\xi . \end{aligned}$$

Corollary 3.9

Let \(D_{(1)}=\frac{\partial }{\partial t}\) and let g be as above. Then

$$\begin{aligned} \Vert T^{t+h}g-T^tg\Vert _{p,\mu }\le h M_T\Vert D_{(2)}g\Vert _{p,\mu }. \end{aligned}$$
(3.6)

Proof

$$\begin{aligned} T^{t+h}g(x)-T^tg(x)=h\frac{\partial }{\partial t}T^{t+\eta }g(x)=hD_{(2),x}T^{t+\eta }g(x)=hT^{t+\eta }D_{(2)}g(x). \end{aligned}$$

Taking into consideration (3.1), we get the estimation.

Lemma 3.10

The translation given in Example 2.3 is regular and \(M_T=1\).

Proof

The boundedness of the operator by a constant C is proved in [41, Theorem 2.6]. It can be shown similarly that \(M_T=1\). Indeed, according to Mehler’s formula and [28, Remark 2.5]

$$\begin{aligned} 0<W_t(x,y)= & {} (2\pi \sinh 2t)^{-\frac{d}{2}}e^{-\frac{|x-y|^2}{2}\coth 2t-\langle x, y\rangle \tanh t}\nonumber \\\le & {} (2\pi t)^{-\frac{d}{2}}e^{-\frac{|x-y|^2}{4t}}=K_t(x-y). \end{aligned}$$
(3.7)

Thus

$$\begin{aligned} |T^tf|\le |\tilde{f}|*K_t, \end{aligned}$$

and since \(\Vert K_t\Vert _1=1\)

$$\begin{aligned} \Vert T^tf\Vert _p\le \Vert |\tilde{f}|*K_t\Vert _p\le \Vert \tilde{f}\Vert _p, \end{aligned}$$

\(1\le p \le \infty \).

Let \({\mathcal {E}}_1= C^2_0\subset L^p({\mathbb {R}}^d)\). In view of (3.6), (T1) is fulfilled.

We prove property (T2) by similar arguments. Considering the uniform convergence of (2.5) for a \(t>0\), the recurrence and derivation formulae for Hermite functions imply that

$$\begin{aligned} \frac{\partial }{\partial x_j}W_t(x,y)=(y_je^{-2t}-x_j)W_t(x,y)-e^{-2t}\frac{\partial }{\partial y_j}W_t(x,y). \end{aligned}$$

Thus

$$\begin{aligned}{} & {} W_t(x+\chi ,y)-W_t(x,y)\\{} & {} \quad =\langle e^{-2t}y-(x+\xi ),\chi \rangle W_t(x+\xi ,y)-e^{-2t}\langle \nabla _yW_t(x+\xi ,y), \chi \rangle . \end{aligned}$$

Let \(f\in C^1_0({\mathbb {R}}^d)\), \({\textrm{supp}}f\subset B_r\), \(x\in B_R\)

$$\begin{aligned}{} & {} \left| \frac{1}{a}\int _0^{a}\int _{{\mathbb {R}}^d}(W_t(x+\chi ,y)-W_t(x,y))\tilde{f}(y)\textrm{d}y\textrm{d}t\right| \\{} & {} \quad \le \frac{1}{a}\int _0^{a}\left( 2|\chi |(r+R)\Vert T^t|f|\Vert _p+\int _{{\mathbb {R}}^d}W_t(x+\chi ,y)\langle \chi , \nabla _y \tilde{f}(y)\rangle \textrm{d}y\right) \textrm{d}t\\{} & {} \quad \le |\chi |\left( 2(r+R)\Vert \tilde{f}\Vert _p+\Vert \nabla _y \tilde{f}\Vert _p\right) . \end{aligned}$$

To prove (T3), let us recalling (3.7). Then, we have

$$\begin{aligned} \left| \frac{1}{a}\int _0^{a}T^tf\textrm{d}t\right| \le \frac{1}{a}\int _0^{a}\int _{{\mathbb {R}}^d}| \tilde{f}(y)|K_t(x-y)\textrm{d}y\textrm{d}t\le \Vert \tilde{f}\Vert _p. \end{aligned}$$

\(\square \)

Lemma 3.11

The translation given in Example 2.8 is regular and \(M_T=1\).

Proof

For boundedness, see [18, Lemma 5.2].

As it is mentioned above, in [11], it is pointed out that a translation generated by a hyperbolic equation is necessarily symmetric. It can be expressed by an integral transform whose kernel is symmetric in its three variables; see [18]. Thus, according to Remark 3.2, it is enough to prove property (T1). Since f is even, we can take \(I={\mathbb {R}}_+\). Let \({\mathcal {D}}=C^1_0({\mathbb {R}}_+)\subset L^p_\mu ({\mathbb {R}}_+)\) the dense set in question, where

$$\begin{aligned} \textrm{d}\mu (x)=\frac{1}{\sqrt{2\pi }}w(x)\textrm{d}x; \end{aligned}$$
(3.8)

cf. (2.13). Let \(f\in C^1_0({\mathbb {R}}_+)\) and let \(f(z)=g(\cosh 2z)\). In view of [18, (5.1)] and [18, (4.16)]

$$\begin{aligned} T^tf(x)=\int _{{\mathbb {R}}_+}f(z)K(x,t,z)\textrm{d}\mu (z)=\int _0^1\int _0^\pi g(x,t,r,\psi )\textrm{d}m(r,\psi ), \end{aligned}$$

where

$$\begin{aligned}{} & {} (x,t,r,\psi )\\{} & {} \quad =2(\cosh ^2 x\cosh ^2 t+r^2\sinh ^2 x \sinh ^2 t +2r\cos \psi \sinh x\sinh t \cosh x\cosh t)-1, \end{aligned}$$

and

$$\begin{aligned} \textrm{d}m(r,\psi )=\frac{2\Gamma (\alpha +1)}{\sqrt{\pi }\Gamma (\alpha -\beta )\Gamma (\beta +\frac{1}{2})} (1-r^2)^{\alpha -\beta -1}r^{2\beta +1}\sin ^{2\beta }\psi \textrm{d}r\textrm{d}\psi . \end{aligned}$$

It can be readily seen that

$$\begin{aligned} 2\cosh ^2(x-t)-1\le (x,t,r,\psi )\le 2\cosh ^2(x+t)-1. \end{aligned}$$

Since \(g\in C^1_0({\mathbb {R}}_+)\) too, suppose that \({\textrm{supp}}g(z)\subset B_R\). Recalling that \(0<t<M_0\) \({\textrm{supp}}g_{t,r,\psi }(x)\subset B_{ C(M_0,R)}\). Considering that \(\frac{\partial }{\partial y}g(x,t,r,\psi )\) is bounded and \(m(r,\psi )\) is a bounded measure, one can conclude that (T1) fulfils with \(\delta = c \varepsilon \).\(\square \)

Lemma 3.12

The translations given by Examples 2.4 and 2.5 are regular and \(M_T=1\) in both cases.

Proof

The proof is based on the standard convolution structure.

First, let us observe that the kernel functions are positive and \(\int _{\mathbb {R}}K_{e,p}(x,t)\textrm{d}x=1\) for all \(t>0\), which gives the operator norm.

In both cases, let \({\mathcal {E}}_1={\mathcal {E}}_2=C_0({\mathbb {R}})\). In harmonic case by the standard substitution \(\frac{\xi -x}{t+h}=\tan \varphi \), where \(h>0\) and \(h=0\) in the different integrals, we have

$$\begin{aligned}{} & {} \left\| T^{t+h} g(x)-T^{t} g(x)\right\| _{p}\\{} & {} \quad =\left( \int _{\mathbb {R}}\left| \frac{1}{\pi }\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}(g(x+(t+h)\tan \varphi ) -g(x+t\tan \varphi ))\textrm{d}\varphi \right| ^p\textrm{d}x\right) ^{\frac{1}{p}}\\{} & {} \quad \le \left( \int _{\mathbb {R}}\left| \frac{1}{\pi }\int _{-\frac{\pi }{2}}^{-\alpha }(\cdot ))\textrm{d}\varphi \right| ^p\textrm{d}x\right) ^{\frac{1}{p}}+\left( \int _{\mathbb {R}}\left| \frac{1}{\pi } \int _{-\alpha }^{\alpha }(\cdot ))\textrm{d}\varphi \right| ^p\textrm{d}x\right) ^{\frac{1}{p}}\\{} & {} \qquad +\left( \int _{\mathbb {R}}\left| \frac{1}{\pi }\int _{\alpha }^{\frac{\pi }{2}} (\cdot ))\textrm{d}\varphi \right| ^p\textrm{d}x\right) ^{\frac{1}{p}}\\{} & {} \quad =I+II+III.\\{} & {} I, \hspace{4pt}III \le 2 \Vert g\Vert _p\frac{\frac{\pi }{2}-\alpha }{\pi }, \end{aligned}$$

which is small if \(\alpha \) is close enough to \(\frac{\pi }{2}\).

Since g is uniformly continuous for fixed \(\alpha \) and \(\varepsilon \) we can choose h, \(|h|<1\), such that \(|h\tan \varphi |\) is small enough to be the integrand in II small, less than \(\varepsilon \), say.

Then, recalling that g is compactly supported and \(|t| \le M_0\), there is an R, such that the support of \(g(x+(t+h)\tan \varphi )-g(x+t\tan \varphi )\) is in \([-R,R]\). , \(II\le \varepsilon (2R)^{\frac{1}{p}}.\)

Similarly, in parabolic case, let \(u=\frac{\xi -x}{2b\sqrt{t+h}}\). Then

$$\begin{aligned}{} & {} \left\| T^{t+h} g(x)-T^{t} g(x)\right\| _{p}\\{} & {} \quad =\left( \int _{\mathbb {R}}\left| \frac{1}{\sqrt{\pi }}\int _{{\mathbb {R}}}(g(x+2b\sqrt{t+h}u) -g(x+2b\sqrt{t}u))e^{-u^2}\textrm{d}u\right| ^p\textrm{d}x\right) ^{\frac{1}{p}}\\{} & {} \quad \le \frac{1}{\sqrt{\pi }}\int _{|u|>R}e^{-u^2}\left( \int _{\mathbb {R}}|g(x+2b\sqrt{t+h}u) -g(x+2b\sqrt{t}u)|^p\textrm{d}x\right) ^{\frac{1}{p}}\textrm{d}u\\{} & {} \qquad +\frac{1}{\sqrt{\pi }}\int _{|u|\le R}(\cdot )=I+II. \end{aligned}$$

Recalling that g is compactly supported, the p-norm inside is bounded; thus, if R is large enough, I is small. The uniform continuity of g implies that II is small if h is small enough.

Turning to the proof of (3.3) and (3.4), we define \(\nu (t)\). If \(1<p<\infty \) and in parabolic case also for \(p=1\), let \(\textrm{d}\nu (t)=\textrm{d}t\), and if \(p=1\), in the elliptic case, let \(\textrm{d}\nu (t)=\sqrt{t}\textrm{d}t\). Then, the same replacements imply the results. Indeed, let us see the elliptic case first

$$\begin{aligned}{} & {} |M_ag(x+h))-M_ag(x)|\\{} & {} \quad \le \frac{1}{A\pi }\int _{B_a} \left( \int _{-\frac{\pi }{2}}^{-\alpha }(\cdot ) +\int _{-\alpha }^{\alpha }|g(x+(t+h)\tan \varphi )-g(x+t\tan \varphi )|\textrm{d}\varphi \right. \\{} & {} \qquad \left. + \int _{\alpha }^{\frac{\pi }{2}}(\cdot )\right) \textrm{d}\nu (t). \end{aligned}$$

In the first and third integrals, the boundedness of the integrand, in the second one, the uniform continuity of g ensures the required estimates.

To prove (3.4), let \(1<p<\infty \)

$$\begin{aligned} |M_af(x)|\le \frac{1}{A}\int _{B_a}\Vert f\Vert _p\frac{1}{\pi t}\left( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}} \cos ^{2(p'-1)}(\varphi )t\textrm{d}\varphi \right) ^{\frac{1}{p'}} \textrm{d}t \le c a^{-\frac{1}{p}}\Vert f\Vert _p. \end{aligned}$$

If \(p=1\)

$$\begin{aligned} |M_af(x)|\le \Vert f\Vert _1\frac{1}{A}\int _{B_a}\sqrt{t}\frac{1}{t}\textrm{d}t=\frac{c}{a}\Vert f\Vert _1. \end{aligned}$$

In the parabolic case

$$\begin{aligned}{} & {} |M_ag(x+h))-M_ag(x)|\\{} & {} \quad \le \frac{1}{a\sqrt{\pi }}\int _{B_a} \left( \int _{|u|>R} (\cdot )+ \int _{|u|\le R}|g(x+h+2b\sqrt{t}u) -g(x+2b\sqrt{t}u)|e^{-u^2}\textrm{d}u\right) \textrm{d}t. \end{aligned}$$

Again, in the first integral, we refer to boundedness; in the second one to the uniform continuity of g.

Let \(1 \le p<\infty \). Then

$$\begin{aligned} |M_af(x)|\le \frac{1}{\sqrt{\pi }} \Vert f\Vert _p\Vert e^{-u^2}\Vert _{p'}\frac{1}{A}\int _{B_a}(2b\sqrt{t})^{\frac{1}{p'}-1}\textrm{d}t \le c a^{-\frac{1}{2p}}\Vert f\Vert _p, \end{aligned}$$

which is (3.4).\(\square \)

Lemma 3.13

The translation defined in Example 2.6 is regular and \(M_T=1\).

Proof

For boundedness, see [35, Theorem 2].

Let our dense sets \({\mathcal {E}}={\mathcal {E}}_i\), \(i=1,2\) be the polynomials, that is

$$\begin{aligned} {\mathcal {E}}:=\left\{ \sum _{k=0}^N a_k P_k^{(\lambda )}(\cos \vartheta ): a_k\in {\mathbb {R}}, \hspace{4pt}\hspace{4pt}N\in {\mathbb {N}}\right\} . \end{aligned}$$

Let \(P\in {\mathcal {E}}\).

$$\begin{aligned}{} & {} \Vert T^{r+h}P(\vartheta )-T^rp(\vartheta )\Vert _{\mu ,p}=\left\| \sum _{k=0}^N a_k ((r+h)^k-r^k)P_k^{(\lambda )}(\cos \vartheta )\right\| _{\mu ,p}\\{} & {} \quad \le h c(N)\sum _{k=0}^N |a_k| \Vert P_k^{(\lambda )}(\cos \vartheta )\Vert _{\mu ,p}=C(P) h, \end{aligned}$$

which is (T1). (T2) can be shown similarly. Let \(\nu (r)=\lambda (r)\), the Lebesgue measure on [0, 1) and \(P\in {\mathcal {D}}\)

$$\begin{aligned}{} & {} |M_{a,R}P(\vartheta +h)-M_{a,R}P(\vartheta )|\\{} & {} \quad =\left| \sum _{k=0}^N a_k \left( P_k^{(\lambda )}(\cos (\vartheta +h))-P_k^{(\lambda )}(\cos \vartheta )\right) \frac{1}{a}\int _0^a r^k\textrm{d}r\right| \\{} & {} \quad =\left| -2h\lambda \sin \tilde{\vartheta }\sum _{k=0}^N \frac{a_k}{k+1}a^kP_{k-1}^{(\lambda +1)}(\cos \tilde{\vartheta })\right| \le c(P,a)|h|. \end{aligned}$$

In view of (2.9) \(P(r,\vartheta ,\varphi )>0\), and if \(r\le \frac{1}{8}\), say, then it is also bounded. Thus, if \(a\le \frac{1}{8}\), \(1\le p<\infty \)

$$\begin{aligned} |M_af(\vartheta )|\le \frac{1}{a}\int _0^a\Vert f\Vert _{\mu ,p}\Vert P(r,\vartheta ,\varphi )\Vert _{\mu ,p'}\textrm{d}r\le c\Vert f\Vert _{\mu ,p}, \end{aligned}$$

where \(\Vert \cdot \Vert _{\mu ,\infty }:=\Vert \cdot \Vert _{\infty }\).\(\square \)

Lemma 3.14

The translation defined in Example 2.7 is regular and bounded.

Proof

For boundedness, see [12, Theorem 2.2].

The dense sets are the set of polynomials. By the notation (2.10), it is as follows:

$$\begin{aligned} {\mathcal {E}}_{1,2}={\mathcal {E}}=\left\{ \sum _{k=0}^n c_k \varphi _k^{(\alpha )}: \hspace{4pt}n\in {\mathbb {N}}, \hspace{4pt}c_k\in {\mathbb {R}}\right\} . \end{aligned}$$

Let \(f\in {\mathcal {E}}\), \(f(x)=\sum _{k=0}^n c_k \varphi _k^{(\alpha )}(x)\).

Then, \(T^tf(x)=\sum _{k=0}^n e^{-(2k+\alpha +1)t}c_k \varphi _k^{(\alpha )}(x)\). Thus

$$\begin{aligned} \left\| T^{t+h} f(x)-T^{t} f(x)\right\| _{p}=\left\| \sum _{k=0}^n e^{-(2k+\alpha +1)t}\left( e^{-(2k+\alpha +1)h}-1\right) c_k \varphi _k^{(\alpha )}(x)\right\| _p, \end{aligned}$$

that is h can be chosen appropriately depending on f.

Similarly, since \(x\in B_R\)

$$\begin{aligned} |M_{a,R}f(x+h)-M_{a,R}f(x)|=\left| \sum _{k=0}^n \frac{e^{-(2k+\alpha +1)a}-1}{(2k+\alpha +1)a}c_k \left( \varphi _k^{(\alpha )}(x+h)-\varphi _k^{(\alpha )}(x)\right) \right| \end{aligned}$$

is small with an appropriate h depending on f.

To prove the third property, we consider the next estimation of the kernel function; see [16, (2.4)]

$$\begin{aligned} 0\ge W_t^\alpha (x,y)\le C\frac{e^{-c\frac{(x-y)^2}{t}}}{\sqrt{t}}. \end{aligned}$$

Let \(g\in L^p(I)\)

$$\begin{aligned}{} & {} |M_{a,R}f(x)|\le \left| \int _0^{x-1}g(y)\frac{1}{a}\int _0^aW_t^\alpha (x,y)\textrm{d}t\textrm{d}y\right| +\left| \int _{x-1}^{x+1}(\cdot )\right| +\left| \int _{x+1}^\infty (\cdot )\right| \\{} & {} \quad =I+II+III.\\{} & {} \quad II\le \Vert g\Vert _p\frac{c(p)}{\sqrt{a}}.\\{} & {} \quad I,III\le \Vert g\Vert _p\frac{1}{\sqrt{a}}\left\| e^{-c\frac{(x-\cdot )^2}{a}}\right\| _{p'}. \end{aligned}$$

The consequence of Theorem 3.5 is the next one.

Corollary 3.15

Let \(K\subset L^p_\mu (I)\) be bounded. K is precompact if and only if it is equivanishing and equicontinuous in mean with T defined in Examples 2.3, 2.4, 2.5, 2.6, 2.7 and 2.8.

4 Regularity with respect to an integral transform I: Pego-type theorems

In [36], Pego characterized the compact sets of \(L^2({\mathbb {R}}^d)\) by Fourier transform. This result was extended to different spaces and to Abelian groups; see, e.g., [22, 23] and the references therein. A Pego-type theorem by Laplace transform was proved in [33]. These results based on standard translation. The extension of Pego’s theorem to Bessel and Laguerre translations and transforms is given in [31]. The main ingredients of this type of theorems are a translation, a generated convolution, and a corresponding integral transform.

4.1 Convolution

The study of convolution structures with general translations dates back to the 70s and it is a widely studied topic; see, e.g., [6, 9, 18,19,20] for Jacobi, [26] and [32]. These general convolutions similarly to the standard ones are defined as follows. For f and g, appropriate functions

$$\begin{aligned} f*g(t):=\int _IT^tf(x)g(x)\textrm{d}\mu (x). \end{aligned}$$

All the convolutions in the listed papers are based on symmetric translations which are related to hyperbolic equations. This symmetry ensures algebraic structures with respect to the convolution in question, since the symmetry of the kernel of the translation in its three variables implies the commutativity and associativity of the convolution; see the references above. In parabolic and elliptic cases, the kernels of the translations are symmetric only in two variables; thus, the associativity fails.

To prove Pego-type theorems, the next relations of convolution, integral transform, and translation are necessary

$$\begin{aligned} {\mathcal {I}}(f*g)={\mathcal {I}}(f){\mathcal {I}}(g), \end{aligned}$$

and

$$\begin{aligned} {\mathcal {I}}(T^tf)(z)=\psi (t,z){\mathcal {I}}(f)(z), \hspace{4pt}\hspace{4pt}T^t{\mathcal {I}}(f)(z)={\mathcal {I}}(\psi f)(z), \end{aligned}$$

where \({\mathcal {I}}\) is the integral transform and \(\psi \) is an appropriate function. The translation is regular with respect to the integral transform if it possesses the properties above.

4.2 Pego-type theorems by standard convolution

Pego-type characterization of compactness can be given in cases of Examples 2.4 and 2.5.

In elliptic case, we characterize compactness of sets in \(L^2({\mathbb {R}}_+)\) and consider cosine transformation; in parabolic case, the sets are in \(L^2({\mathbb {R}})\) and the corresponding transformation is the standard Fourier one.

The transformation pairs are normalized as follows:

$$\begin{aligned} {\mathcal {I}}(f)(z):=\hat{f}(z)=\int _{{\mathbb {R}}} f(x)k(xz) \textrm{d}x, \hspace{4pt}\hspace{4pt}\hspace{4pt}{\mathcal {J}}(\hat{f})(x)=\frac{1}{2\pi }\int _{{\mathbb {R}}} \hat{f}(z)k(xz) \textrm{d}z, \end{aligned}$$

where \({\mathcal {I}}=C\) or \({\mathcal {I}}={\mathcal {F}}\), the cosine or the Fourier transform with \(k(xz)=\cos (xz)\) or \(k(xz)=e^{-ixz}\), respectively. First, the actions between translations and transformations are derived. Suppose that f is in the Schwartz class, and in elliptic case, we assume that it is even

$$\begin{aligned} T^tC(f)(z)= & {} \frac{1}{\pi }\int _{{\mathbb {R}}}2\int _{{\mathbb {R}}_+} f(u)\cos u\eta \textrm{d}u \frac{t}{(\eta -z)^2+t^2}\textrm{d}\eta \\= & {} 2\int _{{\mathbb {R}}_+} f(u)\frac{1}{\pi }\int _{{\mathbb {R}}} \cos u\eta \textrm{d}u \frac{t}{(\eta -z)^2+t^2}\textrm{d}\eta \textrm{d}u\\ {}= & {} 2\int _{{\mathbb {R}}_+} e^{-ut}f(u)\cos uz \textrm{d}u\\= & {} C(e^{-|\cdot |t}f(\cdot ))(z); \end{aligned}$$

cf. [8, 1.2 (13)]. By similar calculation, supposing that \(z\ge 0\)

$$\begin{aligned} C(T^tf(x))(z)=e^{-zt}C(f)(z). \end{aligned}$$

In the parabolic case, we have

$$\begin{aligned} T^t{\mathcal {F}}(f)(z)= & {} \int _{{\mathbb {R}}}\int _{{\mathbb {R}}} \frac{e^{-\frac{(z-\eta )^2}{4b^2t}}}{2b\sqrt{\pi }\sqrt{t}}e^{-i\eta u}f(u)\textrm{d}u\textrm{d}\eta =\int _{{\mathbb {R}}}T^t_z(e^{-i(\cdot )u})f(u)\textrm{d}u\\= & {} \int _{{\mathbb {R}}}f(u)e^{-b^2tu^2}e^{-izu}\textrm{d}u={\mathcal {F}}\left( f(\cdot )e^{-b^2t(\cdot )^2}\right) (z). \end{aligned}$$

On the other hand

$$\begin{aligned} {\mathcal {F}}(T^tf(x))(z)=e^{-b^2tz^2}{\mathcal {F}}(f)(z). \end{aligned}$$
(4.1)

As it is well known

$$\begin{aligned} {\mathcal {I}}: \hspace{4pt}L^p \longrightarrow L^{p'}, \hspace{4pt}\hspace{4pt}\hspace{4pt}1\le p \le 2. \end{aligned}$$

Let \(K\subset L^p(I)\) (\(1\le p \le 2\)). Denote \(\hat{K}\subset L^{'}p(0,\infty )\), \(\hat{K}={\mathcal {I}}(K)\). With this notation, we have the next Pego-type theorem.

Theorem 4.1

Let \(1\le p \le 2\), \(K\subset L^p(I)\) a bounded set.

Let us consider Examples 2.4 and 2.5.

Assume that K satisfies \(\mathbf{P_a}\). Then, \(\hat{K}\) fulfils \(\mathbf{P_{b_1}}\) and \(\mathbf{P_{b_2}}\).

On the other hand, if K satisfies \(\mathbf{P_{b_1}}\), then \(\hat{K}\) fulfils \(\mathbf{P_a}\).

Proof

Let \(g(\cdot ,t)=e^{-|\cdot |t}\) or \(g(t,\cdot )=e^{-b^2t(\cdot )^2}\), and f as above

$$\begin{aligned}{} & {} \Vert T^{t+h}{\mathcal {I}}(f)(z)-T^t{\mathcal {I}}(f)(z)\Vert _{p', x}= \Vert {\mathcal {I}}(g(\cdot ,t+h)f(\cdot ))(z)-{\mathcal {I}}(g(\cdot ,t)f(\cdot ))(z)\Vert _{p',z}\\{} & {} \quad \le c \Vert g(x,t+h)f(x)-g(x,t)f(x)\Vert _p\\{} & {} \quad \le \left( \int _{B_R}|f(x)|^p\left| g(x,t+h)-g(x,t)\right| ^p\textrm{d}x\right) ^{\frac{1}{p}}+2\left( \int _R^\infty |f(x)|^p\textrm{d}x\right) ^{\frac{1}{p}}. \end{aligned}$$

In view of \(\mathbf{P_a}\), the second term is small. Since \(x \in B_R\), \(g(x,\cdot )\) is uniformly continuous. Thus, considering that K is bounded, \(\mathbf{P_{b_1}}\) is satisfied for dense set uniformly and so for \(\hat{K}\)

$$\begin{aligned}{} & {} \left| M_{a,R}{\mathcal {I}}(f)(z+h)-M_{a,R}{\mathcal {I}}(f)(z)\right| \\{} & {} \quad =\left| \frac{1}{A}\int _{B_a}\int _{{\mathbb {R}}}g(u,t)f(u)\left( k (u(z+h))-k(uz)\right) \textrm{d}u\textrm{d}\nu (t)\right| \\{} & {} \quad \le \int _{{\mathbb {R}}}\frac{1}{A}\int _{B_a}g(u,t)\textrm{d}\nu (t) |f(u)|\left| k (u(z+h))-k (uz)\right| \textrm{d}u\\{} & {} \quad \le c\left( \int _{B_R}h(u)|f(u)|\left| k(u(z+h))-k (uz)\right| \textrm{d}u+\int _{{\mathbb {R}}\setminus B_R}(\cdot )\right) =c(III+IV), \end{aligned}$$

where \(h(u)=\frac{1}{A}\int _{B_a}g(u,t)\textrm{d}\nu (t)\) is uniformly bounded in u, and if \(u\in {\mathbb {R}}\setminus B_R\)

$$\begin{aligned} h(u)\le \frac{c(a)}{|u|} \end{aligned}$$

in both cases. Thus

$$\begin{aligned} IV\le \left( \int _{{\mathbb {R}}\setminus B_R} |f|^p\right) ^{\frac{1}{p}}c(a)\left( \int _{{\mathbb {R}}\setminus B_R} u^{-p'}\textrm{d}u\right) ^{\frac{1}{p'}}, \end{aligned}$$

which is small by \(\mathbf{P_a}\), considering that \(p'>1\), and

$$\begin{aligned} III\le c\Vert f\Vert _p\Vert k(u(z+h))-k(uz)\Vert _{p',[0,R]}. \end{aligned}$$

Since the interval is bounded, h can be chosen, such that the third factor be small and so, as above, \(\hat{K}\) fulfils \(\mathbf{P_{b_2}}\).

To prove the opposite direction, we use the convolution below in cosine transform case, and the standard one in Fourier transform case. For cosine transform, supposing that f, g are even

$$\begin{aligned} f*g(x):=\int _{{\mathbb {R}}}f(\xi )\frac{1}{2}(g(x+\xi )+g(x-\xi ))\textrm{d}\xi . \end{aligned}$$

Assume that K fulfils \(\mathbf{P_{b_1}}\). Let \(k_t(x)=K_{t,e}(x)\) or \(k_t(x)=K_{t,p}(x)\), \(x\in {\mathbb {R}}\), \(y> 0\). Choose R so large that \(|{\mathcal {I}}(k_t)(\eta )|\le \frac{1}{2}\) if \(\eta \in {\mathbb {R}}{\setminus } B_R\)

$$\begin{aligned}{} & {} \left( \int _{{\mathbb {R}}\setminus B_R}|{\mathcal {I}}(f)(\eta )|^{p'}\textrm{d}\eta \right) ^{\frac{1}{p'}}\le 2\left( \int _{{\mathbb {R}}\setminus B_R}|{\mathcal {I}}(f)(\eta )(1-{\mathcal {I}}(k_t)(\eta )|^{p'}\textrm{d}\eta \right) ^{\frac{1}{p'}}\\{} & {} \quad \le c \Vert f-f*k_t\Vert _p= c\Vert f -T^tf(x)\Vert _p. \end{aligned}$$

Thus, \(\mathbf{P_{b_1}}\) implies \(\mathbf{P_a}\).\(\square \)

The next corollary implies that equicontinuity in Examples 2.4 and 2.5 sense is equivalent with the standard one from this point of view.

Corollary 4.2

Let \(K\subset L^2(I)\) be a bounded set and \(\hat{K}\) the Fourier or the cosine transform of K, respectively. Then, K is precompact if and only if K and \(\hat{K}\) are equicontinuous in mean in sense of Examples 2.5 and 2.4, respectively.

4.3 A Pego-type theorem by general convolution

In Laguerre and Bessel cases, Pego-type theorems are proved in [31]. Below, we prove a similar theorem with Jacobi transform. As the translation in the original and dual spaces is different, this situation is more complex than the Bessel case.

Jacobi transform is defined as it follows; see [17, Proposition 3]. Recalling the measure \(\textrm{d}\mu (x)=\frac{1}{\sqrt{2\pi }}w(x)\textrm{d}x\) (see (2.13) and (3.8)), let \(f\in L^2_\mu ({\mathbb {R}}_+)\) and \(\lambda \in {\mathbb {R}}_+\)

$$\begin{aligned} {J}(f)(\lambda ):=\hat{f}(\lambda )=\int _0^\infty f(x)\varphi _\lambda ^{(\alpha ,\beta )}(x)\text {d}\mu (x), \end{aligned}$$

where the integral is convergent in \(L^2_\nu ({\mathbb {R}}_+)\) with \(\textrm{d}\nu (\lambda )=\frac{1}{\sqrt{2\pi }}v(\lambda )\textrm{d}\lambda \), where

$$\begin{aligned} v(\lambda )=\left| \frac{2^{\varrho -i\lambda }\Gamma (i\lambda )\Gamma (\alpha +1)}{\Gamma \left( \frac{\varrho +i\lambda }{2}\right) \Gamma \left( \frac{\varrho +i\lambda }{2}-\beta \right) }\right| ^{-2}. \end{aligned}$$

\({J}: f\mapsto \hat{f}\) is a linear and norm-preserving map of \(L^2_\mu ({\mathbb {R}}_+)\) onto \(L^2_\nu ({\mathbb {R}}_+)\). The inverse transform is given by

$$\begin{aligned} {J}^{-1}(f)(x)=f(x)=\int _0^\infty \hat{f}(\lambda )\varphi _\lambda ^{(\alpha ,\beta )}(x)\text {d}\nu (\lambda ), \end{aligned}$$

where the integral is convergent in \(L^2_\mu ({\mathbb {R}}_+)\).

Since J maps \(L^1_\mu ({\mathbb {R}}_+)\) into \(L^\infty ({\mathbb {R}}_+)\), see [18, (3.2)], Riesz–Thorin theorem ensures that J maps \(L^p_\mu ({\mathbb {R}}_+)\) to \(L^{p'}_\nu ({\mathbb {R}}_+)\) continuously, where \(1\le p\le 2\).

As the range of the Jacobi transform is different from its domain, introduction of a dual translation is necessary, see [19]. It is given by the next formula

$$\begin{aligned} T_d^\eta g(\xi )=\int _0^\infty a(\xi ,\eta ,\zeta )g(\zeta )\textrm{d}\nu (\zeta ), \end{aligned}$$
(4.2)

where the dual kernel is

$$\begin{aligned} a(\xi ,\eta ,\zeta )={\mathcal {J}}(\varphi _\xi ^{(\alpha ,\beta )}\varphi _\eta ^{(\alpha ,\beta )})(\zeta )=\int _0^\infty \varphi _\xi ^{(\alpha ,\beta )}(x)\varphi _\eta ^{(\alpha ,\beta )}(x)\varphi _\zeta ^{(\alpha ,\beta )}(x)\textrm{d}\mu (x); \end{aligned}$$

see [19, (4.14)]. That is the kernel is symmetric in its three variables. Moreover

$$\begin{aligned} a(\xi ,\eta ,\zeta )\ge 0, \hspace{4pt}(\xi ,\eta ,\zeta )\in {\mathbb {R}}^3; \hspace{4pt}\hspace{4pt}\int _0^\infty a(\xi ,\eta ,\zeta )\textrm{d}\nu (\zeta )=1; \end{aligned}$$

see [19, Theorem 4.4 and (4.17)]. This ensures, in standard way, that

$$\begin{aligned} \Vert T_d^\eta g\Vert _{p,\nu }\le \Vert g\Vert _{p,\nu }, \hspace{4pt}\hspace{4pt}1\le p \le \infty . \end{aligned}$$
(4.3)

Subsequently, we need the next properties (see also [21, Section 2]).

Lemma 4.3

Let \(f\in C_0({\mathbb {R}}_+)\). Then

$$\begin{aligned} {J}(T^t f)(\lambda )=\varphi _\lambda ^{(\alpha ,\beta )}(t){J}(f)(\lambda ) \end{aligned}$$
(4.4)

and

$$\begin{aligned} T_d^\eta \hat{f}(\lambda )={J}(f\varphi _\eta ^{(\alpha ,\beta )})(\lambda ). \end{aligned}$$
(4.5)

Proof

By [18, (4.2)]

$$\begin{aligned} \int _0^\infty K(x,t,z)\varphi _\lambda ^{(\alpha ,\beta )}(z)\textrm{d}\mu (z) =\varphi _\lambda ^{(\alpha ,\beta )}(t)\varphi _\lambda ^{(\alpha ,\beta )}(x). \end{aligned}$$

Thus

$$\begin{aligned} {J}(T^t f)(\lambda )=&{} \int _0^\infty \int _0^\infty f(z)K(x,t,z)\text {d}\mu (z)\varphi _\lambda ^{(\alpha ,\beta )}(x)\text {d}\mu (x)\\=&{} \int _0^\infty f(z)\varphi _\lambda ^{(\alpha ,\beta )}(t)\varphi _\lambda ^{(\alpha ,\beta )}(z)\text {d}\mu (z), \end{aligned}$$

where by the assumption on f Fubini theorem can be applied and the last integral is convergent. It proves (4.4).

Similarly, in view of [19, (4.16)]

$$\begin{aligned} \int _0^\infty a(\xi ,\eta ,\zeta )\varphi _\zeta ^{(\alpha ,\beta )}(x)\textrm{d}\nu (\zeta )=\varphi _\xi ^{(\alpha ,\beta )}(x)\varphi _\eta ^{(\alpha ,\beta )}(x), \end{aligned}$$

which implies that

$$\begin{aligned} T_d^\eta \hat{f}(\lambda )= & {} \int _0^\infty \int _0^\infty f(x)a(\lambda ,\eta ,\zeta )\varphi _\zeta ^{(\alpha ,\beta )}(x)\textrm{d}\mu (x)\textrm{d}\nu (\zeta )\\= & {} \int _0^\infty f(x)\varphi _\eta ^{(\alpha ,\beta )}(x)\varphi _\lambda ^{(\alpha ,\beta )}(x)\textrm{d}\mu (x), \end{aligned}$$

which, with the remark above, ensures (4.5).\(\square \)

Before stating the next theorem, we introduce the corresponding convolution. For appropriate functions

$$\begin{aligned} f*g(t):=\int _0^\infty T^tf(x)g(x)\textrm{d}\mu (x) \end{aligned}$$

and

$$\begin{aligned} {J}(f*g)={J}(f){J}(g); \end{aligned}$$

furthermore, the convolution is commutative and associative; see [18, (5.3), (5.4)].

Theorem 4.4

Let \(1\le p \le 2\), \(K\subset L^p_\mu ({\mathbb {R}}_+)\) a bounded set. Let \(\hat{K}:={J}(K)\).

Let us consider Example 2.8.

If K satisfies \(\mathbf{P_a}\), then \(\hat{K}\) fulfils \(\mathbf{P_{b_1}}\) and \(\mathbf{P_{b_2}}\).

If K satisfies \(\mathbf{P_{b_1}}\), then \(\hat{K}\) fulfils \(\mathbf{P_a}\).

Proof

Let \(f\in K\). In view of (4.5)

$$\begin{aligned} \Vert T_d^{\eta +h}\hat{f}(\lambda )-T_d^\eta \hat{f}(\lambda )\Vert _{p',\nu }=&{} \Vert {J}(f(\varphi _{\eta +h}^{(\alpha ,\beta )}-\varphi _\eta ^{(\alpha ,\beta )}))\Vert _{p',\nu }\\\le&{} c\Vert f(\varphi _{\eta +h}^{(\alpha ,\beta )}-\varphi _\eta ^{(\alpha ,\beta )})\Vert _{p,\mu }. \end{aligned}$$

According to [17, Theorem 2, (ib)], if \(\eta \in {\mathbb {R}}\)

$$\begin{aligned} \frac{\partial }{\partial \eta }\varphi _{\eta }^{(\alpha ,\beta )}(x)\le K(1+x)^2e^{-\varrho x}. \end{aligned}$$

Thus, since \(\varrho >0\), \(\Vert T_d^{\eta +h}\hat{f}(\lambda )-T_d^\eta \hat{f}(\lambda )\Vert _{p',\nu }\le c(\varrho )h\Vert f\Vert _{p,\mu }\), so \(\mathbf{P_{b_1}}\) is satisfied for \(\hat{K}\).

Repeating the standard arguments, for an arbitrary \(\delta >0\), take a function \(g_\delta \), such that \({\textrm{supp}}g_\delta \subset [-\delta ,\delta ]\), \(\int _{{\mathbb {R}}_+}g_\delta (t)\textrm{d}\mu (t)=1\), \(\hat{g}_\delta >0\), cf. [19, Lemma 4.2]. As \(\hat{g}_\delta (\lambda )\) tends to zero when \(\lambda \) tends to infinity, choose R so large that \(\hat{g}_\delta (\lambda )<\frac{1}{2}\) if \(\lambda >R\). Then

$$\begin{aligned} \left( \int _R^\infty |\hat{f}|^{p'}\textrm{d}\nu \right) ^{\frac{1}{p'}}\le & {} \left( \int _R^\infty |\hat{f}(1-\hat{g}_\delta )|^{p'}\textrm{d}\nu \right) ^{\frac{1}{p'}} \le c(p)\Vert f-f*g_\delta \Vert _{p,\mu }\\= & {} c(p) \left( \int _0^\infty \left| \int _0^\infty (f(x)-T^x f(t))g_{\delta }(t)\textrm{d}\mu (t)\right| ^p\textrm{d}\mu (x)\right) ^{\frac{1}{p}}\\\le & {} c(p)\left( \int _0^\infty \int _0^\infty \left| f(x)-T^tf(x)\right| ^p \textrm{d}\mu (x)g_\delta (t)\textrm{d}\mu (t)\right) ^{\frac{1}{p}}\\\le & {} c(p)\sup _{0\le |t|\le \delta } \Vert f(x)-T^tf(x)\Vert _{p,\mu }. \end{aligned}$$

5 Modulus of smoothness and K-functional: regularity with respect to an integral transform II

As it is well known, the modulus of smoothness generated by the standard translation is equivalent with the Peetre’s K-functional; see, e.g., [14, page 171]. This property is extended to Bessel translation; see [37]. Below, we derive the same equivalence for heat-diffusion semigroup, where the semigroup property implies arguments very similar to the standard case, and for Example 2.5, where the corresponding integral transform proves to be the right tool.

Recalling the notation of (3.5), we can define the Sobolev space generated by \(D_{(2)}\) as follows:

$$\begin{aligned} W_{D,r,\mu }^p=\{f(x)\in L^p_\mu : D_{(2),x}^l f\in L^p_\mu , \hspace{4pt}l=1, \dots , r\}. \end{aligned}$$

Then, the corresponding K-functional is

$$\begin{aligned} K_r(f,t,L^p_\mu , W_{D,r,\mu }^p):=K_r(f,t)_p=\inf _{g\in W_{D,r,\mu }^p}\{\Vert f-g\Vert _{p,\mu }+t\Vert (D_{(2)}^r g\Vert _{p,\mu }\}. \end{aligned}$$

The moduli of smoothness generated by general translations are defined as follows.

Definition 5.1

Let

$$\begin{aligned} \Delta _tf(x):=T^tf(x)-f(x), \hspace{4pt}\hspace{4pt}\hspace{4pt}\Delta _t^rf(x):=\Delta _t(\Delta _t^{r-1}f)(x). \end{aligned}$$

Let \(1\le p\le \infty \). The p-modulus of smoothness is

$$\begin{aligned} \omega _r(f,t)_p:=\sup _{0<h\le t}\Vert \Delta _h^rf\Vert _{p, \mu }. \end{aligned}$$

Remark 5.2

$$\begin{aligned} \Delta _t^rf(x)=(I-T^t)^rf=\sum _{k=0}^r(-1)^{r-k}\left( {\begin{array}{c}r\\ k\end{array}}\right) (T^t)^kf(x), \end{aligned}$$
(5.1)

where \((T^t)^k=T^t\circ \dots \circ T^t\). For the standard and the heat-diffusion translation semigroups, \((T^t)^k\) becomes \(T^{kt}\).

Theorem 5.3

There are positive constants \(M=M(r,p)\) independent of f, such that for Examples 2.3 and 2.5

$$\begin{aligned} \frac{1}{M}\omega _r(f,t)_p \le K_r(f,t^{cr})_p\le M\omega _r(f,t)_p, \end{aligned}$$

\(r\in {\mathbb {N}}\), \(1\le p\le \infty \), \(c={\textrm{ord}}(D_{(1)})\).

Remark 5.4

In our parabolic cases \(c=1\), in Bessel case \(c=2\); cf. [37].

Proof

By standard arguments

$$\begin{aligned} \omega _r(f,t)_p \le \omega _r(f-g,t)_p+\omega _r(g,t)_p \le 2^r\Vert f-g\Vert _p + t^r\Vert D_{(2)} g\Vert _p. \end{aligned}$$

In view of (3.6) and Remark 5.2

$$\begin{aligned} \Vert \Delta _t^rg\Vert _p\le t^r\Vert D_{(2)} g\Vert _p. \end{aligned}$$

The second estimation needs different methods in the two cases.

First, we deal with the heat semigroup. Let us introduce the next notation

$$\begin{aligned} g_{r,t}(x):= & {} f(x)+(-1)^{r+1}r^r\int _{(0, \frac{1}{r})^r}\Delta _{t(\sum _{i=1}^rz_i)}^rf(x)\textrm{d}z_1\dots \textrm{d}z_r\nonumber \\= & {} \quad r^r\int _{(0, \frac{1}{r})^r}\sum _{k=1}^r(-1)^{k+1}\left( {\begin{array}{c}r\\ k\end{array}}\right) u\left( x,kt\left( \sum _{i=1}^rz_i\right) \right) \textrm{d}z_1\dots \textrm{d}z_r. \end{aligned}$$
(5.2)

Thus, according to (2.6)

$$\begin{aligned} D_{(2),x}^rg_{r,t}(x)=\frac{r^r}{t^r}\sum _{k=1}^r(-1)^{k+1}\left( {\begin{array}{c}r\\ k\end{array}}\right) \frac{1}{k^r}\Delta _{\frac{kt}{r}}^rf(x). \end{aligned}$$
(5.3)

Considering the operator norm, (5.2) and (5.3) \(g_{r,t} \in W^p_{r,L}\). By Minkowski’s inequality

$$\begin{aligned} t^r\Vert D_{(2),x}^rg_{r,t}(x)\Vert _p\le (2r)^r\omega _r(f,t)_p. \end{aligned}$$

Recalling the definition of \(g_{r,t}\), as above

$$\begin{aligned} \Vert f-g_{r,t}\Vert _p= \left\| r^r\int _{(0, \frac{1}{r})^r}\Delta _{t(\sum _{i=1}^ru_i)}^rf(x)\textrm{d}u_1\dots \textrm{d}u_r\right\| _p\le M(r,p)\omega _r(f,t)_p. \end{aligned}$$

To prove the second inequality for Example 2.5, we apply the integral transform method again. According to (5.1) and (4.1)

$$\begin{aligned} ({\mathcal {F}}(\Delta _t^rf))(z)=(1-e^{-b^2tz^2})^r\hat{f}(z). \end{aligned}$$

Let \(I_1:=I\cap [-1,1]\) and \(I_2:=I{\setminus } (-2,2)\). Let us define

$$\begin{aligned} \eta (x)= \left\{ \begin{array}{ll} 1, \hspace{4pt}x\in I_1,\\ 0, \hspace{4pt}x\in I_2\\ \in C_0^\infty \text{ otherwise },\end{array}\right. \end{aligned}$$

such that \(0\le \eta \le 1\). Let \(\varepsilon >0\) and define \(\eta _\varepsilon (x):=\eta \left( \frac{x}{\varepsilon }\right) \). Let

$$\begin{aligned} \varrho _\varepsilon :={\mathcal {F}}^{-1}\left( \eta _\varepsilon \right) . \end{aligned}$$

It is enough to take an \(f\in L^1\cap L^p\). Define

$$\begin{aligned} g:=g_\varepsilon :=(\varrho _\varepsilon *f). \end{aligned}$$

Let \(0<t<\frac{1}{b^2\varepsilon ^2}\). Then, we have

$$\begin{aligned} {\mathcal {F}}(D_{(2)}^rg_\varepsilon )(z)= & {} (-1)^rb^{2r}z^{2r}\eta _\varepsilon (z)\hat{g}(z)\\= & {} \frac{1}{t^r}\frac{t^r(-1)^rb^{2r}z^{2r}\eta _\varepsilon (z)}{(1-e^{-b^2tz^2})^r} ({\mathcal {F}}(\Delta _t^rg))(z)\\ {}= & {} \frac{1}{t^r}\frac{t^r(-1)^rb^{2r}z^{2r}\eta (bz\sqrt{t})}{(1-e^{-b^2tz^2})^r}({\mathcal {F}}(\Delta _t^rg))(z). \end{aligned}$$

Let

$$\begin{aligned} h(u)= \left( \frac{-u^2}{1-e^{-u^2}}\right) ^r\eta (u). \end{aligned}$$

Then

$$\begin{aligned} D_{(2)}^rg_\varepsilon =\frac{1}{t^r}\Delta _t^rg{\mathcal {F}}^{-1}(h(bz\sqrt{t})). \end{aligned}$$

Since

$$\begin{aligned} \Vert {\mathcal {F}}^{-1}(h(bz\sqrt{t}))\Vert _1=\Vert {\mathcal {F}}^{-1}(h)\Vert _1=:c, \end{aligned}$$

we have

$$\begin{aligned} \Vert D_{(2)}^rg_\varepsilon \Vert _p\le c\frac{1}{t^r}\Vert \Delta _t^rg\Vert _p. \end{aligned}$$
(5.4)

We proceed similarly to estimate the first term of \(K_r(f,t^{r})_p\)

$$\begin{aligned} {\mathcal {F}}(f-g_\varepsilon )(z)=\left( 1-\eta _\varepsilon \right) \hat{g}(z) =\frac{1-\eta _\varepsilon }{(1-e^{-b^2tz^2})^r}\widehat{\Delta _{t}^rg}(z). \end{aligned}$$

Let

$$\begin{aligned} \frac{1}{(1-x)^r}=\sum _{k=0}^\infty a_kx^k=Q_N(x)+\sum _{k=N+1}^\infty a_kx^k, \hspace{4pt}\hspace{4pt}|x|<1. \end{aligned}$$

We decompose

$$\begin{aligned} \frac{1}{(1-e^{-b^2tz^2})^r}=\left( \frac{1}{(1-e^{-b^2tz^2})^r} -Q_N\left( e^{-b^2tz^2}\right) \right) +Q_N\left( e^{-b^2tz^2}\right) , \end{aligned}$$

that is

$$\begin{aligned}{} & {} {\mathcal {F}}(f-g_\varepsilon )(z)=U_1+U_2\\{} & {} \quad =\left( 1-\eta _\varepsilon \right) Q_N\left( e^{-b^2tz^2}\right) \widehat{\Delta _{t}^rg}(z)\\{} & {} \qquad +\left( 1-\eta _\varepsilon \right) \left( \frac{1}{(1-e^{-b^2tz^2})^r}-Q_N\left( e^{-b^2tz^2}\right) \right) \widehat{\Delta _{t}^rg}(z). \end{aligned}$$

Let \(\varepsilon =\frac{1}{b\sqrt{t}}\) and

$$\begin{aligned} k_N(z)=(1-\eta (z))\left( \frac{1}{1-e^{-z^2}}-Q_N(z)\right) . \end{aligned}$$

Then

$$\begin{aligned} {\mathcal {F}}^{-1}U_2={\mathcal {F}}^{-1}\left( k_N(bz\sqrt{t})\right) *\Delta _{t}^rg. \end{aligned}$$

If N is large enough, we have

$$\begin{aligned} \Vert {\mathcal {F}}^{-1}U_2\Vert _p\le \Vert {\mathcal {F}}^{-1}k_N\Vert _1\Vert \Delta _{t}^rg\Vert _p. \end{aligned}$$
(5.5)

Turning our attention to the first term, we have

$$\begin{aligned} {\mathcal {F}}^{-1}U_1=\sum _{k=0}^Na_k\left( \left( T^t\right) ^k(\Delta _{t}^rg) -\varrho _{\frac{1}{bz\sqrt{t}}}*(\left( T^t\right) ^k(\Delta _{t}^rg)\right) . \end{aligned}$$

Since \(T^t\) is bounded

$$\begin{aligned} \Vert {\mathcal {I}}^{-1}U_1\Vert _p\le C(Q)(1+\Vert \varrho _1\Vert _1)\Vert \Delta _{t}^rf\Vert _p. \end{aligned}$$
(5.6)

Thus, according to (5.5) and (5.6)

$$\begin{aligned} \Vert f-g_\varepsilon \Vert _p\le c \omega _r(f,t)_p. \end{aligned}$$

Recalling (5.4), the second inequality is proved.\(\square \)

Remark 5.5

This integral transform method works, because considering the two expressions \({\mathcal {I}}(\Delta _tf)=(1-\psi (t,\cdot ))\hat{f}\) and \({\mathcal {I}}D_{(2)}f=h(\cdot )\hat{f}\), we have

$$\begin{aligned} h(\cdot ) \sim 1-\psi (t,\cdot ) \end{aligned}$$

around zero. This property is the second regularity property of translation with respect to the integral transform. This is the situation, e.g., in Bessel case; see [37]. For instance, Example 2.4 does not possess second regularity property with respect to the cosine transform.