1 Introduction and Notation

The Korovkin theorem for the space of all continuous \(2\pi \)-periodic functions on the real line asserts that if \(L_{n}:C_{2\pi }\left( \mathbb {R} \right) \rightarrow C_{2\pi }\left( \mathbb {R}\right) \) is a sequence of positive linear operators, then \(\lim \nolimits _{n\rightarrow \infty }L_{n}\left( f\right) =f\) uniformly on \(\mathbb {R}\) for all \(f\in C_{2\pi }\left( \mathbb {R}\right) \) if and only if \(\lim \nolimits _{n\rightarrow \infty }L_{n}\left( f\right) =f\) uniformly on \(\mathbb {R}\) for all \(f\in \left\{ \textbf{1},\cos ,\sin \right\} \), see [8]. For this result and related questions the reader can consult the book of Korovkin [8] and the books of Butzer and Nessel [3] and Altomare and Campiti [1]. In the present paper we will prove that in the case of singular integrals the test set of convergence consists only of two functions \(\textbf{1}\) and \(\cos \), Theorem 2 and moreover, a multivariate version of this result, see Theorem 3. We use Theorem 3 to extend an old result of Korovkin in the univariate case and a result of A. A. Fomin in the bivariate case, to the so called multivariate rectangular and triangular summation, see Theorems 4 and 5. Our approach in the multivariate case is different than that of A. A. Fomin. Let us fix some notation and notions. Let k be a natural number. A function \(f:\mathbb {R}^{k}\rightarrow \mathbb { R}\) is called \(2\pi \)-periodic with respect to every variable if for all \( \left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}\) we have \(f\left( t_{1}+2\pi ,\ldots ,t_{k}\right) =f\left( t_{1},\ldots ,t_{k}\right) \), ..., \( f\left( t_{1},\ldots ,t_{k-1},t_{k}+2\pi \right) =f\left( t_{1},\ldots ,t_{k-1},t_{k}\right) \). We write \(C_{2\pi }\left( \mathbb {R} ^{k}\right) \) (\(C_{2\pi }\left( \mathbb {R}\right) \) for \(k=1\)) to denote the real Banach space of the all continuous functions \(f:\mathbb {R} ^{k}\rightarrow \mathbb {R}\) (\(f:\mathbb {R}\rightarrow \mathbb {R}\) for \(k=1\)) which are \(2\pi \)-periodic with respect to every variable endowed with the norm \(\left\| f\right\| =\sup \nolimits _{t\in \left[ -\pi ,\pi \right] ^{k}}\left| f\left( t\right) \right| =\sup \nolimits _{t\in \mathbb {R} ^{k}}\left| f\left( t\right) \right| \) (\(\left\| f\right\| =\sup \nolimits _{t\in \left[ -\pi ,\pi \right] }\left| f\left( t\right) \right| =\sup \nolimits _{t\in \mathbb {R}}\left| f\left( t\right) \right| \) for \(k=1\)). If \(f\in C_{2\pi }\left( \mathbb {R}^{k}\right) \) and \(\left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}\) we define \( f_{t_{1},\ldots ,t_{k}}:\mathbb {R}^{k}\rightarrow \mathbb {R}\) by

$$\begin{aligned} f_{t_{1},\ldots ,t_{k}}\left( \theta _{1},\ldots ,\theta _{k}\right) =f\left( t_{1}-\theta _{1},\ldots ,t_{k}-\theta _{k}\right) . \end{aligned}$$

Note that \(f_{t_{1},\ldots ,t_{k}}\in C_{2\pi }\left( \mathbb {R}^{k}\right) \). A function \(f\in C_{2\pi }\left( \mathbb {R}^{k}\right) \) is called positive and we write, as usual, \(f\ge 0\) if \(f\left( t_{1},\ldots ,t_{k}\right) \ge 0\) , \(\forall \left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}\) and also if \( f,g\in C_{2\pi }\left( \mathbb {R}^{k}\right) \) the notation \(f\le g\) means \( g-f\ge 0\). An operator \(L:C_{2\pi }\left( \mathbb {R}^{k}\right) \rightarrow C_{2\pi }\left( \mathbb {R}^{k}\right) \) is called positive if \(f\ge 0\) implies \(L\left( f\right) \ge 0\). By \(\textbf{1}\) we denote the constant function \(\textbf{1}:\mathbb {R}^{k}\rightarrow \mathbb {R}\), \(\textbf{1} \left( t_{1},\ldots ,t_{k}\right) =1\). For every \(1\le i\le k\), \(pr_{i}: \mathbb {R}^{k}\rightarrow \mathbb {R}\) are the canonical projections \( pr_{i}\left( x_{1},\ldots ,x_{k}\right) =x_{i}\). All notation and notion used and not defined are standard, see [1].

2 A New Convergence Result

We need the following result, see [12, Corollary 1] and also [9, 11]. The case \(k=1\) can be found in [8, page 15–16 and 24].

Lemma 1

Let \(f\in C_{2\pi }\left( \mathbb {R}^{k}\right) \). Then for all \(\varepsilon >0\) there exists \(\eta _{\varepsilon }>0\) such that for all \(\left( s_{1},\ldots ,s_{k}\right) \in \mathbb {R}^{k}\), all \( \left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}\) we have

$$\begin{aligned} \left| f\left( s_{1},\ldots ,s_{k}\right) -f\left( t_{1},\ldots ,t_{k}\right) \right| <\varepsilon +\eta _{\varepsilon }\sum \limits _{j=1}^{k}\left[ \textbf{1}-\cos \left( s_{j}-t_{j}\right) \right] . \end{aligned}$$

The following technical lemma is the key ingredient for the proof of the main result of this paper.

Lemma 2

Let \(f\in C_{2\pi }\left( \mathbb {R}^{k}\right) \). Then for all \(\varepsilon >0\) there exists \(\eta _{\varepsilon }>0\) such that for all linear positive functionals \(x^{*}:C_{2\pi }\left( \mathbb {R}^{k}\right) \rightarrow \mathbb {R}\) and all \(\left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R} ^{k}\) the following relation holds

$$\begin{aligned}{} & {} \left| x^{*}\left( f_{t_{1},\ldots ,t_{k}}\right) -f\left( t_{1},\ldots ,t_{k}\right) \right| \\{} & {} \quad \le \varepsilon x^{*}\left( \textbf{1}\right) +\eta _{\varepsilon }\sum \limits _{j=1}^{k}\left[ x^{*}\left( \textbf{1}\right) -x^{*}\left( \cos \circ pr_{j}\right) \right] +\left\| f\right\| \left| x^{*}\left( \textbf{1}\right) -1\right| . \end{aligned}$$

Proof

Let \(\varepsilon >0\). Then from Lemma 1 there exists \(\eta _{\varepsilon }>0\) such that for all \(\left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}\), all \(\left( \theta _{1},\ldots ,\theta _{k}\right) \in \mathbb {R}^{k}\) we have

$$\begin{aligned} \left| f\left( t_{1}-\theta _{1},\ldots ,t_{k}-\theta _{k}\right) -f\left( t_{1},\ldots ,t_{k}\right) \right| <\varepsilon +\eta _{\varepsilon }\sum \limits _{j=1}^{k}\left( 1-\cos \theta _{j}\right) \end{aligned}$$

that is for all \(\left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}\),

$$\begin{aligned} \left| f_{t_{1},\ldots ,t_{k}}-f\left( t_{1},\ldots ,t_{k}\right) \textbf{1} \right| <\varepsilon \textbf{1}+\eta _{\varepsilon }\sum \limits _{j=1}^{k}\left( \textbf{1}-\cos \circ pr_{j}\right) \text { in } C_{2\pi }\left( \mathbb {R}^{k}\right) . \end{aligned}$$

If \(x^{*}:C_{2\pi }\left( \mathbb {R}^{k}\right) \rightarrow \mathbb {R}\) is a linear positive functional we deduce that

$$\begin{aligned} \left| x^{*}\left( f_{t_{1},\ldots ,t_{k}}\right) -f\left( t_{1},\ldots ,t_{k}\right) x^{*}\left( \textbf{1}\right) \right| \le \varepsilon x^{*}\left( \textbf{1}\right) +\eta _{\varepsilon }\sum \limits _{j=1}^{k}\left[ x^{*}\left( \textbf{1}\right) -x^{*}\left( \cos \circ pr_{j}\right) \right] . \end{aligned}$$

Then

$$\begin{aligned}{} & {} \left| x^{*}\left( f_{t_{1},\ldots ,t_{k}}\right) -f\left( t_{1},\ldots ,t_{k}\right) \right| \\{} & {} \quad \le \left| x^{*}\left( f_{t_{1},\ldots ,t_{k}}\right) -f\left( t_{1},\ldots ,t_{k}\right) x^{*}\left( \textbf{1}\right) \right| +\left| f\left( t_{1},\ldots ,t_{k}\right) x^{*}\left( \textbf{1}\right) -f\left( t_{1},\ldots ,t_{k}\right) \right| \\{} & {} \quad \le \varepsilon x^{*}\left( \textbf{1}\right) +\eta _{\varepsilon }\sum \limits _{j=1}^{k}\left[ x^{*}\left( \textbf{1}\right) -x^{*}\left( \cos \circ pr_{j}\right) \right] +\left\| f\right\| \left| x^{*}\left( \textbf{1}\right) -1\right| . \end{aligned}$$

\(\square \)

The next result is the main result of this paper.

Theorem 1

Let \(x_{n}^{*}:C_{2\pi }\left( \mathbb {R} ^{k}\right) \rightarrow \mathbb {R}\) be a sequence of linear positive functionals. Then the following assertions are equivalent:

  1. (i)

    For all \(f\in C_{2\pi }\left( \mathbb {R}^{k}\right) \) we have \( \lim \nolimits _{n\rightarrow \infty }x_{n}^{*}\left( f_{t_{1},\ldots ,t_{k}}\right) =f\left( t_{1},\ldots ,t_{k}\right) \) uniformly with respect to \(\left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}\).

  2. (ii)

    For all \(f\in C_{2\pi }\left( \mathbb {R}^{k}\right) \) and all \(\left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}\) we have \(\lim \nolimits _{n \rightarrow \infty }x_{n}^{*}\left( f_{t_{1},\ldots ,t_{k}}\right) =f\left( t_{1},\ldots ,t_{k}\right) \).

  3. (iii)

    \(\lim \nolimits _{n\rightarrow \infty }x_{n}^{*}\left( \textbf{1} \right) =1\) and \(\lim \nolimits _{n\rightarrow \infty }x_{n}^{*}\left( \cos \circ pr_{j}\right) =1\) for all \(j=1\), ..., k.

Proof

(i)\(\Rightarrow \)(ii) is trivial. (ii)\(\Rightarrow \)(iii). From (ii) we have, in particular, \(\lim \nolimits _{n\rightarrow \infty }x_{n}^{*}\left( f_{0,\ldots ,0}\right) =f\left( 0,\ldots ,0\right) \) for all \(f\in C_{2\pi }\left( \mathbb {R}^{k}\right) \). Then since for \(f\in \{ \textbf{1},\cos \circ pr_{1},\ldots ,\cos \circ pr_{k}\} \) we have \(f_{0,\ldots ,0}=f\) and \(f\left( 0,\ldots ,0\right) =1\) we get (iii).

(iii)\(\Rightarrow \)(i) Let \(f\in C_{2\pi }\left( \mathbb {R}^{k}\right) \) and \(\varepsilon >0\). From Lemma 2 there exists \(\delta _{\varepsilon }>0\) such that for all \(n\in \mathbb {N}\) and all \(\left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}\) the following relation holds

$$\begin{aligned}{} & {} \left| x_{n}^{*}\left( f_{t_{1},\ldots ,t_{k}}\right) -f\left( t_{1},\ldots ,t_{k}\right) \right| \\{} & {} \qquad \le \frac{\varepsilon }{6}x_{n}^{*}\left( \textbf{1}\right) +\delta _{\varepsilon }\sum \limits _{j=1}^{k}\left[ x_{n}^{*}\left( \textbf{1} \right) -x_{n}^{*}\left( \cos \circ pr_{j}\right) \right] \\{} & {} \qquad \quad +\left\| f\right\| \left| x_{n}^{*}\left( \textbf{1}\right) -1\right| . \end{aligned}$$

Since \(\lim \nolimits _{n\rightarrow \infty }x_{n}^{*}\left( \textbf{1} \right) =1<2\) there exists \(n_{0}\in \mathbb {N}\) such that for all \(n\ge n_{0}\) we have \(x_{n}^{*}\left( \textbf{1}\right) <2\) and there exists \( mss_{\varepsilon }\in \mathbb {N}\) such that for all \(n\ge m_{\varepsilon }\) we have \(\left| x_{n}^{*}\left( \textbf{1}\right) -1\right| < \frac{\varepsilon }{3\left( \left\| f\right\| +1\right) }\). From (iii) \(\lim \nolimits _{n\rightarrow \infty }\sum \nolimits _{j=1}^{k}\left[ x_{n}^{*}\left( \textbf{1}\right) -x_{n}^{*}\left( \cos \circ pr_{j}\right) \right] =0\), hence there exists \(p_{\varepsilon }\in \mathbb {N}\) such that for all \(n\ge p_{\varepsilon }\) we have \(\sum \nolimits _{j=1}^{k}\left[ x_{n}^{*}\left( \textbf{1}\right) -x_{n}^{*}\right. \left. \left( \cos \circ pr_{j}\right) \right] <\frac{\varepsilon }{3\delta _{\varepsilon }}\). We deduce that for all \(n\ge \max \left( n_{0},m_{\varepsilon },p_{\varepsilon }\right) \) and all \(\left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}\) we have \(\left| x_{n}^{*}\left( f_{t_{1},\ldots ,t_{k}}\right) -f\left( t_{1},\ldots ,t_{k}\right) \right| <\varepsilon \) that is (i). \(\square \)

Let us state as an explicit result the univariate case of Theorem 1.

Corollary 1

Let \(x_{n}^{*}:C_{2\pi }\left( \mathbb {R}\right) \rightarrow \mathbb {R}\) be a sequence of linear positive functionals. Then the following assertions are equivalent:

  1. (i)

    For all \(f\in C_{2\pi }\left( \mathbb {R}\right) \) we have \( \lim \nolimits _{n\rightarrow \infty }x_{n}^{*}\left( f_{t}\right) =f\left( t\right) \) uniformly with respect to \(t\in \mathbb {R}\).

  2. (ii)

    For all \(f\in C_{2\pi }\left( \mathbb {R}\right) \) and all \(t\in \mathbb { R}\) we have \(\lim \nolimits _{n\rightarrow \infty }x_{n}^{*}\left( f_{t}\right) =f\left( t\right) \).

  3. (iii)

    \(\lim \nolimits _{n\rightarrow \infty }x_{n}^{*}\left( \textbf{1} \right) =1\) and \(\lim \nolimits _{n\rightarrow \infty }x_{n}^{*}\left( \cos \right) =1\).

We prove in the sequel a new convergence result for univariate singular integrals. The novelty of this result is that the test set of convergence consists only of two functions \(\textbf{1}\) and \(\cos \). Let us mention that in all the results which appear in the literature the test set of convergence consist of the three functions \(\textbf{1}\), \(\cos \) and \(\sin \) , see the book of Korovkin [8], or Butzer and Nessel [3, page 58, Theorem 1.3.7 and Corollary 1.3.8]. For an operator version of the classical Korovkin theorem we recommend the reader to consult our very recent paper [13].

Theorem 2

Let \(K_{n}:\mathbb {R}\rightarrow \left[ 0,\infty \right) \) be a sequence of continuous functions. Then the following assertions are equivalent:

  1. (i)

    For all \(f\in C_{2\pi }\left( \mathbb {R}\right) \) we have

    $$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{1}{2\pi }\int _{-\pi }^{\pi }K_{n}\left( \theta \right) f\left( t-\theta \right) \textrm{d}\theta =f\left( t\right) \text { uniformly with respect to }t\in \mathbb {R}. \end{aligned}$$
  2. (ii)

    For all \(f\in C_{2\pi }\left( \mathbb {R}\right) \) and all \(t\in \mathbb { R}\) we have

    $$\begin{aligned}{} & {} \lim \limits _{n\rightarrow \infty }\frac{1}{2\pi }\int _{-\pi }^{\pi }K_{n}\left( \theta \right) f\left( t-\theta \right) \textrm{d}\theta \nonumber \\ {}{} & {} =f\left( t\right) . \end{aligned}$$
  3. (iii)

    \(\lim \nolimits _{n\rightarrow \infty }\frac{1}{2\pi }\int _{-\pi }^{\pi }K_{n}\left( t\right) \textrm{d}t=1\) and \(\lim \nolimits _{n\rightarrow \infty }\frac{1}{ 2\pi }\int _{-\pi }^{\pi }K_{n}\left( t\right) \cos tdt=1\).

Proof

Let us define \(x_{n}^{*}:C_{2\pi }\left( \mathbb {R}\right) \rightarrow \mathbb {R}\) by \(x_{n}^{*}\left( f\right) =\frac{1}{2\pi }\int _{-\pi }^{\pi }K_{n}\left( \theta \right) f\left( \theta \right) \textrm{d}\theta \) and note that \(x_{n}^{*}\left( f_{t}\right) =\frac{1}{2\pi }\int _{-\pi }^{\pi }K_{n}\left( \theta \right) f_{t}\left( \theta \right) \textrm{d}\theta =\frac{1}{ 2\pi }\int _{-\pi }^{\pi }K_{n}\left( \theta \right) f\left( t-\theta \right) \textrm{d}\theta \). Then the equivalences from the statement follow from Corollary 1. \(\square \)

Our next objective is to prove a multivariate version of Theorem 2. Let us introduce some common notations.

Definition 1

Let \(k\ge 2\) be a natural number and \(K: \mathbb {R}^{k}\rightarrow \mathbb {R}\) a continuous function. We define \( K^{\left\langle 1\right\rangle },\ldots ,K^{\left\langle k\right\rangle }: \mathbb {R}\rightarrow \mathbb {R}\) by

$$\begin{aligned}{} & {} K^{\left\langle 1\right\rangle }\left( t\right) =\frac{1}{\left( 2\pi \right) ^{k-1}}\int _{\left[ -\pi ,\pi \right] ^{k-1}}K\left( t,\theta _{2},\ldots ,\theta _{k}\right) \textrm{d}\theta _{2}\cdot \cdot \cdot \textrm{d}\theta _{k},\\{} & {} K^{\left\langle 2\right\rangle }\left( t\right) =\frac{1}{\left( 2\pi \right) ^{k-1}}\int _{\left[ -\pi ,\pi \right] ^{k-1}}K\left( \theta _{1},t,\theta _{3},\ldots ,\theta _{k}\right) \textrm{d}\theta _{1}\textrm{d}\theta _{3}\cdot \cdot \cdot \textrm{d}\theta _{k}{,\ldots ,}\\{} & {} K^{\left\langle k\right\rangle }\left( t\right) =\frac{1}{\left( 2\pi \right) ^{k-1}}\int _{\left[ -\pi ,\pi \right] ^{k-1}}K\left( \theta _{1},\ldots ,\theta _{k-1},t\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{k-1}. \end{aligned}$$

Let us note that by Fubini’s theorem for every \(j=1\), ..., k we have

$$\begin{aligned} \frac{1}{2\pi }\int _{-\pi }^{\pi }K^{\left\langle j\right\rangle }\left( t\right) \textrm{d}t=\frac{1}{\left( 2\pi \right) ^{k}}\int _{\left[ -\pi ,\pi \right] ^{k}}K\left( \theta _{1},\ldots ,\theta _{k}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{k}. \end{aligned}$$

Theorem 3

Let \(k\ge 2\) be a natural number and \(K_{n}:\mathbb {R} ^{k}\rightarrow \left[ 0,\infty \right) \) a sequence of continuous functions. Then the following assertions are equivalent:

  1. (i)

    For all \(f\in C_{2\pi }\left( \mathbb {R}^{k}\right) \) we have

    $$\begin{aligned}{} & {} \lim \limits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{k}}\int _{ \left[ -\pi ,\pi \right] ^{k}}K_{n}\left( \theta _{1},\ldots ,\theta _{k}\right) f\left( t_{1}-\theta _{1},\ldots ,t_{k}-\theta _{k}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{k} \\{} & {} \quad =f\left( t_{1},\ldots ,t_{k}\right) \text { uniformly with respect to }\left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}. \end{aligned}$$
  2. (ii)

    For all \(f\in C_{2\pi }\left( \mathbb {R}^{k}\right) \) and all \(\left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}\) we have

    $$\begin{aligned}{} & {} \lim \limits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{k}}\int _{ \left[ -\pi ,\pi \right] ^{k}}K_{n}\left( \theta _{1},\ldots ,\theta _{k}\right) f\left( t_{1}-\theta _{1},\ldots ,t_{k}-\theta _{k}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{k}\\{} & {} \qquad =f\left( t_{1},\ldots ,t_{k}\right) .\text { } \end{aligned}$$
  3. (iii)
    $$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{k}}\int _{ \left[ -\pi ,\pi \right] ^{k}}K_{n}\left( \theta _{1},\ldots ,\theta _{k}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{k}=1 \end{aligned}$$

    and

    $$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{1}{2\pi }\int _{-\pi }^{\pi }K_{n}^{\left\langle j\right\rangle }\left( \theta _{j}\right) \cos \theta _{j}\textrm{d}\theta _{j}=1\text { for every }j=1,\ldots ,k. \end{aligned}$$

Proof

Let us define \(x_{n}^{*}:C_{2\pi }\left( \mathbb {R}^{k}\right) \rightarrow \mathbb {R}\) by \(x_{n}^{*}\left( f\right) =\frac{1}{\left( 2\pi \right) ^{k}}\int _{\left[ -\pi ,\pi \right] ^{k}}K_{n}\left( \theta _{1},\ldots ,\theta _{k}\right) f\left( \theta _{1},\ldots ,\theta _{k}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{k}\) and note that for all \(\left( t_{1},\ldots ,t_{k}\right) \in \mathbb {R}^{k}\) we have

$$\begin{aligned} x_{n}^{*}\left( f_{t_{1},\ldots ,t_{k}}\right)= & {} \frac{1}{\left( 2\pi \right) ^{k}}\int _{\left[ -\pi ,\pi \right] ^{k}}K_{n}\left( \theta _{1},\ldots ,\theta _{k}\right) f_{t_{1},\ldots ,t_{k}}\left( \theta _{1},\ldots ,\theta _{k}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{k} \\= & {} \frac{1}{\left( 2\pi \right) ^{k}}\int _{\left[ -\pi ,\pi \right] ^{k}}K_{n}\left( \theta _{1},\ldots ,\theta _{k}\right) f\left( t_{1}-\theta _{1},\ldots ,t_{k}-\theta _{k}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{k}. \end{aligned}$$

Then the equivalences from the statement follow from Theorem 1. \(\square \)

3 Some Calculations

In the rest of the paper we will apply the general result in Theorem 3 in the study of convergence of some multivariate operators, see Theorems 4 and 5. This kind of operators are multivariate versions of the operators studied by Korovkin in the paper [7, Theorem 1] and in the bivariate case by Fomin in [4, Theorem 1]; related questions are treated in [2,3,4,5,6].

Proposition 1

Let \(h:\left\{ 0,\ldots ,n\right\} ^{2}\rightarrow \mathbb {C}\) and \(f:\left\{ -n,\ldots ,-1,0,1,\ldots ,n\right\} \rightarrow \mathbb {C}\). Then:

  1. (i)

    \(\sum \nolimits _{k,l=0,k>l}^{n}h\left( k,l\right) f\left( k-l\right) =\nu _{1}f\left( 1\right) +\nu _{2}f\left( 2\right) +\nu _{3}f\left( 3\right) +\cdot \cdot \cdot +\nu _{n}f\left( n\right) \) where

    $$\begin{aligned} \nu _{1}= & {} h\left( 1,0\right) +\ h\left( 2,1\right) +\ h\left( 3,2\right) +\cdot \cdot \cdot +h\left( n-1,n-2\right) +h\left( n,n-1\right) , \\ \nu _{2}= & {} h\left( 2,0\right) +\ h\left( 3,1\right) +\ h\left( 4,2\right) +\cdot \cdot \cdot +h\left( n,n-2\right) ,\ldots , \\ \nu _{n-1}= & {} h\left( n-1,0\right) +h\left( n,1\right) ,\nu _{n}=h\left( n,0\right) . \end{aligned}$$
  2. (ii)

    \(\sum \nolimits _{k,l=0,k<l}^{n}h\left( k,l\right) f\left( k-l\right) =\lambda _{1}f\left( -1\right) +\lambda _{2}f\left( -2\right) +\lambda _{3}f\left( -3\right) +\cdot \cdot \cdot +\lambda _{n}f\left( -n\right) \), where

    $$\begin{aligned} \lambda _{1}= & {} h\left( 0,1\right) +\ h\left( 1,2\right) +\ h\left( 2,3\right) +\cdot \cdot \cdot +h\left( n-2,n-1\right) +h\left( n-1,n\right) ,\\ \lambda _{2}= & {} h\left( 0,2\right) +\ h\left( 1,3\right) +\ h\left( 2,4\right) +\cdot \cdot \cdot +h\left( n-2,n\right) ,\ldots , \\ \lambda _{n-1}= & {} h\left( 0,n-1\right) +h\left( 1,n\right) ,\lambda _{n}=h\left( 0,n\right) . \end{aligned}$$

Proof

(i) We have

$$\begin{aligned}{} & {} \sum \limits _{k,l=0,k>l}^{n}h\left( k,l\right) f\left( k-l\right) =\sum \limits _{l=0}^{n-1}\sum \limits _{k=l+1}^{n}h\left( k,l\right) f\left( k-l\right) \\{} & {} \quad =\sum \limits _{l=0}^{n-1}\left( \sum \limits _{k=1}^{n-l}h\left( l+k,l\right) f\left( k\right) \right) =\sum \limits _{k=1}^{n}h\left( k,0\right) f\left( k\right) +\sum \limits _{k=1}^{n-1}h\left( k+1,1\right) f\left( k\right) \\{} & {} \quad \quad +\sum \limits _{k=1}^{n-2}h\left( k+2,2\right) f\left( k\right) +\cdot \cdot \cdot +\sum \limits _{k=1}^{2}h\left( n-2+k,n-2\right) f\left( k\right) +h\left( n,n-1\right) f\left( 1\right) \\{} & {} \quad =h\left( 1,0\right) f\left( 1\right) +h\left( 2,0\right) f\left( 2\right) +h\left( 3,0\right) f\left( 3\right) +\cdot \cdot \cdot \ +h\left( n-1,0\right) \\{} & {} \quad \quad f\left( n-1\right) +h\left( n,0\right) f\left( n\right) +\ h\left( 2,1\right) f\left( 1\right) +h\left( 3,1\right) f\left( 2\right) +h\left( 4,1\right) \\ {}{} & {} \qquad \times f\left( 3\right) +\cdot \cdot \cdot \ +\ h\left( n,1\right) f\left( n-1\right) \\{} & {} \quad \quad +\ h\left( 3,2\right) f\left( 1\right) +h\left( 4,2\right) f\left( 2\right) +h\left( 5,2\right) f\left( 3\right) +\cdot \cdot \cdot +h\left( n,2\right) f\left( n-2\right) +\cdot \cdot \cdot \\{} & {} \quad \quad +h\left( n-1,n-2\right) f\left( 1\right) +h\left( n,n-2\right) f\left( 2\right) +h\left( n,n-1\right) f\left( 1\right) \\{} & {} \quad =\nu _{1}f\left( 1\right) +\nu _{2}f\left( 2\right) +\nu _{3}f\left( 3\right) +\cdot \cdot \cdot +\nu _{n}f\left( n\right) . \end{aligned}$$

(ii) Let us define \(g\left( k,l\right) =h\left( l,k\right) \) and \(\varphi \left( k\right) =f\left( -k\right) \). Then \(\sum \nolimits _{k,l=0,k<l}^{n}h \left( k,l\right) f\left( k-l\right) =\sum \nolimits _{k,l=0,l>k}^{n}g\left( l,k\right) \varphi \left( l-k\right) \) and by (i) \(\sum \nolimits _{k,l=0,l>k}^{n}g\left( l,k\right) \varphi \left( l-k\right) =\lambda _{1}\varphi \left( 1\right) +\lambda _{2}\varphi \left( 2\right) +\cdot \cdot \cdot +\lambda _{n}\varphi \left( n\right) =\lambda _{1}f\left( -1\right) +\lambda _{2}f\left( -2\right) +\lambda _{3}f\left( -3\right) +\cdot \cdot \cdot +\lambda _{n}f\left( -n\right) \) where

$$\begin{aligned} \lambda _{1}= & {} g\left( 1,0\right) +\ g\left( 2,1\right) +\ g\left( 3,2\right) +\cdot \cdot \cdot +g\left( n-1,n-2\right) +g\left( n,n-1\right) \\= & {} h\left( 0,1\right) +\ h\left( 1,2\right) +\ h\left( 2,3\right) +\cdot \cdot \cdot +h\left( n-2,n-1\right) +h\left( n-1,n\right) , \\ \lambda _{2}= & {} g\left( 2,0\right) +\ g\left( 3,1\right) +\ g\left( 4,2\right) +\cdot \cdot \cdot +g\left( n,n-2\right) \\= & {} h\left( 0,2\right) +\ h\left( 1,3\right) +\ h\left( 2,4\right) +\cdot \cdot \cdot +h\left( n-2,n\right) ,\ldots , \\ \lambda _{n-1}= & {} g\left( n-1,0\right) + g\left( n,1\right) = h\left( 0,n-1\right) +h\left( 1,n\right) ,\lambda _{n}=g\left( n,0\right) =h\left( 0,n\right) . \end{aligned}$$

\(\square \)

Corollary 2

(i) Let \(h:\left\{ 0,\ldots ,n\right\} ^{2}\rightarrow \mathbb {C}\) be such that \(h\left( k,l\right) =h\left( l,k\right) \) for all \(\left( k,l\right) \in \left\{ 0,\ldots ,n\right\} ^{2}\) and \(f:\left\{ -n,\ldots ,-1,0,1,\ldots ,n\right\} \rightarrow \mathbb {C}\) an even function. Then

$$\begin{aligned} \sum \limits _{k,l=0,k\ne l}^{n}h\left( k,l\right) f\left( k-l\right) =2\lambda _{1}f\left( 1\right) +2\lambda _{2}f\left( 2\right) +2\lambda _{3}f\left( 3\right) +\cdot \cdot \cdot +2\lambda _{n}f\left( n\right) \end{aligned}$$

where \(\lambda _{1}=h\left( 0,1\right) +h\left( 1,2\right) +\cdot \cdot \cdot +h\left( n-1,n\right) \), \(\lambda _{2}=h\left( 0,2\right) +h\left( 1,3\right) +\cdot \cdot \cdot +h\left( n-2,n\right) \), ..., \(\lambda _{n-1}=h\left( 0,n-1\right) +h\left( 1,n\right) \), \(\lambda _{n}=h\left( 0,n\right) \).

(ii) Let \(h:\left\{ 0,\ldots ,n\right\} ^{2}\rightarrow \mathbb {C}\) be such that \( h\left( k,l\right) =-h\left( l,k\right) \) for all \(\left( k,l\right) \in \left\{ 0,\ldots ,n\right\} ^{2}\) and \(f:\left\{ -n,\ldots ,-1,0,1,\ldots ,n\right\} \rightarrow \mathbb {C}\) an odd function. Then

$$\begin{aligned} \sum \limits _{k,l=0,k\ne l}^{n}h\left( k,l\right) f\left( k-l\right) =2\nu _{1}f\left( 1\right) +2\nu _{2}f\left( 2\right) +2\nu _{3}f\left( 3\right) +\cdot \cdot \cdot +2\nu _{n}f\left( n\right) \end{aligned}$$

where \(\nu _{1}=h\left( 1,0\right) +\ h\left( 2,1\right) +\ h\left( 3,2\right) +\cdot \cdot \cdot +h\left( n-1,n-2\right) +h\left( n,n-1\right) \), \(\nu _{2}=h\left( 2,0\right) +\ h\left( 3,1\right) +\ h\left( 4,2\right) +\cdot \cdot \cdot +h\left( n,n-2\right) \), ..., \(\nu _{n-1}=h\left( n-1,0\right) +h\left( n,1\right) \), \(\nu _{n}=h\left( n,0\right) \).

Proof

(i) Since f is even, by Proposition 1(ii)\( \sum \nolimits _{k,l=0,k<l}^{n}h\left( k,l\right) f\left( k-l\right) =\lambda _{1}f\left( 1\right) +\lambda _{2}f\left( 2\right) +\cdot \cdot \cdot +\lambda _{n}f\left( n\right) \). By Proposition 1(i) \( \sum \nolimits _{k,l=0,k>l}^{n}h\left( k,l\right) f\left( k-l\right) =\nu _{1}f\left( 1\right) +\nu _{2}f\left( 2\right) +\cdot \cdot \cdot +\nu _{n}f\left( n\right) \). Since \(h\left( k,l\right) =h\left( l,k\right) \), \( \forall \left( k,l\right) \in \left\{ 0,\ldots ,n\right\} ^{2}\) we deduce that \( \nu _{1}=\lambda _{1}\), ..., \(\nu _{n}=\lambda _{n}\) and hence \( \sum \nolimits _{k,l=0,k>l}^{n}h\left( k,l\right) f\left( k-l\right) =\lambda _{1}f\left( 1\right) +\lambda _{2}f\left( 2\right) +\cdot \cdot \cdot +\lambda _{n}f\left( n\right) \). To finish the proof let us note that \( \sum \nolimits _{k,l=0,k\ne l}^{n}h\left( k,l\right) f\left( k-l\right) =\sum \nolimits _{k,l=0,k>l}^{n}h\left( k,l\right) f\left( k-l\right) +\sum \nolimits _{k,l=0,k<l}^{n}h\left( k,l\right) f\left( k-l\right) \).

(ii) Since f is odd from Proposition 1(ii)

\(\sum \nolimits _{k,l=0,k<l}^{n}h\left( k,l\right) f\left( k-l\right) =-\left( \lambda _{1}f\left( 1\right) +\lambda _{2}f\left( 2\right) +\cdot \cdot \cdot +\lambda _{n}f\left( n\right) \right) \). Since \(h\left( k,l\right) =-h\left( l,k\right) \) we get \(\lambda _{1}=-\nu _{1}\), ..., \(\lambda _{n}=-\nu _{n}\) and hence \(\sum \nolimits _{k,l=0,k<l}^{n}h\left( k,l\right) f\left( k-l\right) =\nu _{1}f\left( 1\right) +\nu _{2}f\left( 2\right) +\cdot \cdot \cdot +\nu _{n}f\left( n\right) \). Also by Proposition 1(i) \(\sum \nolimits _{k,l=0,k>l}^{n}h\left( k,l\right) f\left( k-l\right) =\nu _{1}f\left( 1\right) +\nu _{2}f\left( 2\right) +\cdot \cdot \cdot +\nu _{n}f\left( n\right) \) and to finish the proof let us note that \( \sum \nolimits _{k,l=0,k\ne l}^{n}h\left( k,l\right) f\left( k-l\right) =\sum \nolimits _{k,l=0,k>l}^{n}h\left( k,l\right) f\left( k-l\right) +\sum \nolimits _{k,l=0,k<l}^{n}h\left( k,l\right) f\left( k-l\right) \). \(\square \)

If \(z\in \mathbb {C}\) then \(\mathfrak {R}\left( z\right) \) denotes the real part of z and \(\mathfrak {I}m\left( z\right) \) denotes the imaginary part of z.

Proposition 2

Let \(\varphi _{n}:\left\{ 0,\ldots ,n\right\} \rightarrow \mathbb {C}\). Then for all \(\theta \in \mathbb {R}\) the following equality holds

$$\begin{aligned}{} & {} \left| \sum \limits _{k=0}^{n}\varphi _{n}\left( k\right) e^{ik\theta }\right| ^{2}=\alpha _{n0}+2\alpha _{n1}\cos \theta +2\alpha _{n2}\cos 2\theta +\cdot \cdot \cdot +2\alpha _{nn-1}\cos \left( n{-}1\right) \\{} & {} \quad \theta +2\alpha _{nn}\cos n\theta \\{} & {} \qquad +2\beta _{n1}\sin \theta +2\beta _{n2}\sin 2\theta +\cdot \cdot \cdot +2\beta _{nn-1}\sin \left( n-1\right) \theta +2\beta _{nn}\sin n\theta \end{aligned}$$

where \(\alpha _{n0}=\sum \nolimits _{k=0}^{n}\left| \varphi _{n}\left( k\right) \right| ^{2}\), \(\alpha _{n1}=\sum \nolimits _{k=0}^{n-1}\mathfrak {R} \left( \varphi _{n}\left( k\right) \overline{\varphi _{n}\left( k+1\right) } \right) \), ..., \(\alpha _{nn-1}=\mathfrak {R}\left( \varphi _{n}\left( 0\right) \overline{\varphi _{n}\left( n-1\right) }\right) +\mathfrak {R} \left( \varphi _{n}\left( 1\right) \overline{\varphi _{n}\left( n\right) } \right) \), \(\alpha _{nn}=\mathfrak {R}\left( \varphi _{n}\left( 0\right) \overline{\varphi _{n}\left( n\right) }\right) \), \(\beta _{n1}=\sum \nolimits _{k=0}^{n-1}\mathfrak {I}m\left( \varphi _{n}\left( k+1\right) \overline{\varphi _{n}\left( k\right) }\right) \), ..., \(\beta _{nn-1}=\mathfrak {I}m\left( \varphi _{n}\left( n-1\right) \overline{\varphi _{n}\left( 0\right) }\right) +\mathfrak {I}m\left( \varphi _{n}\left( n\right) \overline{\varphi _{n}\left( 1\right) }\right) \), \(\beta _{nn}= \mathfrak {I}m\left( \varphi _{n}\left( n\right) \overline{\varphi _{n}\left( 0\right) }\right) \).

Proof

We have

$$\begin{aligned} \left| \sum \limits _{k=0}^{n}\varphi _{n}\left( k\right) e^{ik\theta }\right| ^{2}= & {} \left( \sum \limits _{k=0}^{n}\varphi _{n}\left( k\right) e^{ik\theta }\right) \left( \overline{\sum \limits _{l=0}^{n}\varphi _{n}\left( l\right) e^{il\theta }}\right) \nonumber \\ {}= & {} \sum \limits _{k,l=0}^{n}\varphi _{n}\left( k\right) \overline{\varphi _{n}\left( l\right) }e^{i\left( k-l\right) \theta } \end{aligned}$$

and since the left member is real we deduce that

$$\begin{aligned} \left| \sum \limits _{k=0}^{n}\varphi _{n}\left( k\right) e^{ik\theta }\right| ^{2}=\sum \limits _{k,l=0}^{n}\mathfrak {R}\left( \varphi _{n}\left( k\right) \overline{\varphi _{n}\left( l\right) }e^{i\left( k-l\right) \theta }\right) . \end{aligned}$$

From the equality \(\mathfrak {R}\left( z\overline{w}\right) =\left( \mathfrak { R}z\right) \left( \mathfrak {R}w\right) +\left( \mathfrak {I}mz\right) \left( \mathfrak {I}mw\right) \), \(z,w\in \mathbb {C}\) we get

$$\begin{aligned} \mathfrak {R}\left( \varphi _{n}\left( k\right) \overline{\varphi _{n}\left( l\right) }e^{i\left( k-l\right) \theta }\right)= & {} \mathfrak {R}\left( \varphi _{n}\left( k\right) \overline{\varphi _{n}\left( l\right) }\right) \cos \left( k-l\right) \theta \\{} & {} +\mathfrak {I}m\left( \varphi _{n}\left( k\right) \overline{\varphi _{n}\left( l\right) }\right) \sin \left( k-l\right) \theta \end{aligned}$$

and thus

$$\begin{aligned} \left| \sum \limits _{k=0}^{n}\varphi _{n}\left( k\right) e^{ik\theta }\right| ^{2}= & {} \sum \limits _{k=0}^{n}\left| \varphi _{n}\left( k\right) \right| ^{2}+\sum \limits _{k,l=0,k\ne l}^{n}\mathfrak {R}\left( \varphi _{n}\left( k\right) \overline{\varphi _{n}\left( l\right) }\right) \cos \left( k-l\right) \theta \\{} & {} +\sum \limits _{k,l=0,k\ne l}^{n}\mathfrak {I}\left( \varphi _{n}\left( k\right) \overline{\varphi _{n}\left( l\right) }\right) \sin \left( k-l\right) \theta . \end{aligned}$$

Since \(h\left( k,l\right) =\mathfrak {R}\left( \varphi _{n}\left( k\right) \overline{\varphi _{n}\left( l\right) }\right) \) has the property \(h\left( k,l\right) =h\left( l,k\right) \) and \(\cos \) is an even function, from Corollary 2(i)

$$\begin{aligned} \sum \limits _{k,l=0,k\ne l}^{n}\mathfrak {R}\left( \varphi _{n}\left( k\right) \overline{\varphi _{n}\left( l\right) }\right) \cos \left( k-l\right) \theta= & {} 2\alpha _{n1}\cos \theta +2\alpha _{n2}\cos 2\theta +\cdot \cdot \cdot \\{} & {} +2\alpha _{nn}\cos n\theta . \end{aligned}$$

Since \(h\left( k,l\right) =\mathfrak {I}m\left( \varphi _{n}\left( k\right) \overline{\varphi _{n}\left( l\right) }\right) \) has the property \(h\left( k,l\right) =-h\left( l,k\right) \) (use \(\mathfrak {I}m\left( z\overline{w} \right) =\left( \mathfrak {I}mz\right) \left( \mathfrak {R}w\right) -\left( \mathfrak {R}z\right) \left( \mathfrak {I}mw\right) \), \(z,w\in \mathbb {C}\)) and \(\sin \) is an odd function, from Corollary 2(ii) we get

$$\begin{aligned} \sum \limits _{k,l=0,k\ne l}^{n}\mathfrak {I}m\left( \varphi _{n}\left( k\right) \overline{\varphi _{n}\left( l\right) }\right) \sin \left( k-l\right) \theta= & {} 2\beta _{n1}\sin \theta +2\beta _{n2}\sin 2\theta +\cdot \cdot \cdot \\{} & {} +2\beta _{nn}\sin n\theta . \end{aligned}$$

\(\square \)

The real case of Proposition 2, that is \(\varphi _{n}\) takes the real values, is a well-known result, see [10, Problem 39 page 77 with solution at pages 258–259].

We need the following well-known result. For the sake of completeness we include its proof.

Proposition 3

Let m be a natural number, \(\beta _{n},\gamma _{n}:\left\{ 0,1,\ldots ,n\right\} ^{m}\rightarrow \mathbb {C}\) and \( B_{n},C_{n}:\mathbb {R}^{m}\rightarrow \mathbb {C}\) defined by

$$\begin{aligned} B_{n}\left( \theta _{1},\ldots ,\theta _{m}\right)= & {} \sum \limits _{k_{1},\ldots ,k_{m}=0}^{n}\beta _{n}\left( k_{1},\ldots ,k_{m}\right) e^{ik_{1}\theta _{1}}\cdot \cdot \cdot e^{ik_{m}\theta _{m}}, \\ C_{n}\left( \theta _{1},\ldots ,\theta _{m}\right)= & {} \sum \limits _{k_{1},\ldots ,k_{m}=0}^{n}\gamma _{n}\left( k_{1},\ldots ,k_{m}\right) e^{ik_{1}\theta _{1}}\cdot \cdot \cdot e^{ik_{m}\theta _{m}}. \end{aligned}$$

Then

$$\begin{aligned}{} & {} \frac{1}{\left( 2\pi \right) ^{m}}\int _{\left[ -\pi ,\pi \right] ^{m}}B_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) \overline{C_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) }\textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m} \\{} & {} \qquad =\sum \limits _{k_{1},\ldots ,k_{m}=0}^{n}\beta _{n}\left( k_{1},\ldots ,k_{m}\right) \overline{\gamma _{n}\left( k_{1},\ldots ,k_{m}\right) }. \end{aligned}$$

In particular,

$$\begin{aligned} \frac{1}{\left( 2\pi \right) ^{m}}\int _{\left[ -\pi ,\pi \right] ^{m}}\left| B_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) \right| ^{2}\textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m}=\sum \limits _{k_{1},\ldots ,k_{m}=0}^{n}\left| \beta _{n}\left( k_{1},\ldots ,k_{m}\right) \right| ^{2}. \end{aligned}$$

Proof

From \(B_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) \overline{C_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) }= \sum \nolimits _{k_{1},\ldots ,k_{m}=0,l_{1},\ldots ,l_{m}=0}^{n}\beta _{n}\left( k_{1},\ldots ,k_{m}\right) \overline{\gamma _{n}\left( l_{1},\ldots ,l_{m}\right) } e^{i\left( k_{1}-l_{1}\right) \theta _{1}}\cdot \cdot \cdot e^{i\left( k_{m}-l_{m}\right) \theta _{m}}\) and Fubini’s theorem we have

$$\begin{aligned}{} & {} \frac{1}{\left( 2\pi \right) ^{m}}\int _{\left[ -\pi ,\pi \right] ^{m}}B_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) \overline{C_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) }\textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m}\\{} & {} \qquad =\sum \limits _{k_{1},\ldots ,k_{m}=0,l_{1},\ldots ,l_{m}=0}^{n}\beta _{n}\left( k_{1},\ldots ,k_{m}\right) \overline{\gamma _{n}\left( l_{1},\ldots ,l_{m}\right) } \delta _{k_{1}l_{1}}\cdot \cdot \cdot \delta _{k_{m}l_{m}}\\{} & {} \qquad =\sum \limits _{k_{1},\ldots ,k_{m}=0}^{n}\beta _{n}\left( k_{1},\ldots ,k_{m}\right) \overline{\gamma _{n}\left( k_{1},\ldots ,k_{m}\right) }. \end{aligned}$$

Above \(\delta _{kl}=1\) if \(k=l\); 0 if \(k\ne l\). \(\square \)

4 A Technical Lemma

The following technical lemma will be essential in what follows. We recall that the scalar product in the space \(L_{2}\left[ -\pi ,\pi \right] \) is defined by \(\left\langle f,g\right\rangle _{L_{2}\left[ -\pi ,\pi \right] }= \frac{1}{2\pi }\int _{-\pi }^{\pi }f\left( t\right) \overline{g\left( t\right) }\textrm{d}t\). Moreover, we use that the system of functions \(e_{k}\left( t\right) =e^{ikt}\) is orthonormal in \(L_{2}\left[ -\pi ,\pi \right] \), \( \left\langle e_{k},e_{l}\right\rangle _{L_{2}\left[ -\pi ,\pi \right] }=\delta _{kl}\).

Lemma 3

Let \(m\ge 2\) be a natural number, \(\varphi _{n}:\left\{ 0,1,\ldots ,n\right\} ^{m}\rightarrow \mathbb {R}\). For every \(j=1\), ..., m and all \(k_{j}=0,\ldots ,n\) let \(A_{k_{j}}:\mathbb {R}^{m-1}\rightarrow \mathbb {C}\) be defined by

$$\begin{aligned}{} & {} A_{k_{j}}\left( \theta _{1},\ldots ,\theta _{j-1},\theta _{j+1},\ldots ,\theta _{m}\right) \\{} & {} \qquad =\sum \limits _{k_{1},\ldots ,k_{j-1},k_{j+1},\ldots ,k_{m}=0}^{n}\varphi _{n}\left( k_{1},k_{2},\ldots ,k_{m}\right) \\{} & {} \qquad \quad e^{ik_{1}\theta _{1}}\cdot \cdot \cdot e^{ik_{j-1}\theta _{j-1}}e^{ik_{j+1}\theta _{j+1}}\cdot \cdot \cdot e^{ik_{m}\theta _{m}} \end{aligned}$$

and \(\alpha _{n\text { }1}^{\left( j\right) }:\mathbb {R}^{m-1}\rightarrow \mathbb {R}\) defined by

$$\begin{aligned}{} & {} \alpha _{n\text { }1}^{\left( j\right) }\left( \theta _{1},\ldots ,\theta _{j-1},\theta _{j+1},\ldots ,\theta _{m}\right) \\{} & {} \quad =\sum \limits _{k_{j}=0}^{n-1}\mathfrak {R}\left( A_{k_{j}}\left( \theta _{1},\ldots ,\theta _{j-1},\theta _{j+1},\ldots ,\theta _{m}\right) \overline{ A_{k_{j}+1}\left( \theta _{1},\ldots ,\theta _{j-1},\theta _{j+1},\ldots ,\theta _{m}\right) }\right) . \end{aligned}$$

Then

$$\begin{aligned}{} & {} \frac{1}{\left( 2\pi \right) ^{m-1}}\int _{\left[ -\pi ,\pi \right] ^{m-1}}\alpha _{n\text { }1}^{\left( j\right) }\left( \theta _{1},\ldots ,\theta _{j-1},\theta _{j+1},\ldots ,\theta _{m}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{j-1}\textrm{d}\theta _{j+1}\cdot \cdot \cdot \textrm{d}\theta _{m}\\{} & {} \qquad =\sum \limits _{k_{j}=0}^{n-1}\left( \sum \limits _{k_{1},\ldots ,k_{j-1},k_{j+1},\ldots ,k_{m}=0}^{n}\varphi _{n}\left( k_{1},\ldots ,k_{j},\ldots ,k_{m}\right) \right. \\{} & {} \quad \quad \left. \varphi _{n}\left( k_{1},\ldots ,k_{j}+1,\ldots ,k_{m}\right) \right) . \end{aligned}$$

Proof

We prove the case \(j=m\), the others being similar. We have

$$\begin{aligned}{} & {} \frac{1}{\left( 2\pi \right) ^{m-1}}\int _{\left[ -\pi ,\pi \right] ^{m-1}}\alpha _{n\text { }1}^{\left( m\right) }\left( \theta _{1},\ldots ,\theta _{m-1}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-1} \nonumber \\{} & {} \qquad =\mathfrak {R}\left( \sum \limits _{k_{m}=0}^{n-1}\frac{1}{\left( 2\pi \right) ^{m-1}}\int _{\left[ -\pi ,\pi \right] ^{m-1}}A_{k_{m}}\left( \theta _{1},\ldots ,\theta _{m-1}\right) \right. \nonumber \\{} & {} \quad \quad \quad \left. \overline{A_{k_{m}+1}\left( \theta _{1},\ldots ,\theta _{m-1}\right) }\textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-1}\right) . \end{aligned}$$
(1)

For \(m=2\) from the Proposition 3 we have

$$\begin{aligned} \frac{1}{2\pi }\int _{\left[ -\pi ,\pi \right] }A_{k_{2}}\left( \theta _{1}\right) \overline{A_{k_{2}+1}\left( \theta _{1}\right) }\textrm{d}\theta _{1}=\sum \limits _{k_{1}=0}^{n}\varphi _{n}\left( k_{1},k_{2}\right) \varphi _{n}\left( k_{1},k_{2}+1\right) \end{aligned}$$

and by (1) we get the equality stated.

If \(m\ge 3\) then, by Fubini’s theorem

$$\begin{aligned}{} & {} \frac{1}{\left( 2\pi \right) ^{m-1}}\int _{\left[ -\pi ,\pi \right] ^{m-1}}A_{k_{m}}\left( \theta _{1},\ldots ,\theta _{m-1}\right) \overline{ A_{k_{m}+1}\left( \theta _{1},\ldots ,\theta _{m-1}\right) }\textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-1} \nonumber \\{} & {} \qquad =\frac{1}{\left( 2\pi \right) ^{m-2}}\int _{\left[ -\pi ,\pi \right] ^{m-2}}\left\langle A_{k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2},\cdot \right) ,\right. \nonumber \\{} & {} \left. \quad A_{k_{m}+1}\left( \theta _{1},\ldots ,\theta _{m-2},\cdot \right) \right\rangle _{L_{2}\left[ -\pi ,\pi \right] }\textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-2}. \end{aligned}$$
(2)

Let us fix \(\left( \theta _{1},\ldots ,\theta _{m-2}\right) \in \left[ -\pi ,\pi \right] ^{m-2}\). For all \(\theta _{m-1}\in \left[ -\pi ,\pi \right] \) we have

$$\begin{aligned} A_{k_{m}}\left( \theta _{1},\ldots ,\theta _{m-1}\right) =\sum \limits _{k_{m-1}=0}^{n}A_{k_{m-1}k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2}\right) e^{ik_{m-1}\theta _{m-1}} \end{aligned}$$

where

$$\begin{aligned}{} & {} A_{k_{m-1}k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2}\right) \\{} & {} \qquad =\sum \limits _{k_{1},\ldots ,k_{m-2}=0}^{n}\varphi _{n}\left( k_{1},\ldots ,k_{m-2},k_{m-1},k_{m}\right) e^{ik_{1}\theta _{1}}\cdot \cdot \cdot e^{ik_{m-2}\theta _{m-2}}. \end{aligned}$$

Thus \(A_{k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2},\cdot \right) =\sum \nolimits _{k_{m-1}=0}^{n}A_{k_{m-1}k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2}\right) e_{k_{m-1}}\). The orthonormality of the functions \(\left( e_{k_{m-1}}\right) _{0\le k_{m-1}\le n}\) gives us that

$$\begin{aligned}{} & {} \left\langle A_{k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2},\cdot \right) ,A_{k_{m}+1}\left( \theta _{1},\ldots ,\theta _{m-2},\cdot \right) \right\rangle _{L_{2}\left[ -\pi ,\pi \right] }\\{} & {} \qquad =\left\langle \sum \limits _{k_{m-1}=0}^{n}A_{k_{m-1}k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2}\right) e_{k_{m-1}},\right. \\{} & {} \quad \quad \left. \sum \limits _{k_{m-1}=0}^{n}A_{k_{m-1}k_{m}+1}\left( \theta _{1},\ldots ,\theta _{m-2}\right) e_{k_{m-1}}\right\rangle _{L_{2}\left[ -\pi ,\pi \right] }\\{} & {} \qquad =\sum \limits _{k_{m-1}=0}^{n}A_{k_{m-1}k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2}\right) \overline{A_{k_{m-1}k_{m}+1}\left( \theta _{1},\ldots ,\theta _{m-2}\right) }. \end{aligned}$$

Then

$$\begin{aligned}{} & {} \frac{1}{\left( 2\pi \right) ^{m-2}}\int _{\left[ -\pi ,\pi \right] ^{m-2}}\left\langle A_{k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2},\cdot \right) ,\right. \nonumber \\{} & {} \quad \quad \left. A_{k_{m}+1}\left( \theta _{1},\ldots ,\theta _{m-2},\cdot \right) \right\rangle _{L_{2}\left[ -\pi ,\pi \right] }\textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-2}\nonumber \\{} & {} \quad \sum \limits _{k_{m-1}=0}^{n}\frac{1}{\left( 2\pi \right) ^{m-2}}\int _{\left[ -\pi ,\pi \right] ^{m-2}}A_{k_{m-1}k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2}\right) \nonumber \\{} & {} \qquad \quad \overline{A_{k_{m-1}k_{m}+1}\left( \theta _{1},\ldots ,\theta _{m-2}\right) }\textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-2}. \end{aligned}$$
(3)

Since

$$\begin{aligned}{} & {} A_{k_{m-1}k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2}\right) \\{} & {} \qquad =\sum \limits _{k_{1},\ldots ,k_{m-2}=0}^{n}\varphi _{n}\left( k_{1},\ldots ,k_{m-2},k_{m-1},k_{m}\right) e^{ik_{1}\theta _{1}}\cdot \cdot \cdot e^{ik_{m-2}\theta _{m-2}}, \\{} & {} A_{k_{m-1}k_{m}+1}\left( \theta _{1},\ldots ,\theta _{m-2}\right) \\{} & {} \qquad =\sum \limits _{k_{1},\ldots ,k_{m-2}=0}^{n}\varphi _{n}\left( k_{1},\ldots ,k_{m-2},k_{m-1},k_{m}+1\right) e^{ik_{1}\theta _{1}}\cdot \cdot \cdot e^{ik_{m-2}\theta _{m-2}} \end{aligned}$$

the Proposition 3 gives us that

$$\begin{aligned}{} & {} \frac{1}{\left( 2\pi \right) ^{m-2}}\int _{\left[ -\pi ,\pi \right] ^{m-2}}A_{k_{m-1}k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2}\right) \overline{ A_{k_{m-1}k_{m}+1}\left( \theta _{1},\ldots ,\theta _{m-2}\right) }\nonumber \\{} & {} \qquad \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-2} =\sum \limits _{k_{1},\ldots ,k_{m-2}=0}^{n}\varphi _{n}\left( k_{1},\ldots ,k_{m-2},k_{m-1},k_{m}\right) \nonumber \\{} & {} \qquad \varphi _{n}\left( k_{1},\ldots ,k_{m-2},k_{m-1},k_{m}+1\right) . \end{aligned}$$
(4)

From the relations (3) and (4) we deduce that

$$\begin{aligned}{} & {} \frac{1}{\left( 2\pi \right) ^{m-2}}\int _{\left[ -\pi ,\pi \right] ^{m-2}}\nonumber \\{} & {} \quad \left\langle A_{k_{m}}\left( \theta _{1},\ldots ,\theta _{m-2},\cdot \right) ,A_{k_{m}+1}\left( \theta _{1},\ldots ,\theta _{m-2},\cdot \right) \right\rangle _{L_{2}\left[ -\pi ,\pi \right] }\textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-2}\nonumber \\{} & {} \qquad =\sum \limits _{k_{1},\ldots ,k_{m-1}=0}^{n}\varphi _{n}\left( k_{1},\ldots ,k_{m-1},k_{m}\right) \varphi _{n}\left( k_{1},\ldots ,k_{m-1},k_{m}+1\right) . \end{aligned}$$
(5)

To finish the proof just combine the relations (1), (2) and (5). \(\square \)

5 The Case of Rectangular Summation

In this section, we apply the above general results to find the necessary and sufficient conditions that some natural multivariate operators associated to some summation methods to be convergent. For various other methods of summation of trigonometric series the reader can consult the book of Weiss [14].

Theorem 4

Let \(m\ge 2\) be a natural number, \( \varphi _{n}:\left\{ 0,1,\ldots ,n\right\} ^{m}\rightarrow \mathbb {R}\), \(K_{n}: \mathbb {R}^{m}\rightarrow \left[ 0,\infty \right) \) defined by

$$\begin{aligned} K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) =\left| \sum \limits _{k_{1},\ldots ,k_{m}=0}^{n}\varphi _{n}\left( k_{1},\ldots ,k_{m}\right) e^{ik_{1}\theta _{1}}\cdot \cdot \cdot e^{ik_{m}\theta _{m}}\right| ^{2}. \end{aligned}$$

Then the following assertions are equivalent:

  1. (i)

    For all \(f\in C_{2\pi }\left( \mathbb {R}^{m}\right) \) we have

    $$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{m}}\int _{ \left[ -\pi ,\pi \right] ^{m}}K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) f\left( t_{1}{-}\theta _{1},\ldots ,t_{m}-\theta _{m}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m}{=}f\left( t\right) \end{aligned}$$

    uniformly with respect to \(t=\left( t_{1},\ldots ,t_{m}\right) \in \mathbb {R} ^{m} \).

  2. (ii)

    For all \(f\in C_{2\pi }\left( \mathbb {R}^{m}\right) \) and all \(t=\left( t_{1},\ldots ,t_{m}\right) \in \mathbb {R}^{m}\) we have

    $$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{m}}\int _{ \left[ -\pi ,\pi \right] ^{m}}K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) f\left( t_{1}{-}\theta _{1},\ldots ,t_{m}{-}\theta _{m}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m}{=}f\left( t\right) . \end{aligned}$$
  3. (iii)

    \(\lim \nolimits _{n\rightarrow \infty }\sum \nolimits _{k_{1},\ldots ,k_{m}=0}^{n} \left[ \varphi _{n}\left( k_{1},\ldots ,k_{m}\right) \right] ^{2}=1\) and

    $$\begin{aligned}{} & {} \lim \limits _{n\rightarrow \infty }\sum \limits _{k_{1}=0}^{n-1}\left( \sum \limits _{k_{2},\ldots ,k_{m}=0}^{n}\varphi _{n}\left( k_{1},k_{2},\ldots ,k_{m}\right) \varphi _{n}\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) \right) =1,\\{} & {} \lim \limits _{n\rightarrow \infty }\sum \limits _{k_{2}=0}^{n-1}\left( \sum \limits _{k_{1},k_{3},\ldots ,k_{m}=0}^{n}\varphi _{n}\left( k_{1},k_{2},k_{3},\ldots ,k_{m}\right) \varphi _{n}\left( k_{1},k_{2}+1,k_{3},\ldots ,k_{m}\right) \right) \\{} & {} \quad =1,\ldots ,\\{} & {} \lim \limits _{n\rightarrow \infty }\sum \limits _{k_{m}=0}^{n-1}\left( \sum \limits _{k_{1},\ldots ,k_{m-1}=0}^{n}\varphi _{n}\left( k_{1},\ldots ,k_{m-1},k_{m}\right) \varphi _{n}\left( k_{1},\ldots ,k_{m-1},k_{m}+1\right) \right) = 1. \end{aligned}$$

Proof

By Theorem 3 the conditions (i), (ii) are equivalent to \( \lim \nolimits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{m}}\int _{ \left[ -\pi ,\pi \right] ^{m}}K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m}=1\) and \(\lim \nolimits _{n\rightarrow \infty }\frac{1}{2\pi }\int _{-\pi }^{\pi }K_{n}^{\left\langle j\right\rangle }\left( \theta _{j}\right) \cos \theta _{j}\textrm{d}\theta _{j}=1\) for every \( j=1,\ldots ,m\). Since by Proposition 3

$$\begin{aligned} \frac{1}{\left( 2\pi \right) ^{m}}\int _{\left[ -\pi ,\pi \right] ^{m}}\left[ K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) \right] ^{2}\textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m}=\sum \limits _{k_{1},\ldots ,k_{m}=0}^{n}\left[ \varphi _{n}\left( k_{1},\ldots ,k_{m}\right) \right] ^{2} \end{aligned}$$

we obtain \(\lim \nolimits _{n\rightarrow \infty }\sum \nolimits _{k_{1},\ldots ,k_{m}=0}^{n}\left[ \varphi _{n}\left( k_{1},\ldots ,k_{m}\right) \right] ^{2}=1\). For all \(k_{m}=0\), ..., n let us define \(A_{k_{m}}:\mathbb {R}^{m-1}\rightarrow \mathbb {C}\) by \( A_{k_{m}}\left( \theta _{1},\ldots ,\theta _{m-1}\right) =\sum \nolimits _{k_{1},\ldots ,k_{m-1}=0}^{n}\varphi _{n}( k_{1},\ldots ,k_{m-1},k_{m}) e^{ik_{1}\theta _{1}}\cdot \cdot \cdot e^{ik_{m-1}\theta _{m-1}}\). By Proposition 2 we have

$$\begin{aligned}{} & {} K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) =\left| \sum \limits _{k_{m}=0}^{n}A_{k_{m}}\left( \theta _{1},\ldots ,\theta _{m-1}\right) e^{ik_{m}\theta _{m}}\right| ^{2}\\{} & {} \quad =\alpha _{n\text { }0}^{\left( m\right) }\left( \theta _{1},\ldots ,\theta _{m-1}\right) +2\alpha _{n\text { }1}^{\left( m\right) }\left( \theta _{1},\ldots ,\theta _{m-1}\right) \cos \theta _{m}+\cdot \cdot \cdot \\{} & {} \qquad +2\alpha _{n \text { }n}^{\left( m\right) }\left( \theta _{1},\ldots ,\theta _{m-1}\right) \cos n\theta _{m}\\{} & {} \qquad 2\beta _{n\text { }1}^{\left( m\right) }\left( \theta _{1},\ldots ,\theta _{m-1}\right) \sin \theta _{m}+2\beta _{n\text { }2}^{\left( m\right) }\left( \theta _{1},\ldots ,\theta _{m-1}\right) \sin 2\theta _{m}+\cdot \cdot \cdot \\{} & {} \qquad +2\beta _{n\text { }n}^{\left( m\right) }\left( \theta _{1},\ldots ,\theta _{m-1}\right) \sin n\theta _{m} \end{aligned}$$

where \(\alpha _{n\text { }1}^{\left( m\right) }\left( \theta _{1},\ldots ,\theta _{m-1}\right) {=}\sum \nolimits _{k_{m}=0}^{n}\mathfrak {R}\left( A_{k_{m}}\left( \theta _{1},\ldots ,\theta _{m-1}\right) \right. \left. \overline{A_{k_{m}+1}\left( \theta _{1},\ldots ,\theta _{m-1}\right) }\right) \). We deduce that

$$\begin{aligned}{} & {} K_{n}^{\left\langle m\right\rangle }\left( \theta _{m}\right) =\frac{1}{ \left( 2\pi \right) ^{m-1}}\int _{\left[ -\pi ,\pi \right] ^{m-1}}K_{n}\left( \theta _{1},\ldots ,\theta _{m-1},\theta _{m}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-1}\\{} & {} \quad =\nu _{n\text { }0}^{\left( m\right) }+2\nu _{n\text { }1}^{\left( m\right) }\cos \theta _{m}+2\nu _{n\text { }2}^{\left( m\right) }\cos 2\theta _{m}+\cdot \cdot \cdot +2\nu _{n\text { }n}^{\left( m\right) }\cos n\theta _{m}\\{} & {} \qquad +2\gamma _{n\text { }1}^{\left( m\right) }\sin \theta _{m}+2\gamma _{n\text { } 2}^{\left( m\right) }\sin 2\theta _{m}+\cdot \cdot \cdot +2\gamma _{n\text { } n}^{\left( m\right) }\sin n\theta _{m} \end{aligned}$$

where

$$\begin{aligned} \nu _{n\text { }j}^{\left( m\right) }= & {} \frac{1}{\left( 2\pi \right) ^{m-1}} \int _{\left[ -\pi ,\pi \right] ^{m-1}}\alpha _{n\text { }j}^{\left( m\right) }\left( \theta _{1},\ldots ,\theta _{m-1}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-1},j=0,\ldots ,n \\ \gamma _{n\text { }j}^{\left( m\right) }= & {} \frac{1}{\left( 2\pi \right) ^{m-1}}\int _{\left[ -\pi ,\pi \right] ^{m-1}}\beta _{n\text { }j}^{\left( m\right) }\left( \theta _{1},\ldots ,\theta _{m-1}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-1},j=1,\ldots ,n. \end{aligned}$$

Let us note that for all \(n\in \mathbb {N}\) we have \(\frac{1}{2\pi } \int _{-\pi }^{\pi }K_{n}^{\left\langle m\right\rangle }\left( \theta _{m}\right) \cos \theta _{m}\textrm{d}\theta _{m}=\nu _{n\text { }1}^{\left( m\right) } \). The condition of the convergence for \(K_{n}^{\left\langle m\right\rangle } \), that is

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{1}{2\pi }\int _{-\pi }^{\pi }K_{n}^{\left\langle m\right\rangle }\left( \theta _{m}\right) \cos \theta _{m}\textrm{d}\theta _{m}=1 \end{aligned}$$

becomes

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{m-1}}\int _{ \left[ -\pi ,\pi \right] ^{m-1}}\alpha _{n\text { }1}^{\left( m\right) }\left( \theta _{1},\ldots ,\theta _{m-1}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-1}=1. \end{aligned}$$

But, since from the Lemma 3

$$\begin{aligned}{} & {} \frac{1}{\left( 2\pi \right) ^{m-1}}\int _{\left[ -\pi ,\pi \right] ^{m-1}}\alpha _{n\text { }1}^{\left( m\right) }\left( \theta _{1},\ldots ,\theta _{m-1}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m-1}\\{} & {} \quad =\sum \limits _{k_{m}=0}^{n-1}\left( \sum \limits _{k_{1},\ldots ,k_{m-1}=0}^{n}\varphi _{n}\left( k_{1},\ldots ,k_{m-1},k_{m}\right) \varphi _{n}\left( k_{1},\ldots ,k_{m-1},k_{m}+1\right) \right) \end{aligned}$$

we deduce that the condition of the convergence for \(K_{n}^{\left\langle m\right\rangle }\) is equivalent to

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\sum \limits _{k_{m}=0}^{n-1}\left( \sum \limits _{k_{1},\ldots ,k_{m-1}=0}^{n}\varphi _{n}\left( k_{1},\ldots ,k_{m-1},k_{m}\right) \varphi _{n}\left( k_{1},\ldots ,,k_{m-1},k_{m}+1\right) \right) {=}1. \end{aligned}$$

Similar, from the conditions of the convergence for \(K_{n}^{\left\langle 1\right\rangle }\), ..., \(K_{n}^{\left\langle m-1\right\rangle }\) we get the remaining conditions stated in (iii). \(\square \)

6 The Case of Triangular Summation

We use now Theorem 4 to prove a similar result for the triangular summation.

Theorem 5

Let \(m\ge 2\) be a natural number, \(\varphi _{n}:\left\{ 0,1,\ldots ,n\right\} ^{m}\rightarrow \mathbb {R}\), \(K_{n}:\mathbb {R} ^{m}\rightarrow \left[ 0,\infty \right) \) defined by

$$\begin{aligned} K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) =\left| \sum \limits _{k_{1}+\cdot \cdot \cdot +k_{m}\le n}\varphi _{n}\left( k_{1},\ldots ,k_{m}\right) e^{ik_{1}\theta _{1}}\cdot \cdot \cdot e^{ik_{m}\theta _{m}}\right| ^{2}. \end{aligned}$$

Then the following assertions are equivalent:

  1. (i)

    For all \(f\in C_{2\pi }\left( \mathbb {R}^{m}\right) \) we have

    $$\begin{aligned}{} & {} \lim \limits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{m}}\int _{ \left[ -\pi ,\pi \right] ^{m}}K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) f\left( t_{1}-\theta _{1},\ldots ,t_{m}-\theta _{m}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m}\\{} & {} \qquad =f\left( t_{1},\ldots ,t_{m}\right) \end{aligned}$$

    uniformly with respect to \(\left( t_{1},\ldots ,t_{m}\right) \in \mathbb {R}^{m}\).

  2. (ii)

    For all \(f\in C_{2\pi }\left( \mathbb {R}^{m}\right) \) and all \(t=\left( t_{1},\ldots ,t_{m}\right) \in \mathbb {R}^{m}\) we have

    $$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{m}}\int _{ \left[ -\pi ,\pi \right] ^{m}}K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) f\left( t_{1}{-}\theta _{1},\ldots ,t_{m}-\theta _{m}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m}{=}f\left( t\right) . \end{aligned}$$
  3. (iii)

    \(\lim \nolimits _{n\rightarrow \infty }\sum \nolimits _{k_{1}+\cdot \cdot \cdot +k_{m}\le n}\left[ \varphi _{n}\left( k_{1},\ldots ,k_{m}\right) \right] ^{2}=1\) and

    $$\begin{aligned}{} & {} \lim \limits _{n\rightarrow \infty }\sum \limits _{k_{1}=0}^{n-1}\left( \sum \limits _{k_{2}+\cdot \cdot \cdot +k_{m}\le n-k_{1}-1}\varphi _{n}\left( k_{1},k_{2},\ldots ,k_{m}\right) \varphi _{n}\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) \right) =1, \\{} & {} \lim \limits _{n\rightarrow \infty }\sum \limits _{k_{2}=0}^{n-1}\left( \sum \limits _{k_{1}+k_{3}+\cdot \cdot \cdot +k_{m}\le n-k_{2}-1}\varphi _{n}\left( k_{1},k_{2},k_{3},\ldots ,k_{m}\right) \varphi _{n}\left( k_{1},k_{2}+1,k_{3},\ldots ,k_{m}\right) \right) \\{} & {} \qquad =1,\ldots ,\\{} & {} \lim \limits _{n\rightarrow \infty }\sum \limits _{k_{m}=0}^{n-1}\left( \sum \limits _{k_{1}+\cdot \cdot \cdot +k_{m-1}\le n-k_{m}-1}\varphi _{n}\left( k_{1},\ldots ,k_{m-1},k_{m}\right) \varphi _{n}\left( k_{1},\ldots ,k_{m-1},k_{m}+1\right) \right) \\{} & {} \qquad =1. \end{aligned}$$

Proof

For every \(n\in \mathbb {N}\) let \(\Delta _{n}=\{ \left( k_{1},\ldots ,k_{m}\right) \in \left\{ 0,1,\ldots ,n\right\} ^{m}\mid k_{1}+\cdot \cdot \cdot +k_{m}\le n\} \) and \(\beta _{n}:\left\{ 0,1,\ldots ,n\right\} ^{m}\rightarrow \mathbb {R}\) defined by

$$\begin{aligned} \beta _{n}\left( k_{1},\ldots ,k_{m}\right) =\left\{ \begin{array}{c} \varphi _{n}\left( k_{1},\ldots ,k_{m}\right) \text { if }\left( k_{1},\ldots ,k_{m}\right) \in \Delta _{n} \\ \text { }0\text { if }\left( k_{1},\ldots ,k_{m}\right) \notin \Delta _{n} \end{array} \right. . \end{aligned}$$

Then \(K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) =\left| \sum \nolimits _{k_{1},\ldots ,k_{m}=0}^{n}\beta _{n}\left( k_{1},\ldots ,k_{m}\right) e^{ik_{1}\theta _{1}}\cdot \cdot \cdot e^{ik_{m}\theta _{m}}\right| ^{2}\) . From Theorem 4 (i), (ii) are equivalent to \(\lim \nolimits _{n\rightarrow \infty }\sum \nolimits _{k_{1},\ldots ,k_{m}=0}^{n} \left[ \beta _{n}\left( k_{1},\ldots ,k_{m}\right) \right] ^{2}=1\) and

$$\begin{aligned}{} & {} \lim \limits _{n\rightarrow \infty }\sum \limits _{k_{1}=0}^{n-1}\left( \sum \limits _{k_{2},\ldots ,k_{m}=0}^{n}\beta _{n}\left( k_{1},k_{2},\ldots ,k_{m}\right) \beta _{n}\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) \right) =1,\\{} & {} \lim \limits _{n\rightarrow \infty }\sum \limits _{k_{2}=0}^{n-1}\left( \sum \limits _{k_{1},k_{3},\ldots ,k_{m}=0}^{n}\beta _{n}\left( k_{1},k_{2},k_{3},\ldots ,k_{m}\right) \beta _{n}\left( k_{1},k_{2}+1,k_{3},\ldots ,k_{m}\right) \right) \\{} & {} \qquad =1,\ldots ,\\{} & {} \lim \limits _{n\rightarrow \infty }\sum \limits _{k_{m}=0}^{n-1}\left( \sum \limits _{k_{1},\ldots ,k_{m-1}=0}^{n}\beta _{n}\left( k_{1},\ldots ,k_{m-1},k_{m}\right) \beta _{n}\left( k_{1},\ldots ,k_{m-1},k_{m}+1\right) \right) \\{} & {} \qquad =1. \end{aligned}$$

Since

$$\begin{aligned}{} & {} \sum \limits _{k_{1},\ldots ,k_{m}=0}^{n}\left[ \beta _{n}\left( k_{1},\ldots ,k_{m}\right) \right] ^{2}=\sum \limits _{\left( k_{1},\ldots ,k_{m}\right) \in \Delta _{n}}\left[ \beta _{n}\left( k_{1},\ldots ,k_{m}\right) \right] ^{2}\\{} & {} \qquad +\sum \limits _{\left( k_{1},\ldots ,k_{m}\right) \notin \Delta _{n}}\left[ \beta _{n}\left( k_{1},\ldots ,k_{m}\right) \right] ^{2}=\sum \limits _{k_{1}+\cdot \cdot \cdot +k_{m}\le n}\left[ \varphi _{n}\left( k_{1},\ldots ,k_{m}\right) \right] ^{2} \end{aligned}$$

we get the first condition in (iii). Now let us note the equality

$$\begin{aligned}{} & {} \sum \limits _{k_{2},\ldots ,k_{m}=0}^{n}\beta _{n}\left( k_{1},k_{2},\ldots ,k_{m}\right) \beta _{n}\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) \\{} & {} \qquad =\sum \limits _{k_{2}+\cdot \cdot \cdot +k_{m}\le n-k_{1}-1}\beta _{n}\left( k_{1},k_{2},\ldots ,k_{m}\right) \beta _{n}\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) \\{} & {} \quad \quad \quad +\sum \limits _{k_{2}+\cdot \cdot \cdot +k_{m}\ge n-k_{1}}\beta _{n}\left( k_{1},k_{2},\ldots ,k_{m}\right) \beta _{n}\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) . \end{aligned}$$

If \(k_{2}+\cdot \cdot \cdot +k_{m}\ge n-k_{1}\) then \(\left( k_{1}+1\right) +k_{2}+\cdot \cdot \cdot +k_{m}\ge n+1\) i.e., \(\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) \notin \Delta _{n}\), so \(\beta _{n}\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) =0\) and

$$\begin{aligned}{} & {} \sum \limits _{k_{2},\ldots ,k_{m}=0}^{n}\beta _{n}\left( k_{1},k_{2},\ldots ,k_{m}\right) \beta _{n}\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) \\{} & {} \qquad =\sum \limits _{k_{2}+\cdot \cdot \cdot +k_{m}\le n-k_{1}-1}\beta _{n}\left( k_{1},k_{2},\ldots ,k_{m}\right) \beta _{n}\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) \\{} & {} \qquad =\sum \limits _{k_{2}+\cdot \cdot \cdot +k_{m}\le n-k_{1}-1}\varphi _{n}\left( k_{1},k_{2},\ldots ,k_{m}\right) \varphi _{n}\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) . \end{aligned}$$

We deduce that

$$\begin{aligned}{} & {} \lim \limits _{n\rightarrow \infty }\sum \limits _{k_{1}=0}^{n-1}\left( \sum \limits _{k_{2}+\cdot \cdot \cdot +k_{m}\le n-k_{1}-1}\varphi _{n}\left( k_{1},k_{2},\ldots ,k_{m}\right) \varphi _{n}\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) \right) \\{} & {} \qquad =1. \end{aligned}$$

In the same way we get the remaining conditions in (ii). \(\square \)

7 Two Examples

Proposition 4

Let \(m\ge 2\) be a natural number, \( \varphi :\left[ 0,1\right] ^{m}\rightarrow \mathbb {R}\) a continuous function and \(K_{n}:\mathbb {R}^{m}\rightarrow \left[ 0,\infty \right) \) defined by

$$\begin{aligned} K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) =\frac{1}{n^{m}}\left| \sum \limits _{k_{1},\ldots ,k_{m}=0}^{n}\varphi \left( \frac{k_{1}}{n},\ldots ,\frac{ k_{m}}{n}\right) e^{ik_{1}\theta _{1}}\cdot \cdot \cdot e^{ik_{m}\theta _{m}}\right| ^{2}. \end{aligned}$$

Then the following assertions are equivalent:

  1. (i)

    For all \(f\in C_{2\pi }\left( \mathbb {R}^{m}\right) \) we have

    $$\begin{aligned}{} & {} \lim \limits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{m}}\int _{ \left[ -\pi ,\pi \right] ^{m}}K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) f\left( t_{1}-\theta _{1},\ldots ,t_{m}-\theta _{m}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m}\\{} & {} \qquad =f\left( t\right) \end{aligned}$$

    uniformly with respect to \(t=\left( t_{1},\ldots ,t_{m}\right) \in \mathbb {R} ^{m} \).

  2. (ii)

    For all \(f\in C_{2\pi }\left( \mathbb {R}^{m}\right) \) and all \(t=\left( t_{1},\ldots ,t_{m}\right) \in \mathbb {R}^{m}\) we have

    $$\begin{aligned}{} & {} \lim \limits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{m}}\int _{ \left[ -\pi ,\pi \right] ^{m}}K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) f\left( t_{1}-\theta _{1},\ldots ,t_{m}-\theta _{m}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m}\\{} & {} \qquad =f\left( t\right) . \end{aligned}$$
  3. (iii)

    \(\int _{\left[ 0,1\right] ^{m}}\varphi ^{2}\left( t_{1},\ldots ,t_{m}\right) \textrm{d}t_{1}\cdot \cdot \cdot \textrm{d}t_{m}=1\).

Proof

Let us define \(\varphi _{n}:\left\{ 0,1,\ldots ,n\right\} ^{m}\rightarrow \mathbb {R}\) by \(\varphi _{n}\left( k_{1},\ldots ,k_{m}\right) =\frac{1}{\sqrt{ n^{m}}}\varphi \left( \frac{k_{1}}{n},\ldots ,\frac{k_{m}}{n}\right) \). We denote \(S_{n,0}=\sum \nolimits _{k_{1},\ldots ,k_{m}=0}^{n}\left[ \varphi _{n}\left( k_{1},\ldots ,k_{m}\right) \right] ^{2}\),

$$\begin{aligned} S_{n,1}=\sum \limits _{k_{1}=0}^{n-1}\left( \sum \limits _{k_{2},\ldots ,k_{m}=0}^{n}\varphi _{n}\left( k_{1},k_{2},\ldots ,k_{m}\right) \varphi _{n}\left( k_{1}+1,k_{2},\ldots ,k_{m}\right) \right) . \end{aligned}$$

The condition \(\lim \nolimits _{n\rightarrow \infty }\sum \nolimits _{k_{1},\ldots ,k_{m}=0}^{n}\left[ \varphi _{n}\left( k_{1},\ldots ,k_{m}\right) \right] ^{2}=1\) is equivalent to \(\lim \nolimits _{n \rightarrow \infty }S_{n,0}=\lim \nolimits _{n\rightarrow \infty }\frac{1}{n^{m}} \sum \nolimits _{k_{1},\ldots ,k_{m}=0}^{n}\varphi ^{2}\left( \frac{k_{1}}{n},\ldots , \frac{k_{m}}{n}\right) =1\), that is \(\int _{\left[ 0,1\right] ^{m}}\varphi ^{2}\left( t_{1},\ldots ,t_{m}\right) \textrm{d}t_{1}\cdot \cdot \cdot \textrm{d}t_{m}=1\). Let \( \varepsilon >0\). Since \(\varphi \) is uniformly continuous, there exists \( \delta _{\varepsilon }>0\) such that if \(\max \nolimits _{1\le i\le n}\left| t_{i}-u_{i}\right| <\delta _{\varepsilon }\) we have \( \left| \varphi \left( t_{1},\ldots ,t_{m}\right) -\varphi \left( v_{1},\ldots ,v_{m}\right) \right| <\varepsilon \). There exists \( n_{\varepsilon }\in \mathbb {N}\) such that \(\frac{1}{n}<\delta _{\varepsilon } \) for all \(n\ge n_{\varepsilon }\). Let \(n\ge n_{\varepsilon }\). Then for all \(0\le k_{1}\le n-1\), \(1\le k_{2},\ldots ,k_{m}\le n\) we have

$$\begin{aligned} \left| \varphi \left( \frac{k_{1}+1}{n},\frac{k_{2}}{n},\ldots ,\frac{k_{m}}{ n}\right) -\varphi \left( \frac{k_{1}}{n},\frac{k_{2}}{n},\ldots ,\frac{k_{m}}{n} \right) \right| <\varepsilon . \end{aligned}$$

and hence

$$\begin{aligned} \left| S_{n,1}-S_{n,0}\right|\le & {} \frac{\left\| \varphi \right\| }{n^{m}}\sum \limits _{k_{1}=0}^{n-1}\sum \limits _{k_{2},\ldots ,k_{m}=0}^{n}\left| \varphi \left( \frac{k_{1}+1}{n}, \frac{k_{2}}{n},\ldots ,\frac{k_{m}}{n}\right) {-}\varphi \left( \frac{k_{1}}{n}, \frac{k_{2}}{n},\ldots ,\frac{k_{m}}{n}\right) \right| \\\le & {} \varepsilon \frac{\left\| \varphi \right\| \left( n+1\right) ^{m-1}}{n^{m-1}}<\varepsilon \left\| \varphi \right\| 2^{m-1} \end{aligned}$$

where \(\left\| \varphi \right\| =\sup \nolimits _{\left( t_{1},\ldots ,t_{m}\right) \in \left[ 0,1\right] ^{m}}\left| \varphi \left( t_{1},\ldots ,t_{m}\right) \right| \). We deduce that \(\lim \nolimits _{n \rightarrow \infty }[ S_{n,1}-S_{n,0}] =0\) and so \( \lim \nolimits _{n\rightarrow \infty }S_{n,1}=\lim \nolimits _{n\rightarrow \infty }S_{n,0}=\int _{\left[ 0,1\right] ^{m}}\varphi ^{2}\left( t_{1},\ldots ,t_{m}\right) \textrm{d}t_{1}\cdot \cdot \cdot \textrm{d}t_{m}=1\). By a similar reasoning we verify all the conditions (iii) from Theorem 4(iii). \(\square \)

Proposition 5

Let \(m\ge 2\) be a natural number, \( T_{m}=\{ \left( t_{1},\ldots ,t_{m}\right) \in \left[ 0,1\right] ^{m}\mid t_{1}+\cdot \cdot \cdot +t_{m}\le 1\} \), \(\varphi :T_{m}\rightarrow \mathbb {R}\) a continuous function and \(K_{n}:\mathbb {R}^{m}\rightarrow \left[ 0,\infty \right) \) defined by

$$\begin{aligned} K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) =\frac{1}{n^{m}}\left| \sum \limits _{k_{1}+\cdot \cdot \cdot +k_{m}\le n}\varphi \left( \frac{k_{1} }{n},\ldots ,\frac{k_{m}}{n}\right) e^{ik_{1}\theta _{1}}\cdot \cdot \cdot e^{ik_{m}\theta _{m}}\right| ^{2}. \end{aligned}$$

Then the following assertions are equivalent:

  1. (i)

    For all \(f\in C_{2\pi }\left( \mathbb {R}^{m}\right) \) we have

    $$\begin{aligned}{} & {} \lim \limits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{m}}\int _{ \left[ -\pi ,\pi \right] ^{m}}K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) f\left( t_{1}-\theta _{1},\ldots ,t_{m}-\theta _{m}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m}\\{} & {} \qquad =f\left( t_{1},\ldots ,t_{m}\right) \end{aligned}$$

    uniformly with respect to \(\left( t_{1},\ldots ,t_{m}\right) \in \mathbb {R}^{m}\).

  2. (ii)

    For all \(f\in C_{2\pi }\left( \mathbb {R}^{m}\right) \) and all \(t=\left( t_{1},\ldots ,t_{m}\right) \in \mathbb {R}^{m}\) we have

    $$\begin{aligned}{} & {} \lim \limits _{n\rightarrow \infty }\frac{1}{\left( 2\pi \right) ^{m}}\int _{ \left[ -\pi ,\pi \right] ^{m}}K_{n}\left( \theta _{1},\ldots ,\theta _{m}\right) f\left( t_{1}-\theta _{1},\ldots ,t_{m}-\theta _{m}\right) \textrm{d}\theta _{1}\cdot \cdot \cdot \textrm{d}\theta _{m}\\{} & {} \qquad =f\left( t\right) . \end{aligned}$$
  3. (iii)

    \(\int _{T_{m}}\varphi ^{2}\left( t_{1},\ldots ,t_{m}\right) \textrm{d}t_{1}\cdot \cdot \cdot \textrm{d}t_{m}=1\).

Proof

Let us define \(\varphi _{n}:\left\{ 0,1,\ldots ,n\right\} ^{m}\rightarrow \mathbb {R}\) by \(\varphi _{n}\left( k_{1},\ldots ,k_{m}\right) =\frac{1}{\sqrt{ n^{m}}}\varphi \left( \frac{k_{1}}{n},\ldots ,\frac{k_{m}}{n}\right) \). The condition \(\lim \nolimits _{n\rightarrow \infty }\sum \nolimits _{k_{1}+\cdot \cdot \cdot +k_{m}\le n}\left[ \varphi _{n}\left( k_{1},\ldots ,k_{m}\right) \right] ^{2}=1\) is equivalent to

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{1}{n^{m}}\sum \limits _{k_{1}+\cdot \cdot \cdot +k_{m}\le n}\varphi ^{2}\left( \frac{k_{1}}{n},\ldots ,\frac{k_{m}}{ n}\right) =1 \end{aligned}$$

that is \(\int _{\left[ 0,1\right] ^{m}}\varphi ^{2}\left( t_{1},\ldots ,t_{m}\right) \textrm{d}t_{1}\cdot \cdot \cdot \textrm{d}t_{m}=1\). As in the proof of Proposition 4 we verify all the conditions from Theorem 5(iii).\(\square \)