Proofs of convergence theorems
Auxiliary lemmas
In this section, we present a few lemmas that will be needed in the proofs in the next sections. The lemmas are quite classical results on the convergence of Riemannian sums, but as it is hard to find exactly the same results in other literature, for completeness we prove the lemmas here.
Lemma 9
Let \(\Delta > 0\) and \(\alpha \in [0,1)\) be given constants, \(m = 0,1,2,\ldots \) some nonnegative integer and assume that the \(f(\omega )\) is a bounded integrable function defined on \(\omega \ge m \, \Delta \) with bounded derivative on \(\omega > m \, \Delta \) such that \(\int _{m \, \Delta }^{\infty } |f'(\omega )| \, \mathrm{d}\omega = C^{(m)} < \infty \). Then we have
$$\begin{aligned} \left| \int _{m \, \Delta }^\infty f(\omega ) \, \mathrm{d}\omega - \sum _{j=m+1}^\infty f(j \, \Delta - \alpha \, \Delta ) \, \Delta \right| \le C^{(m)} \, \Delta . \end{aligned}$$
(90)
Furthermore, provided that \(\int _{0}^{\infty } |f'(\omega )| \, \mathrm{d}\omega = C^{(0)} < \infty \), this bound can be made independent of m:
$$\begin{aligned} \left| \int _{m \, \Delta }^\infty f(\omega ) \, \mathrm{d}\omega - \sum _{j=m+1}^\infty f(j \, \Delta - \alpha \, \Delta ) \, \Delta \right| \le C^{(0)} \, \Delta . \end{aligned}$$
(91)
Proof
We can write
$$\begin{aligned} \begin{aligned} \int _{m \, \Delta }^\infty f(\omega ) \, \mathrm{d}\omega = \sum _{j=m+1}^\infty \int _{(j-1) \, \Delta }^{j \, \Delta } f(\omega ) \, \mathrm{d}\omega . \end{aligned} \end{aligned}$$
(92)
By the fundamental theorem of calculus, we get
$$\begin{aligned} \begin{aligned} f(\omega ) = f(j \ \Delta - \alpha \, \Delta ) + \int _{j \, \Delta - \alpha \, \Delta }^\omega f'(\omega ) \, \mathrm{d}\omega , \end{aligned} \end{aligned}$$
(93)
which gives for \(\omega \in ((j-1) \, \Delta , j \, \Delta ]\)
$$\begin{aligned} \begin{aligned} \left| f(\omega ) - f(j \, \Delta - \alpha \, \Delta ) \right|&\le \left| \int _{j \, \Delta - \alpha \, \Delta }^\omega f'(\omega ) \, \mathrm{d}\omega \right| \\&\le \left| \int _{j \, \Delta - \alpha \, \Delta }^\omega \left| f'(\omega ) \right| \mathrm{d}\omega \right| \\&\le \int _{(j-1) \, \Delta }^{j \, \Delta } \left| f'(\omega ) \right| \, \mathrm{d}\omega \end{aligned} \end{aligned}$$
(94)
We now get
$$\begin{aligned} \begin{aligned}&\left| \int _{m \, \Delta }^\infty f(\omega ) \, \mathrm{d}\omega - \sum _{j=m+1}^\infty f(j \, \Delta - \alpha \, \Delta ) \, \Delta \right| \\&\quad = \left| \sum _{j=m+1}^\infty \int _{(j-1) \, \Delta }^{j \, \Delta } [f(\omega ) - f(j \, \Delta - \alpha \, \Delta )] \, \mathrm{d}\omega \right| \\&\quad \le \sum _{j=m+1}^\infty \int _{(j-1) \, \Delta }^{j \, \Delta } \left| f(\omega ) - f(j \, \Delta - \alpha \, \Delta ) \right| \, \mathrm{d}\omega \\&\quad \le \sum _{j=m+1}^\infty \int _{(j-1) \, \Delta }^{j \, \Delta } \left[ \int _{(j-1) \, \Delta }^{j \, \Delta } \left| f'(\omega ) \right| \, \mathrm{d}\omega \right] \, \mathrm{d}\omega \\&\quad = \sum _{j=m+1}^\infty \int _{(j-1) \, \Delta }^{j \, \Delta } \left| f'(\omega ) \right| \, \mathrm{d}\omega \, \Delta \\&\quad = \underbrace{\int _{m \, \Delta }^{\infty } \left| f'(\omega ) \right| \, \mathrm{d}\omega }_{C^{(m)}} \, \Delta \le \underbrace{\int _{0}^{\infty } \left| f'(\omega ) \right| \, \mathrm{d}\omega }_{C^{(0)}} \, \Delta , \end{aligned} \end{aligned}$$
(95)
which concludes the proof. \(\square \)
Lemma 10
Assume that \(f(\omega )\) is a bounded integrable function defined on \(\omega \ge 0\) with bounded derivative on \(\omega > 0\) and \(g(\omega )\) is a bounded function defined on \(\omega \ge 0\) such that \(|g(w)| \le D\). Further assume that \(\int _0^{\infty } |f'(\omega )| \, \mathrm{d}\omega = C < \infty \). Then for any \(\alpha ,\beta \in [0,1)\) and \(\Delta > 0\) we have
$$\begin{aligned} \left| \sum _{j=1}^\infty [f(j \, \Delta ) - f(j \, \Delta - \alpha \, \Delta )] \, g(j \, \Delta - \beta \, \Delta ) \right| \le C \, D. \end{aligned}$$
(96)
Proof
By using (94) with \(\omega = j \, \Delta - \alpha \, \Delta \), we get
$$\begin{aligned} \begin{aligned} \left| f(j \, \Delta ) - f(j \, \Delta - \alpha \, \Delta ) \right| \le \int _{(j-1) \, \Delta }^{j \, \Delta } \left| f'(\omega ) \right| \, \mathrm{d}\omega , \end{aligned} \end{aligned}$$
(97)
and further
$$\begin{aligned} \begin{aligned}&\left| \sum _{j=1}^\infty [f(j \, \Delta ) - f(j \, \Delta - \alpha \, \Delta )] \, g(j \, \Delta - \beta \, \Delta ) \right| \\&\quad \le \sum _{j=1}^\infty \left| f(j \, \Delta ) - f(j \, \Delta - \alpha \, \Delta ) \right| \, |g(j \, \Delta - \beta \, \Delta )| \\&\quad \le \sum _{j=1}^\infty \int _{(j-1) \, \Delta }^{j \, \Delta } \left| f'(\omega ) \right| \, \mathrm{d}\omega \, D \\&\quad = \int _{0}^{\infty } \left| f'(\omega ) \right| \, \mathrm{d}\omega \, D \\&\quad = C \, D. \\ \end{aligned} \end{aligned}$$
(98)
\(\square \)
Lemma 11
Assume that \(f(\omega ) \ge 0\) is a positive bounded integrable function defined on \(\omega \ge 0\) with bounded derivative on \(\omega > 0\) such that \(\int _0^{\infty } f(\omega ) \, \mathrm{d}\omega = C_0 < \infty \) and \(\int _0^{\infty } |f'(\omega )| \, \mathrm{d}\omega = C_1 \le \infty \), and \(g(\omega )\) is a bounded integrable function defined on \(\omega \ge 0\) with bounded derivative on \(\omega > 0\) such that \(|g'(\omega )| \le D\). Then for any \(\alpha ,\beta \in [0,1)\) and \(\Delta > 0\) we have for \(C = C_1 + C_0\):
$$\begin{aligned} \left| \sum _{j=1}^\infty f(j \, \Delta - \alpha \, \Delta ) \, [g(j \, \Delta ) - g(j \, \Delta - \beta \, \Delta )] \right| \le C \, D. \end{aligned}$$
(99)
Proof
By applying the mean value theorem to (97), we get that for some \(\omega ^*_j \in [j \, \Delta - \alpha \, \Delta , j \, \Delta ]\) we have
$$\begin{aligned} \begin{aligned}&\left| g(j \, \Delta ) - g(j \, \Delta - \beta \, \Delta ) \right| \\&\quad \le |g'(\omega ^*_j)| \, \beta \, \Delta \le |g'(\omega ^*_j)| \, \Delta \le D \, \Delta . \end{aligned} \end{aligned}$$
(100)
By using Lemma 9, we get
$$\begin{aligned} \begin{aligned}&\sum _{j=1}^\infty f(j \, \Delta - \alpha \, \Delta ) \, \Delta \\&\quad = \left| \sum _{j=1}^\infty f(j \, \Delta - \alpha \, \Delta ) \, \Delta - \int _0^\infty f(\omega ) \, \mathrm{d}\omega \right. \\&\quad \qquad \left. + \int _0^\infty f(\omega ) \, \mathrm{d}\omega \right| \\&\quad \le \left| \sum _{j=1}^\infty f(j \, \Delta - \alpha \, \Delta ) \, \Delta - \int _0^\infty f(\omega ) \, \mathrm{d}\omega \right| \\&\quad \qquad + \left| \int _0^\infty f(\omega ) \, \mathrm{d}\omega \right| \\&\quad \le C_1 + C_0 = C. \end{aligned} \end{aligned}$$
(101)
Hence,
$$\begin{aligned} \begin{aligned}&\left| \sum _{j=1}^\infty f(j \, \Delta - \alpha \, \Delta ) \, [g(j \, \Delta ) - g(j \, \Delta - \beta \, \Delta )] \right| \\&\quad \le \sum _{j=1}^\infty f(j \, \Delta - \alpha \, \Delta ) \, \left| g(j \, \Delta ) - g(j \, \Delta - \beta \, \Delta ) \right| \, \\&\quad \le \sum _{j=1}^\infty f(j \, \Delta - \alpha \, \Delta ) \, \Delta \, D \\&\quad \le C \, D. \end{aligned} \end{aligned}$$
(102)
\(\square \)
Proof of Theorem 1
The Wiener–Khintchin identity and the symmetry of the spectral density allows us to write
$$\begin{aligned} k(x,x')&= \frac{1}{2\pi } \, \int _{-\infty }^{\infty } S(\omega ) \, \exp (-i\, \omega \, (x - x')) \,{\mathrm {d}}\omega \nonumber \\&= \frac{1}{\pi } \, \int _{0}^{\infty } S(\omega ) \, \cos (\omega \, (x - x')) \,{\mathrm {d}}\omega . \end{aligned}$$
(103)
In a one-dimensional domain \(\varOmega = [-L,L]\) with Dirichlet boundary conditions, we have an m-term approximation of the form
$$\begin{aligned} \begin{aligned} \widetilde{k}_m(x,x')&= \sum _{j=1}^m S\left( \frac{\pi \, j}{2L}\right) \, \frac{1}{L} \, \sin \left( \frac{\pi \, j \, (x + L)}{2L} \right) \\&\quad \times \sin \left( \frac{\pi \, j \, (x' + L)}{2L} \right) . \end{aligned} \end{aligned}$$
(104)
We start by showing the convergence by growing the domain and therefore first consider an approximation with an infinite number of terms \(m=\infty \):
$$\begin{aligned} \begin{aligned} \widetilde{k}_\infty (x,x')&= \sum _{j=1}^\infty S\left( \frac{\pi \, j}{2L}\right) \, \frac{1}{L} \, \sin \left( \frac{\pi \, j \, (x + L)}{2L} \right) \\&\quad \times \sin \left( \frac{\pi \, j \, (x' + L)}{2L} \right) . \end{aligned} \end{aligned}$$
(105)
For that purpose, we rewrite the summation above in (105) as
$$\begin{aligned}&\sum _{j=1}^{\infty } S\left( \frac{\pi \, j}{2L}\right) \, \frac{1}{L} \, \sin \left( \frac{\pi \, j \, (x + L)}{2L} \right) \, \sin \left( \frac{\pi \, j \, (x' + L)}{2L} \right) \nonumber \\&\quad = \sum _{j=1}^\infty S\left( \frac{\pi \, j}{2L} \right) \, \cos \left( \frac{\pi \, j \, (x - x')}{2L} \right) \, \frac{1}{2L} \nonumber \\&\qquad - \frac{1}{2L} \, \sum _{j=1}^\infty \left[ S\left( \frac{\pi \, 2j}{2L} \right) \, - S\left( \frac{\pi \, (2j-1)}{2L} \right) \right] \,\nonumber \\&\qquad \times \cos \left( \frac{\pi \, 2j \, (x + x')}{2L} \right) \nonumber \\&\qquad - \frac{1}{2L} \, \sum _{j=1}^\infty S\left( \frac{\pi \, (2j-1)}{2L} \right) \, \Bigg [ \cos \left( \frac{\pi \, 2j \, (x + x')}{2L} \right) \nonumber \\&\qquad - \cos \left( \frac{\pi \, (2j-1) \, (x + x')}{2L} \right) \Bigg ]. \end{aligned}$$
(106)
and consider the three summations above separately. The analysis of them is done in the next three lemmas.
Lemma 12
Assume that on \(\omega \ge 0\) we have \(S(\omega ) \le B < \infty \) and \(\int _0^\infty S(w) \, \,{\mathrm {d}}\omega = A < \infty \), and on \(\omega > 0\)\(S(\omega )\) has a bounded derivative \(|S'(\omega )| \le D < \infty \) and that \(\int _0^\infty |S'(\omega )| \, \,{\mathrm {d}}\omega = C < \infty \). Then there exists a constant \(D_2\) such that for all \(x,x' \in [-\widetilde{L},\widetilde{L}]\) we have
$$\begin{aligned}&\left| \sum _{j=1}^\infty S\left( \frac{\pi \, j}{2L} \right) \, \cos \left( \frac{\pi \, j \, (x - x')}{2L} \right) \, \frac{1}{2L}\right. \nonumber \\&\left. \quad - \frac{1}{\pi } \, \int _{0}^{\infty } S(\omega ) \, \cos (\omega \, (x - x')) \,{\mathrm {d}}\omega \right| \le \frac{D_2}{L}. \end{aligned}$$
(107)
Proof
By using Lemma 9 with \(\Delta = \frac{\pi }{2L}\), \(f(\omega ) = \frac{1}{\pi } \, S(\omega ) \, \cos (\omega \, (x - x')) \,{\mathrm {d}}\omega \), \(m=0\), and \(\alpha = 0\) as well as the assumptions on \(S(\omega )\) and boundedness of sine and cosine we get that
$$\begin{aligned}&\left| \sum _{j=1}^\infty S\left( \frac{\pi \, j}{2L} \right) \, \cos \left( \frac{\pi \, j \, (x - x')}{2L} \right) \, \frac{1}{2L} \right. \nonumber \\&\left. \qquad -\, \frac{1}{\pi } \int _0^\infty S(\omega ) \, \cos (\omega \, (x - x')) \,{\mathrm {d}}\omega \right| \nonumber \\&\quad \le \frac{1}{\pi } \int \big | S'(w) \, \cos (\omega \, (x - x')) \,{\mathrm {d}}\omega \nonumber \\&\qquad -\, S(w) \, (x - x') \, \sin (\omega \, (x - x')) \big | \, \,{\mathrm {d}}\omega \, \frac{\pi }{2L} \nonumber \\&\quad \le \frac{1}{2L} \int \big | S'(w) \, \cos (\omega \, (x - x')) \big | \,{\mathrm {d}}\omega \\&\qquad +\, \frac{1}{2L} \int \big | S(w) \, (x - x') \, \sin (\omega \, (x - x')) \big | \, \,{\mathrm {d}}\omega \nonumber \\&\quad \le \frac{1}{2L} \int \big | S'(w) \big | \,{\mathrm {d}}\omega + \frac{|x - x'| }{2L} \, \int \big | S(w) \big | \, \,{\mathrm {d}}\omega \nonumber \\&\quad \le \frac{1}{2L} \int \big | S'(w) \big | \,{\mathrm {d}}\omega + \frac{\widetilde{L} }{L} \, \int \big | S(w) \big | \, \,{\mathrm {d}}\omega \nonumber \\&\quad \le \frac{1}{2L} C + \frac{\widetilde{L} }{L} \, A,\nonumber \end{aligned}$$
(108)
which gives the result with \(D_2 = \frac{C}{2} + \widetilde{L} \, A\). \(\square \)
Lemma 13
Assume that for \(\omega \ge 0\), \(S(\omega )\) is a bounded integrable function with a bounded derivative on \(\omega > 0\) such that \(\int _0^\infty |S'(\omega )| \, \,{\mathrm {d}}\omega = C < \infty \), then there exists a constant \(D_3\) such that
$$\begin{aligned} \begin{aligned}&\left| \frac{1}{2L} \, \sum _{j=1}^\infty \left[ S\left( \frac{\pi \, 2j}{2L} \right) \, - S\left( \frac{\pi \, (2j-1)}{2L} \right) \right] \, \right. \\&\qquad \left. \times \cos \left( \frac{\pi \, 2j \, (x + x')}{2L} \right) \right| \\&\quad \le \frac{D_3}{L}. \end{aligned} \end{aligned}$$
(109)
Proof
The result follows by using Lemma 10 with \(\Delta = \frac{\pi }{L}\), \(\alpha = 1/2\), \(\beta = 0\), \(f(\omega ) = S(\omega )\), and \(g(\omega ) = \cos (\omega \, (x + x'))\) and by recalling that \(|\cos (\omega \, (x + x'))| \le 1\), which gives the constant \(D_3 = \frac{C}{2}\). \(\square \)
Lemma 14
Assume that for \(\omega \ge 0\), \(S(\omega )\) is a bounded positive integrable function with bounded derivative on \(\omega > 0\) such that \(\int _0^\infty S(\omega ) \, \,{\mathrm {d}}\omega = A < \infty \) and \(\int _0^\infty |S'(\omega )| \, \,{\mathrm {d}}\omega = C < \infty \). Then there exists a constant \(D_4\) such that
$$\begin{aligned} \begin{aligned}&\left| \frac{1}{2L} \, \sum _{j=1}^\infty S\left( \frac{\pi \, (2j-1)}{2L} \right) \, \Bigg [ \cos \left( \frac{\pi \, 2j \, (x + x')}{2L} \right) \right. \\&\left. \qquad - \cos \left( \frac{\pi \, (2j-1) \, (x + x')}{2L} \right) \right| \le \frac{D_4}{L}. \end{aligned} \end{aligned}$$
Proof
By using Lemma 11 with \(\Delta = \frac{\pi }{L}\), \(\alpha = 1/2\), \(\beta = 1/2\), \(f(\omega ) = S(\omega )\), and \(g(\omega ) = \cos (\omega \, (x + x'))\) we get
$$\begin{aligned} \begin{aligned}&\left| \frac{1}{2L} \, \sum _{j=1}^\infty S\left( \frac{\pi \, (2j-1)}{2L} \right) \, \Bigg [ \cos \left( \frac{\pi \, 2j \, (x + x')}{2L} \right) \right. \\&\left. \qquad - \cos \left( \frac{\pi \, (2j-1) \, (x + x')}{2L} \right) \right| \\&\quad \le \frac{(A + C) \, D'}{2L}, \end{aligned} \end{aligned}$$
where \(D'\) is an upper bound for \(|(x + x') \, \sin (\omega \, (x + x'))|\). We can now select \(D' = 2\widetilde{L}\), which gives \(D_4 = (A + C) \,\widetilde{L}\). \(\square \)
Next, we combine the above lemmas to get the following result.
Lemma 15
Let the assumptions of Lemmas 12, 13, and 14 be satisfied. Then there exists a constant \(D_1\) such that for all \(x,x' \in [-\,\widetilde{L},\widetilde{L}]\) we have
$$\begin{aligned}&\left| \sum _{j=1}^{\infty } S\left( \frac{\pi \, j}{2L}\right) \, \frac{1}{L} \, \sin \left( \frac{\pi \, j \, (x + L)}{2L} \right) \times \sin \left( \frac{\pi \, j \, (x' + L)}{2L} \right) \right. \nonumber \\&\left. \quad - \frac{1}{\pi } \, \int _{0}^{\infty } S(\omega ) \, \cos (\omega \, (x - x')) \,{\mathrm {d}}\omega \right| \le \frac{D_1}{L}. \end{aligned}$$
(110)
That is,
$$\begin{aligned} \left| \widetilde{k}_{\infty }(x,x') - k(x,x') \right| \le \frac{D_1}{L}, \quad \text { for } x,x' \in [-\widetilde{L},\widetilde{L}]. \end{aligned}$$
(111)
Furthermore, the explicit expression for the constant is given as
$$\begin{aligned} D_1 = C + (2 A + C) \, \widetilde{L}. \end{aligned}$$
(112)
Proof
Using triangle inequality to the difference of (106) and \(\frac{1}{\pi } \, \int _{0}^{\infty } S(\omega ) \, \cos (\omega \, (x - x')) \,{\mathrm {d}}\omega \) along with Lemmas 12, 13, and 14 gives
$$\begin{aligned}&\left| \sum _{j=1}^\infty S\left( \frac{\pi \, j}{2L} \right) \, \cos \left( \frac{\pi \, j \, (x - x')}{2L} \right) \, \frac{1}{2L} \right. \nonumber \\&\qquad - \frac{1}{2L} \, \sum _{j=1}^\infty \left[ S\left( \frac{\pi \, 2j}{2L} \right) \, - S\left( \frac{\pi \, (2j-1)}{2L} \right) \right] \,\nonumber \\&\qquad \times \cos \left( \frac{\pi \, 2j \, (x + x')}{2L} \right) \nonumber \\&\qquad - \frac{1}{2L} \, \sum _{j=1}^\infty S\left( \frac{\pi \, (2j-1)}{2L} \right) \, \Bigg [ \cos \left( \frac{\pi \, 2j \, (x + x')}{2L} \right) \nonumber \\&\qquad - \cos \left( \frac{\pi \, (2j-1) \, (x + x')}{2L} \right) \Bigg ] \nonumber \\&\left. \qquad - \frac{1}{\pi } \, \int _{0}^{\infty } S(\omega ) \, \cos (\omega \, (x - x')) \,{\mathrm {d}}\omega \right| \nonumber \\&\quad \le \left| \sum _{j=1}^\infty S\left( \frac{\pi \, j}{2L} \right) \, \cos \left( \frac{\pi \, j \, (x - x')}{2L} \right) \, \frac{1}{2L} \right. \nonumber \\&\qquad - \frac{1}{\pi } \, \int _{0}^{\infty } S(\omega ) \, \cos (\omega \, (x - x')) \,{\mathrm {d}}\omega \Bigg | \nonumber \\&\qquad + \Bigg | \frac{1}{2L} \, \sum _{j=1}^\infty \left[ S\left( \frac{\pi \, 2j}{2L} \right) \, - S\left( \frac{\pi \, (2j-1)}{2L} \right) \right] \,\nonumber \\&\left. \qquad \times \cos \left( \frac{\pi \, 2j \, (x + x')}{2L} \right) \right| \nonumber \\&\qquad + \left| \frac{1}{2L} \, \sum _{j=1}^\infty S\left( \frac{\pi \, (2j-1)}{2L} \right) \, \Bigg [ \cos \left( \frac{\pi \, 2j \, (x + x')}{2L} \right) \right. \nonumber \\&\left. \qquad - \cos \left( \frac{\pi \, (2j-1) \, (x + x')}{2L} \right) \Bigg ] \right| \nonumber \\&\quad \le \frac{D_2}{L} + \frac{D_3}{L} + \frac{D_4}{L} = \frac{D_1}{L}, \end{aligned}$$
(113)
where the explicit values for the constants can be found in the proofs of the lemmas. \(\square \)
Let us now consider what happens when we replace the infinite sum approximation with a finite m number of terms. We are now interested in
$$\begin{aligned} \begin{aligned}&\widetilde{k}_{\infty }(x,x') - \widetilde{k}_m(x,x') \\&\quad = \sum _{j=m+1}^{\infty } S\left( \frac{\pi \, j}{2L}\right) \, \frac{1}{L} \, \sin \left( \frac{\pi \, j \, (x + L)}{2L} \right) \, \\&\qquad \times \sin \left( \frac{\pi \, j \, (x' + L)}{2L} \right) . \end{aligned} \end{aligned}$$
(114)
Lemma 16
Assume that on \(\omega \ge 0\), \(S(\omega )\) is bounded and integrable, on \(\omega > 0\) it has a bounded derivative, and that \(\int _0^\infty |S'(\omega )| \, \,{\mathrm {d}}\omega = C < \infty \). Then there exists a constant \(D_5\) such that for all \(x,x' \in [-\widetilde{L},\widetilde{L}]\) we have
$$\begin{aligned} \left| \widetilde{k}_{\infty }(x,x') - \widetilde{k}_m(x,x') \right| \le \frac{D_5}{L} + \frac{2}{\pi } \, \int _{\frac{\pi \, m}{2L}}^{\infty } S(\omega ) \,{\mathrm {d}}\omega . \end{aligned}$$
(115)
Proof
Because the sinusoidals are bounded by unity, we get
$$\begin{aligned}&\left| \sum _{j=m+1}^{\infty } S\left( \frac{\pi \, j}{2L}\right) \, \frac{1}{L} \, \sin \left( \frac{\pi \, j \, (x + L)}{2L} \right) \, \right. \nonumber \\&\qquad \left. \times \sin \left( \frac{\pi \, j \, (x' + L)}{2L} \right) \right| \nonumber \\&\quad \le \left| \sum _{j=m+1}^{\infty } S\left( \frac{\pi \, j}{2L}\right) \, \frac{1}{L} \right| . \end{aligned}$$
(116)
For the right-hand side, we can now use Lemma 9 with \(f(\omega ) = \frac{2}{\pi } \, S(\omega )\) and \(\Delta = \frac{\pi }{2L}\), which gives
$$\begin{aligned} \left| \sum _{j=m+1}^{\infty } S\left( \frac{\pi \, j}{2L}\right) \, \frac{1}{L} - \frac{2}{\pi } \int _{\frac{\pi \, m}{2L}}^{\infty } S(\omega ) \,{\mathrm {d}}\omega \right| \le C \, \frac{\pi }{2L} = \frac{D_5}{L}. \end{aligned}$$
(117)
Hence by the triangle inequality we get
$$\begin{aligned}&\left| \sum _{j=m+1}^{\infty } S\left( \frac{\pi \, j}{2L}\right) \, \frac{1}{L} \right| \nonumber \\&\quad = \left| \sum _{j=m+1}^{\infty } S\left( \frac{\pi \, j}{2L}\right) \, \frac{1}{L} - \frac{2}{\pi } \int _{\frac{\pi \, m}{2L}}^{\infty } S(\omega ) \,{\mathrm {d}}\omega \right. \nonumber \\&\left. \qquad + \frac{2}{\pi } \int _{\frac{\pi \, m}{2L}}^{\infty } S(\omega ) \,{\mathrm {d}}\omega \right| \nonumber \\&\quad \le \left| \sum _{j=m+1}^{\infty } S\left( \frac{\pi \, j}{2L}\right) \, \frac{1}{L} - \frac{2}{\pi } \int _{\frac{\pi \, m}{2L}}^{\infty } S(\omega ) \,{\mathrm {d}}\omega \right| \nonumber \\&\qquad + \frac{2}{\pi } \int _{\frac{\pi \, m}{2L}}^{\infty } S(\omega ) \,{\mathrm {d}}\omega \nonumber \\&\quad \le \frac{D_5}{L} + \frac{2}{\pi } \int _{\frac{\pi \, m}{2L}}^{\infty } S(\omega ) \,{\mathrm {d}}\omega \end{aligned}$$
(118)
and thus the result follows. \(\square \)
Remark 17
We can also obtain a bit more defined bound by not using an m-independent bound for forming \(D_5\), which under the assumptions of Lemma 16 gives
$$\begin{aligned} \begin{aligned}&\left| \widetilde{k}_{\infty }(x,x') - \widetilde{k}_m(x,x') \right| \\&\quad \le \frac{\pi }{2L} \, \int _{\frac{\pi \, m}{2L}}^{\infty } |S'(\omega )| \,{\mathrm {d}}\omega + \frac{2}{\pi } \, \int _{\frac{\pi \, m}{2L}}^{\infty } S(\omega ) \,{\mathrm {d}}\omega . \end{aligned} \end{aligned}$$
(119)
The lemmas presented in this section can now be combined to a proof of the one-dimensional convergence theorem as follows:
Proof of Theorem 1
The first result follows by combining Lemmas 15 and 16 via the triangle inequality. Because our assumptions imply that
$$\begin{aligned} \lim _{x \rightarrow \infty } \int _{x}^{\infty } S(\omega ) \,{\mathrm {d}}\omega = 0, \end{aligned}$$
(120)
for any fixed L we have
$$\begin{aligned} \lim _{m \rightarrow \infty } \left[ \frac{E}{L} + \frac{2}{\pi } \, \int _{\frac{\pi \, m}{2L}}^{\infty } S(\omega ) \,{\mathrm {d}}\omega \right] \rightarrow \frac{E}{L}. \end{aligned}$$
(121)
If we now take the limit \(L \rightarrow \infty \), the second result in the theorem follows. \(\square \)
Proof of Theorem 4
When \(\mathbf {x} \in \mathbb {R}^d\), the Wiener–Khintchin identity and symmetry of the spectral density imply that
$$\begin{aligned} \begin{aligned} k(\mathbf {x},\mathbf {x}')&= \frac{1}{(2\pi )^d} \, \int _{\mathbb {R}^d} S({\omega }) \, \exp (-i\, {\omega }^\mathsf {T}(\mathbf {x} - \mathbf {x}')) \,{\mathrm {d}}{\omega } \\&= \frac{1}{\pi ^d} \, \int _{0}^{\infty } \cdots \int _{0}^{\infty } S({\omega }) \, \prod _{k=1}^d \cos (\omega _k \, (x_k - x_k')) \,{\mathrm {d}}\omega _1 \cdots \,{\mathrm {d}}\omega _d. \end{aligned} \end{aligned}$$
(122)
The \(m = {{\hat{m}}}^d\) term approximation now has the form
$$\begin{aligned} \widetilde{k}_m(\mathbf {x},\mathbf {x}')&= \sum _{j_1,\ldots ,j_d=1}^{{\hat{m}}} S\left( \frac{\pi \, j_1}{2L_1},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \nonumber \\&\qquad \times \prod _{k=1}^d \frac{1}{L_k} \, \sin \left( \frac{\pi \, j_k \, (x_k + L_k)}{2L_k} \right) \,\nonumber \\&\qquad \times \sin \left( \frac{\pi \, j_k \, (x_k' + L_k)}{2L_k} \right) . \end{aligned}$$
(123)
As in the one-dimensional problem, we start by considering the case where \(\hat{m} = \infty \).
Lemma 18
Let the assumptions of Lemma 15 be satisfied for each \(\omega _j \mapsto S(\omega _1,\ldots ,\omega _d)\) separately. Then there exists a constant \(D_1\) such that for all \(\mathbf {x},\mathbf {x}' \in [-\widetilde{L},\widetilde{L}]^d\) we have
$$\begin{aligned}&\left| \sum _{j_1,\ldots ,j_d=1}^\infty S\left( \frac{\pi \, j_1}{2L_1},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \times \prod _{k=1}^d \frac{1}{L_k} \, \sin \left( \frac{\pi \, j_k \, (x_k + L_k)}{2L_k} \right) \,\right. \nonumber \\&\qquad \times \sin \left( \frac{\pi \, j_k \, (x_k' + L_k)}{2L_k} \right) \nonumber \\&\left. \qquad - \frac{1}{\pi ^d} \, \int _{0}^{\infty } \cdots \int _{0}^{\infty } S({\omega }) \, \prod _{k=1}^d \cos (\omega _k \, (x - x')) \,{\mathrm {d}}\omega _1 \cdots \,{\mathrm {d}}\omega _d \right| \nonumber \\&\quad \le D_1 \sum _{k=1}^d \frac{1}{L_k} \le \frac{D_1 \, d}{L}, \end{aligned}$$
(124)
where \(L = \min _k L_k\). That is, for all \(\mathbf {x},\mathbf {x}' \in [-\widetilde{L},\widetilde{L}]^d\)
$$\begin{aligned} \left| \widetilde{k}_{\infty }(\mathbf {x},\mathbf {x}') - k(\mathbf {x},\mathbf {x}') \right| \le D_1 \sum _{k=1}^d \frac{1}{L_k} \le \frac{D_1 \, d}{L}. \end{aligned}$$
(125)
Proof
We can separate the summation over \(j_1\) as follows:
$$\begin{aligned}&\sum _{j_2,\ldots ,j_d=1}^\infty \left[ \sum _{j_1=1}^{\infty } S\left( \frac{\pi \, j_1}{2L_1},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \, \frac{1}{L_1} \right. \nonumber \\&\left. \qquad \times \sin \left( \frac{\pi \, j_1 \, (x_1 + L_1)}{2L_1} \right) \, \sin \left( \frac{\pi \, j_1 \, (x_1' + L_1)}{2L_1} \right) \right] \nonumber \\&\qquad \times \prod _{k=2}^d \frac{1}{L_k} \, \sin \left( \frac{\pi \, j_k \, (x_k + L_k)}{2L_k} \right) \nonumber \\&\qquad \times \sin \left( \frac{\pi \, j_k \, (x_k' + L_k)}{2L_k} \right) . \end{aligned}$$
(126)
By Lemma 15, there now exists a constant \(D_{1,1}\) such that
$$\begin{aligned}&\left| \sum _{j_1=1}^{\infty } S\left( \frac{\pi \, j_1}{2L_1},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \, \frac{1}{L_1} \right. \nonumber \\&\qquad \times \sin \left( \frac{\pi \, j_1 \, (x_1 + L_1)}{2L_1} \right) \, \sin \left( \frac{\pi \, j_1 \, (x_1' + L_1)}{2L_1} \right) \nonumber \\&\qquad - \frac{1}{\pi } \, \int _{0}^{\infty } S\left( \omega _1,\frac{\pi \, j_2}{2L_2},\ldots ,\frac{\pi \, j_d}{2L_d}\right) \, \nonumber \\&\left. \qquad \times \cos (\omega _1 \, (x_1 - x_1')) \,{\mathrm {d}}\omega _1 \right| \nonumber \\&\quad \le \frac{D_{1,1}}{L_1}. \end{aligned}$$
(127)
The triangle inequality then gives
$$\begin{aligned}&\left| \sum _{j_1,\ldots ,j_d=1}^\infty S\left( \frac{\pi \, j_1}{2L_1},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \times \prod _{k=1}^d \frac{1}{L_k} \, \sin \left( \frac{\pi \, j_k \, (x_k + L_k)}{2L_k} \right) \,\right. \nonumber \\&\qquad \times \sin \left( \frac{\pi \, j_k \, (x_k' + L_k)}{2L_k} \right) \nonumber \\&\left. \qquad - \frac{1}{\pi ^d} \, \int _{0}^{\infty } \cdots \int _{0}^{\infty } S({\omega }) \, \prod _{k=1}^d \cos (\omega _j \, (x_k - x_k')) \,{\mathrm {d}}\omega _1 \cdots \,{\mathrm {d}}\omega _d \right| \nonumber \\&\quad \le \frac{D_{1,1}}{L_1} + \left| \frac{1}{\pi } \, \sum _{j_2,\ldots ,j_d=1}^\infty \int _{0}^{\infty } S\left( \omega _1,\frac{\pi \, j_2}{2L_2}, \ldots , \frac{\pi \, j_d}{2L_d}\right) \right. \nonumber \\&\qquad \times \cos (\omega _1 \, (x_1 - x_1')) \,{\mathrm {d}}\omega _1 \nonumber \\&\qquad \times \prod _{k=2}^d \frac{1}{L_k} \, \sin \left( \frac{\pi \, j_k \, (x_k + L_k)}{2L_k} \right) \, \sin \left( \frac{\pi \, j_k \, (x_k' + L_k)}{2L_k} \right) \nonumber \\&\left. \qquad - \frac{1}{\pi ^d} \, \int _{0}^{\infty } \cdots \int _{0}^{\infty } S({\omega }) \, \prod _{k=1}^d \cos (\omega _k \, (x_k - x_k')) \,{\mathrm {d}}\omega _1 \cdots \,{\mathrm {d}}\omega _d \right| . \end{aligned}$$
(128)
We can now similarly bound with respect to the summations over \(j_2,\ldots ,j_d\) which leads to a bound of the form \(\frac{D_{1,1}}{L_1} + \cdots + \frac{D_{1,d}}{L_d}\). Taking \(D_1 = \max _k D_{1,k}\) leads to the desired result. \(\square \)
Now we can consider what happens in the finite truncation of the series. That is, we analyze the following residual sum
$$\begin{aligned}&\widetilde{k}_{\infty }(\mathbf {x},\mathbf {x}') - \widetilde{k}_m(\mathbf {x},\mathbf {x}') \nonumber \\&\quad = \sum _{j_1,\ldots ,j_d={{\hat{m}}}+1}^{\infty } S\left( \frac{\pi \, j_1}{2L_1},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \nonumber \\&\qquad \times \prod _{k=1}^d \frac{1}{L_k} \, \sin \left( \frac{\pi \, j_k \, (x_k + L_k)}{2L_k} \right) \,\nonumber \\&\qquad \times \sin \left( \frac{\pi \, j_k \, (x_k' + L_k)}{2L_k} \right) . \end{aligned}$$
(129)
Lemma 19
Let assumptions of Lemma 16 be satisfied for each \(\omega _j \mapsto S(\omega _1, \ldots , \omega _d)\). There exists a constant \(D_2\) such that for all \(\mathbf {x},\mathbf {x}' \in [-\widetilde{L},\widetilde{L}]^d\) we have
$$\begin{aligned} \begin{aligned}&\left| \widetilde{k}_{\infty }(\mathbf {x},\mathbf {x}') - \widetilde{k}_m(\mathbf {x},\mathbf {x}') \right| \le \frac{D_2 \, d}{L} + \frac{1}{\pi ^d} \int _{||{\omega }|| \ge \frac{\pi \, \hat{m}}{2L}} S({\omega }) \,{\mathrm {d}}{\omega }, \end{aligned} \end{aligned}$$
(130)
where \(L = \min _k L_k\).
Proof
We can write the following bound
$$\begin{aligned}&\Bigg | \sum _{j_1,\ldots ,j_d={{\hat{m}}}+1}^{\infty } S\left( \frac{\pi \, j_1}{2L_1},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \nonumber \\&\qquad \times \prod _{k=1}^d \frac{1}{L_k} \, \sin \left( \frac{\pi \, j_k \, (x_k + L_k)}{2L_k} \right) \,\nonumber \\&\qquad \times \sin \left( \frac{\pi \, j_k \, (x_k' + L_k)}{2L_k} \right) \Bigg | \nonumber \\&\quad \le \Bigg | \sum _{j_1,\ldots ,j_d={{\hat{m}}}+1}^{\infty } S\left( \frac{\pi \, j_1}{2L_1},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \, \prod _{k=1}^d \frac{1}{L_k} \Bigg |. \end{aligned}$$
(131)
We can now use Lemma 9 with \(f(\omega _1) = \frac{2}{\pi } \, S\left( \omega _1, \frac{\pi \, j_2}{2L_2},\ldots ,\right. \left. \frac{\pi \, j_d}{2L_d} \right) \) and \(\Delta = \frac{\pi }{2L_1}\), which gives
$$\begin{aligned}&\left| \sum _{j_1,\ldots ,j_d={{\hat{m}}}+1}^{\infty } S\left( \frac{\pi \, j_1}{2L_1},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \, \prod _{k=1}^d \frac{1}{L_k} \right. \nonumber \\&\qquad - \frac{2}{\pi } \sum _{j_2,\ldots ,j_d={{\hat{m}}}+1}^{\infty }\nonumber \\&\left. \qquad \times \int _{\frac{\pi \, {{\hat{m}}}}{2L_1}}^{\infty } S\left( \omega _1, \frac{\pi \, j_2}{2L_2},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \,{\mathrm {d}}\omega _1 \, \prod _{k=2}^d \frac{1}{L_k} \right| \nonumber \\&\quad \le \frac{D_{2,1}}{L_1}. \end{aligned}$$
(132)
Using a similar argument again, we get
$$\begin{aligned}&\left| \frac{2}{\pi } \sum _{j_2,\ldots ,j_d={{\hat{m}}}+1}^{\infty } \int _{\frac{\pi \, {{\hat{m}}}}{2L_1}}^{\infty } S\left( \omega _1, \frac{\pi \, j_2}{2L_2},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \,{\mathrm {d}}\omega _1 \, \prod _{k=2}^d \frac{1}{L_k} \right. \nonumber \\&\qquad - \frac{2^2}{\pi ^2} \sum _{j_3,\ldots ,j_d={{\hat{m}}}+1}^{\infty } \int _{\frac{\pi \, {{\hat{m}}}}{2L_1}}^{\infty } \nonumber \\&\qquad \times \int _{\frac{\pi \, {{\hat{m}}}}{2L_2}}^{\infty } S\left( \omega _1, \omega _2, \frac{\pi \, j_3}{2L_3},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \,{\mathrm {d}}\omega _1 \,{\mathrm {d}}\omega _2 \nonumber \\&\left. \qquad \times \prod _{k=3}^d \frac{1}{L_k} \right| \le \frac{D_{2,2}}{L_2}. \end{aligned}$$
(133)
After repeating this for all the indexes, by forming a telescoping sum of the terms and applying the triangle inequality then gives
$$\begin{aligned}&\left| \sum _{j_1,\ldots ,j_d={{\hat{m}}}+1}^{\infty } S\left( \frac{\pi \, j_1}{2L_1},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \, \prod _{k=1}^d \frac{1}{L_k} \right. \nonumber \\&\left. \qquad - \left( \frac{2}{\pi } \right) ^d \int _{\frac{\pi \, {{\hat{m}}}}{2L_1}}^{\infty } \cdots \int _{\frac{\pi \, {{\hat{m}}}}{2L_d}}^{\infty } S( \omega _1, \ldots , \omega _d) \,{\mathrm {d}}\omega _1 \cdots \,{\mathrm {d}}\omega _d \right| \nonumber \\&\quad \le \sum _{k=1}^d \frac{D_{2,k}}{L_k}. \end{aligned}$$
(134)
Applying the triangle inequality again gives
$$\begin{aligned}&\left| \sum _{j_1,\ldots ,j_d={{\hat{m}}}+1}^{\infty } S\left( \frac{\pi \, j_1}{2L_1},\ldots ,\frac{\pi \, j_d}{2L_d} \right) \, \prod _{k=1}^d \frac{1}{L_k} \right| \nonumber \\&\quad \le \sum _{k=1}^d \frac{D_{2,k}}{L_k} + \left( \frac{2}{\pi } \right) ^d \int _{\frac{\pi \, {{\hat{m}}}}{2L_1}}^{\infty } \nonumber \\&\qquad \cdots \int _{\frac{\pi \, {{\hat{m}}}}{2L_d}}^{\infty } S( \omega _1, \ldots , \omega _d) \,{\mathrm {d}}\omega _1 \cdots \,{\mathrm {d}}\omega _d. \end{aligned}$$
(135)
By interpreting the latter integral as being over the positive exterior of a rectangular hypercuboid and bounding it by a integral over exterior of a hypersphere which fits inside the cuboid, we can bound the expression by
$$\begin{aligned} \sum _{k=1}^d \frac{D_{2,k}}{L_k} + \frac{1}{\pi ^d} \int _{||{\omega }|| \ge \frac{\pi \, {{\hat{m}}}}{2L}} S({\omega }) \,{\mathrm {d}}{\omega }. \end{aligned}$$
(136)
The first term can be further bounded by replacing \(L_k\)s with their minimum L and by defining \(D_2 = \max D_{2,k}\) which is d times the maximum of \(D_{2,k}\). This leads to the final form of the result. \(\square \)
Remark 20
Note that analogous to Remark 17 we could tighten the bound for \(D_2\) by letting it depend on \({\hat{m}}\).
Proof of Theorem 4
Analogous to the one-dimensional case, we combine the results of the above lemmas using the triangle inequality. \(\square \)