Efficient function approximation on general bounded domains using splines on a Cartesian grid

Abstract

Functions on a bounded domain in scientific computing are often approximated using piecewise polynomial approximations on meshes that adapt to the shape of the geometry. We study the problem of function approximation using splines on a regular but oversampled grid that is defined on a bounding box. This approach allows the use of high-order and highly structured splines as a basis for piecewise polynomials. The methodology is analogous to that of Fourier extensions, using Fourier series on a bounding box, which leads to spectral accuracy for smooth functions. However, Fourier extension approximations involve solving a highly ill-conditioned linear system, and this is an expensive step. The computational complexity of recent algorithms is $\mathcal {O}\left (N\log ^{2}\left (N\right )\right )$ in 1-D and $\mathcal {O}\left (N^{2}\log ^{2}\left (N\right )\right )$ in 2-D. We show that, compared to Fourier extension, the compact support of B-splines enables improved complexity for multivariate approximations, namely $\mathcal {O}(N)$ in 1-D, $\mathcal {O}\left (N^{3/2}\right )$ in 2-D and more generally $\mathcal {O}\left (N^{3(d-1)/d}\right )$ in d-D with d > 1. By using a direct sparse QR solver for a related linear system, we also observe that the computational complexity can be nearly linear in practice. This comes at the cost of achieving only algebraic rates of convergence. Our statements are corroborated with numerical experiments and Julia code is available.

Appendices

Appendix A. Shift-invariant B-spline spaces

1.1 A.1 Generators of spline spaces

We refer the reader to [6, 7] for an extensive description, in the setting of signal and image processing, of both periodic and non-periodic splines.

The function ϕ(t) is called a generator for S if every f(t) ∈ S can be written as

$$ f(t)=\sum\limits_{k\in\mathbb{Z}}c_{k}\phi\left( t-k\right). $$

There exist multiple generators for one space S. The B-spline β(t) is just one of them. Below we will introduce another which is biorthogonal to B-splines translated over integer shifts.

The 2π-periodic function

$$ u(\omega) \triangleq \sum\limits_{k\in\mathbb{Z}}e^{-i\omega k}\beta(k) $$

is called the characteristic function of the space S [7, Definition (3.2)]. Because of B-spline symmetry and compact support, u is a cosine polynomial. Examples are

$$ u^{2}(\omega) = 1,\quad u^{4}(\omega)=\frac{1 + 2\cos^{2}(\omega/2)}{3},\quad u^{5}(\omega)=\frac{5 + 18\cos^{2}(\omega/2) + \cos^{4}(\omega/2)}{24}. $$

For our B-spline space, the characteristic function is strictly positive.

Another quantity that will be useful is the symbol. The symbol of a function f is defined as

$$ \xi(\omega)=\sum\limits_{k\in\mathbb{Z}}e^{-i\omega k}c(k) $$

(31)

where c(k) are the coefficients of f in the spline basis:

$$ f(t)=\sum\limits_{k\in\mathbb{Z}}c(k)\beta(t-k). $$

The symbol of β(t) is 1.

1.2 A.2 Exponential decay of dual B-splines

We introduce the relation between the symbols of B-splines and their duals, and the characteristic function:

Lemma 1 ([7, Proposition 4.7])

Each generator ϕ(t) of S has its dual counterpart $\tilde \phi (t)$ such that the biorthogonal relations

$$ {\int}_{-\infty}^{\infty} \phi(t-k)\overline{\tilde\phi(t-l)}dt = \delta_{k,l} $$

(32)

hold. The symbols τ(ω) and $\tilde \tau (\omega )$ of the generators ϕ(t) and $\tilde \phi (t)$, respectively, are linked as

$$ \tau(\omega)\overline{\tilde \tau(\omega)} = \frac{1}{u^{2p}(\omega)}, $$

(33)

where u^2p(ω) is the characteristic function of the space S^2p.

The following proof of Theorem 1 is not stated as a theorem in [7], but it follows from the reasoning on page 60 which we repeat here in order to extend the reasonings further on.

Proof of Theorem 1

By Lemma 1 the symbol τ(ω) of $\tilde \beta (t)$ is $\frac {1}{u^{2p}(\omega )}$. Note that its definition is (31)

$$ \tilde\tau(\omega)=\frac{1}{u^{2p}(\omega)} = \sum\limits_{k\in\mathbb{Z}}c(k)e^{-i\omega k}, $$

which shows that c(k) are the Fourier coefficients of $\tilde \tau (\omega )$. The characteristic function u^2p(ω) is a 2π periodic cosine polynomial without zeros on [0, 2π] [31, Lemma 6]. Therefore, $\tilde \tau (\omega )$ is analytic and periodic; and its Fourier coefficients decay exponentially as |k| grows. □

Appendix B. Dual B-splines with respect to discrete sampling

Since we are interested in discrete methods, we also study discrete sequences associated with sampled B-splines on a regular grid. Stability, fast algorithms and applications in this discrete setting were treated in [5, 34,35,36]. These sampled B-splines are not to be confused with another set of discrete B-splines defined by discrete convolutions of rectangular pulses [7]. We consider duality with respect to a discrete inner product involving the sample points. For these dual B-splines we again show exponential decay in Theorem 6.

1.1 B.1 Notation

In this section we adopt the notation of [34]. The centred and shifted sampled B-spline sequences are defined by sampling the centred continuous B-spline with an integer oversampling factor q:

$$ b_{q}(k)=\beta\left( k/q\right),\quad k\in\mathbb{Z}. $$

Sampling the continuous B-splines translated over integer shifts results in translates of the B-spline sequences by corresponding multiples of q,

$$ \beta\left( \tfrac kq-l\right)=b_{q}(k-lq), \quad \forall k,l\in \mathbb{Z}. $$

(34)

Like the B-splines, these discrete sequences have compact support. For p > 0, we find from (1) that their discrete support is

$$ \mathrm{supp} b_{q} = [-Q,Q],\qquad Q = \left\lceil q\frac{p+1}{2}-1\right\rceil. $$

(35)

For p = 0 the support is $\left \lceil -\tfrac {q}2 ,\tfrac {q-1}2 \right \rfloor$ which is not symmetric for even q.

In this text we use the convention that discrete convolution, defined by

$$ \left( a\star b\right)(k)=\sum\limits_{l=-\infty}^{\infty} a(k-l)b(l), $$

takes precedence over evaluation in order to avoid a multitude of brackets and to ease notation. Thus $\left (a\star b\right )(k)=a\star b(k)$. If we introduce the shift operator δ_i(k) such that δ_i ⋆ a(k) = a(k − i) we can write (34) as

$$ \beta\left( \tfrac kq-l\right) = \delta_{q l}\star b_{q}(k), \quad k,l\in\mathbb{Z}. $$

Next, we define upsampling by a factor of q as

$$ [a]_{\uparrow q}(k) = \left\{\begin{array}{ll} a(k^{\prime})& k=q k^{\prime}\\0&\text{otherwise,} \end{array}\right. $$

and downsampling by the same factor as

$$ [a]_{\downarrow q}(k) = a(q k). $$

Analogously to the continuous case above, we can define a discrete shift-invariant space

$$ S_{q}=\text{span}\{{\varPhi}_{q}\}, \quad \text{with} \quad {\varPhi}_{q}=\{\delta_{q k}\star b_{q}\}_{k\in\mathbb{Z}}. $$

We call the sequence g a generator for S_q if every f ∈ S_q can be written as

$$ f(k) = \sum\limits_{l\in\mathbb{Z}}c(l)g(k-q l) $$

which in signal processing notation simplifies to

$$ f = [c]_{\uparrow q}\star g. $$

These spaces are invariant with respect to shifts by integer multiples of q.

1.2 B.2 Discrete duality of sampled splines

We are again interested in a dual generator; but this time based on a discrete inner product on the sequence space S_q,

$$ \langle a, b \rangle = \sum\limits_{k\in\mathbb{Z}} a(k) \overline{b(k)}. $$

(36)

We look for a dual generating sequence $\tilde b_{q}\in S_{q}$ that satisfies

$$ \sum\limits_{i=-\infty}^{\infty} \tilde b_{q}(i - q k) b_{q}(i - q l) =\langle \delta_{q k} \star \tilde b_{q}, \delta_{q l}\star b_{q}\rangle = \delta_{k,l}, ~~~ k,l\in\mathbb{Z},{~~~ \tilde b_{q}\in S_{q}}. $$

(37)

Alternatively, we can describe the duality with respect to an inner product defined on the function space S^q. Since $\tilde b_{q} \in S_{q}$, the sequence corresponds to the samples of a continuous function $\tilde \beta _{q} \in S^{q}$ that has a representation in the basis Φ:

$$ \tilde \beta_{q}(t) \triangleq \sum\limits_{k\in\mathbb{Z}}c_{q}(k)\beta(t-k). $$

(38)

We obtain functions in S^q that are biorthogonal in an oversampled equidistant grid:

$$ \langle\tilde \beta_{q}(\cdot-k), \beta_{q}(\cdot-l)\rangle_{q} = \delta_{k,l}, $$

where we have used a discrete inner product¹

$$ \langle f, g\rangle_{q}=\sum\limits_{k\in\mathbb{Z}}f\left( \tfrac kq\right)\overline{g\left( \tfrac kq\right)}. $$

(39)

The q-shift biorthogonality property results in a reconstruction formula that is similar to the one expressed by (8) in the continuous case. We have that

$$ f(t) = {\sum}_{k \in \mathbb{Z}} \langle f, \tilde{\beta}_{q}\rangle_{q} \phi(t), \qquad f \in S^{p}. $$

For this dual generator, the discretization of (38) yields the expression

$$ \tilde b_{q} = [c_{q}]_{\uparrow q} \star b_{q}. $$

It turns out that the dual generator in S_q is unique. We describe the solution in Sections B.3 and B.4. We will consider other solutions of (37) later on in Section B.5, by relaxing the requirement that the dual sequence lies in S_q.

1.3 B.3 Discrete least squares approximations

In the interpolation problem for a function f on the real line one wants to determine coefficients y(k) such that

$$ f(k) = \sum\limits_{l=-\infty}^{\infty} y(l)\beta(k-l),\quad k\in\mathbb{Z}. $$

This problem is better known as the cardinal B-spline interpolation problem and it is extensively investigated in [31]. The interpolation problem is uniquely solvable.

We are interested in the solution of the more general, oversampled problem

$$ f(k/q) = \sum\limits_{l=-\infty}^{\infty} y(l)\beta(k/q-l),\quad k\in\mathbb{Z} $$

for which a solution not necessarily exists with equality to the samples of f. Thus, we solve the problem in a least squares sense:

$$y = \mathrm{argmin}_{a}\sum\limits_{k=-\infty}^{\infty}|f(k/q)-\sum\limits_{l=-\infty}^{\infty} a(l)\beta(k/q-l)|^{2}. $$

(40)

This problem was investigated in [35].

Written in a convolutional form we want to solve

$$ f_{q}(k) = b_{q}\star [y]_{\uparrow q}(k), $$

(41)

where f_q(k) = f(k/q), in a least squares sense. In section IV.D of [35] the solution y(k) of the least squares problem (41) is written in terms of filters. We restate equations [35, (4.18)–(4.19)]:

$$ y(k) = s_{q}\star [b_{q}\star f_{q}]_{\downarrow q}(k), $$

(42)

where

$$ s_{q}(k) = \left( [b_{q}\star b_{q}]_{\downarrow q}\right)^{-1}(k), $$

(43)

if the inverse of [b_q ⋆ b_q]_↓q exists. The meaning of the inverse in this context is that the sequence s_q satisfies

$$ s_{q} \star [b_{q}\star b_{q}]_{\downarrow q}=\delta_{0}. $$

(44)

The solution can be found by taking Fourier transforms of both sides, writing the convolution as a product of terms, and then moving one factor to the other side by division. The Fourier transform of s_q exists if the Fourier transform of [b_q ⋆ b_q]_↓q does not vanish, else the division introduces poles. The latter Fourier transform is in fact strictly positive, as will be shown in the proof of Theorem 6 later on.

To that end, we define the Fourier transform of a sequence f as $F(\omega ) = \hat F(e^{i\omega })$, where $\hat F(z)$ is the z-transform of f:

$$ \begin{array}{@{}rcl@{}} \hat F(z) = \sum\limits_{k\in\mathbb{Z}}f(k)z^{-k}. \end{array} $$

1.4 B.4 Exponential decay of discrete dual B-splines

We study the dual generators in S_q which are q-shift biorthogonal to b_q in more detail. First, we show that there is a generator that satisfies the conditions (37). The coefficients c(k) of this generator $\tilde b_{q}$ in Φ_q are obtained from the filter defined by (43).

Theorem 1

Provided that (44) is uniquely solvable for s_q, the generator $\tilde b_{q}(k)\in S_{q}$ that is q-shift biorthogonal to b_q (with q-shift biorthogonality defined by (37)) is given by

$$ \tilde b_{q}(k)=[s_{q}]_{\uparrow q}\star b_{q}(k). $$

(45)

Furthermore, the discrete B-spline least squares problem (40) is solved by applying $\tilde b_{q}(k)$ to f_q:

$$ y(k) = [\tilde b_{q}\star f_{q}]_{\downarrow q}(k). $$

Proof

First we show that a sequence in S_q of the form (45) with coefficients equal to s_q is q-shift biorthogonal to b_q. With the notation $a^{\prime }(k) = a(-k)$ for time-reversal and $\overline {a}(k)=\overline {a(k)}$ for complex conjugation of a sequence, we can write the inner product (36) as

$$ \langle a, b\rangle = {\sum}_{k=-\infty}^{\infty} a(k)\overline{b(k)} = a^{\prime}\star \overline{b}(0)=a\star \overline{b}^{\prime}(0). $$

Since b_q(k) = b_q(−k) and b_q(k) is real-valued, we can write for arbitrary g = [d]_↑q ⋆ b_q ∈ S_q

$$\langle \delta_{q k} \star g, \delta_{q l}\star b_{q}\rangle = \delta_{q(k-l)} \star g \star \left( b_{q}\right)'(0)= g \star b_{q}(q(k-l))=\left[[d]_{\uparrow q}\star b_{q}\star b_{q}\right]_{\downarrow q}(k - l). $$

If we use

$$ [[a]_{\uparrow q}\star b]_{\downarrow q}(k) = {\sum}_{i\in\mathbb{Z}}b(q k-q i)a(i) = {\sum}_{l\in\mathbb{Z}}b(q l)a(k-l)=a\star[b]_{\downarrow q}(k) $$

(46)

then we can simplify the expression: we find coefficients of a dual generator by solving

$$ d \star [b_{q}\star b_{q}]_{\downarrow q}=\delta_{0} $$

for d(k). This is exactly the definition of s_q given in (43).

The second equality of the theorem is verified using the definition of s_q, (46) and (42):

$$ [\tilde b_{q}\star f]_{\downarrow q}(k) = [[s_{q}]_{\uparrow q}\star b_{q}\star f]_{\downarrow q}(k) = s_{q}\star [b_{q}\star f]_{\downarrow q}(k)=y(k). $$

□

Knowing the form of the dual generator, we are ready to show that its coefficients s_q decay exponentially. To that end, we study its Fourier transform and those of b_q and

$$ c_{q}(k)=\beta\left( k/q+\tfrac12\right). $$

In the following lemma we first consider the Fourier transforms of b₁ and c₁. The lemma generalizes a result of [5, Prop. 1] and is needed for the proof of the main theorem, Theorem 6, further on.

Lemma 2

Let

$$ \begin{array}{@{}rcl@{}} B_{1}(\omega) &=& \sum\limits_{k\in\mathbb{Z}}b_{1}(k)e^{-i\omega k} = {\sum}_{k=-\infty}^{\infty} \left( \frac{\sin(\tfrac{\omega-2\pi k}2)}{\tfrac{\omega-2\pi k}2}\right)^{p+1}, \end{array} $$

(47)

$$ \begin{array}{@{}rcl@{}} C_{1}(\omega) &=& \sum\limits_{k\in\mathbb{Z}}c_{1}(k)e^{-i\omega k}= {\sum}_{k=-\infty}^{\infty} e^{i\tfrac{\omega-2\pi k}2}\left( \frac{\sin(\tfrac{\omega-2\pi k}2)}{\tfrac{\omega-2\pi k}2}\right)^{p+1}, \end{array} $$

(48)

be the Fourier transforms of b₁(k) and c₁(k) for p ≥ 0 and p > 0, respectively.

Then B₁(ω) and $D(\omega )=e^{-i\tfrac \omega 2}C_{1}(\omega )$ are strictly decreasing in [0,π]. Moreover, B₁(π) > 0 for p ≥ 0 and D(π) = 0 for p > 0.

Proof

Earlier results show that B₁(ω) is strictly positive and decreasing in [0,π]. See, e.g., [31, Lemma 6] and [5, Prop. 1]. This only leaves us to show that $D(\omega )=e^{-i\tfrac \omega 2}C_{1}(\omega )$ is strictly decreasing in [0,π] and has a zero at ω = π for p > 0.

The continuous Fourier transform of β(t) is

$$ B(\omega) = \left( \frac{\sin(\tfrac{\omega}2)}{\tfrac{\omega}2}\right)^{p+1} $$

because of the recursive convolution. The sequences b₁(k) and c₁(k) are the sampled (and for c₁(k), shifted over $\tfrac {1}{2}$) versions of β(t). Therefore, their Fourier transforms can be written as stated in the theorem.

Noting that

$$ \sin\left( \tfrac{\omega-2\pi k}{2}\right)=(-1)^{k}\sin\left( \tfrac{\omega}{2}\right) $$

we simplify D(ω) to

$$ D(\omega)= \left( 2\sin(\tfrac\omega 2)\right)^{p+1}\sum\limits_{k=-\infty}^{\infty} (-1)^{k}(-1)^{k(p+1)}(\omega-2\pi k)^{-(p+1)}. $$

If p is even, D(π) takes the following form:

$$ D(\pi) = \left( \frac{2}{\pi}\right)^{p+1}\sum\limits_{k=-\infty}^{\infty} \frac{1}{(1-2 k)^{p+1}}, $$

which is zero since the summation evaluates to zero:

$$ \begin{array}{@{}rcl@{}} \sum\limits_{k=-\infty}^{-1} \frac{1}{(1-2 k)^{p+1}} &=& \sum\limits_{k=1}^{\infty} \frac{1}{(1+2 k)^{p+1}}\\ &=& \sum\limits_{k=0}^{\infty} \frac{1}{(-1+2 k)^{p+1}} = \sum\limits_{k=0}^{\infty} \frac{(-1)^{p+1}}{(1-2 k)^{p+1}} = -{\sum}_{k=0}^{\infty} \frac{1}{(1-2 k)^{p+1}}. \end{array} $$

For odd p the reasoning is analogous. Therefore, D(ω) vanishes at ω = π for all p > 0.

Next, we expand $\frac {\partial D}{\partial \omega }(\omega )$:

$$ \begin{array}{@{}rcl@{}} \frac{\partial D}{\partial \omega}(\omega) &= (p+1)\sum\limits_{k=-\infty}^{\infty}(-1)^{k} \left( \frac{\sin(\tfrac{\omega-2\pi k}2)}{\tfrac{\omega-2\pi k}2}\right)^{p}\frac{\tfrac12\cos(\tfrac{\omega}{2})(-1)^{k}\tfrac{\omega-2\pi k}{2}-\tfrac12\sin(\tfrac{\omega}{2})(-1)^{k}}{\left( \tfrac{\omega-2\pi k}2\right)^{2}}. \end{array} $$

The expression vanishes at ω = 0. For ω > 0, we split the sum into two parts:

$$ \begin{array}{@{}rcl@{}} \frac2{p+1}\frac{\partial D}{\partial \omega}(\omega) &=& (\cos(\tfrac{\omega}{2})\tfrac{\omega}{2}-\sin(\tfrac{\omega}{2}))\sum\limits_{k=-\infty}^{\infty} \left( \frac{\sin(\tfrac{\omega-2\pi k}2)}{\tfrac{\omega-2\pi k}2}\right)^{p}\frac{1}{\left( \tfrac{\omega-2\pi k}2\right)^{2}} \\ &&-\pi {\sum}_{k=-\infty}^{\infty} \left( \frac{\sin(\tfrac{\omega-2\pi k}2)}{\tfrac{\omega-2\pi k}2}\right)^{p}\frac{k}{\left( \tfrac{\omega-2\pi k}2\right)^{2}} \end{array} $$

(49)

The factor in front of the first summation, $f(\omega )=\cos \limits (\tfrac {\omega }{2})\tfrac {\omega }{2}-\sin \limits (\tfrac {\omega }{2})$, is strictly negative for ω ∈ (0,π] since its derivative $f^{\prime }(\omega )=-\tfrac 14\sin \limits (\tfrac \omega 2)$ is strictly negative in (0,π] and f(0) = 0. Since both prefactors are negative, if both sums are shown to be positive we can conclude that $\frac {\partial D}{\partial \omega }(\omega )$ is negative, hence D(ω) is decreasing. That would conclude the proof.

We consider the range ω ∈ (0,π]. The first sum in (49),

$$ \begin{array}{@{}rcl@{}} S_{1}(\omega) = \sum\limits_{k=-\infty}^{\infty} \left( \frac{\sin(\tfrac{\omega-2\pi k}2)}{\tfrac{\omega-2\pi k}2}\right)^{p}\frac{1}{\left( \tfrac{\omega-2\pi k}2\right)^{2}}, \end{array} $$

has only positive terms when p is even. For p odd, we write $S_{1}(\omega ) = {\sum }_{k=-\infty }^{\infty } (-1)^{k}{f^{p}_{k}}(\omega )$ with

$$ \begin{array}{@{}rcl@{}} {f^{p}_{k}}(\omega)= \left( \frac{\sin(\tfrac{\omega}2)}{\tfrac{\omega-2\pi k}2}\right)^{p}\frac{1}{\left( \tfrac{\omega-2\pi k}2\right)^{2}}. \end{array} $$

Because of the term $\frac 12(\omega - 2\pi k)$ in the denominator, we find that ${f^{p}_{k}}(\omega )>0$ for k ≤ 0 and ${f^{p}_{k}}(\omega )<0$ for k > 0. Furthermore, due to the growth of the denominators of both factors for increasing |k|, it is true that

$$ |f^{p}_{k-1}|<|{f^{p}_{k}}|, \quad\text{for }k\leq0,\qquad |f^{p}_{k+1}|<|{f^{p}_{k}}|, \quad\text{for }k>0. $$

We can then group terms in pairs with alternating signs. For k > 0:

$$ \begin{array}{@{}rcl@{}} &&(-1)^{2k-1}f^{p}_{2k-1}(\omega) + (-1)^{2k}f^{p}_{2k}(\omega)\\ &=& -f^{p}_{2k-1}(\omega) +f^{p}_{2k}(\omega) = |f^{p}_{2k-1}(\omega)| -|f^{p}_{2k}(\omega)| > 0, \end{array} $$

and for k ≤ 0:

$$ \begin{array}{@{}rcl@{}} &&(-1)^{2k-1}f^{p}_{2k-1}(\omega) + (-1)^{2k}f^{p}_{2k}(\omega)\\ &=& -f^{p}_{2k-1}(\omega) +f^{p}_{2k}(\omega) = -|f^{p}_{2k-1}(\omega)| +|f^{p}_{2k}(\omega)| > 0. \end{array} $$

This shows that

$$ S_{1}(\omega) = \sum\limits_{k=-\infty}^{\infty}(-1)^{2k-1}f^{p}_{2k-1}(\omega) + (-1)^{2k}f^{p}_{2k}(\omega) > 0 $$

is also positive for odd p.

For the second sum in (49), we first note that |ω − 2kπ| < |ω + 2kπ| for ω ∈ (0,π], hence

$$ |{f^{p}_{k}}(\omega)| > |f^{p}_{-k}(\omega)|, \quad \text{for }k > 0. $$

(50)

We also set out to show that

$$ (k+1)f^{p}_{k+1}(\omega)-k{f^{p}_{k}}(\omega)\!>\!0\!\quad\text{for } k\!>\!0,\!\quad k{f^{p}_{k}}(\omega) - (k+1)f^{p}_{k+1}(\omega)\!>\!0\quad\text{for } k\!<\!-1. $$

(51)

To that end, we start by rewriting

$$ k{f^{p}_{k}}(\omega) - (k+1)f^{p}_{k+1}(\omega) = \frac{k\left( 1+\frac{1}{ k-\tfrac{\omega}{2\pi}}\right)^{p+2}-(k+1)}{(\omega-2\pi k-2\pi)^{p+2}}. $$

Here, the denominator is negative for k > 0 and positive for k < − 1. For the numerator, we note that

$$ \left( 1+\frac{1}{k-f}\right)^{2}-\frac{k+1}{k} = -\frac{f^{2}-k(k+1)}{k(k-f)^{2}} $$

which is positive for k > 0 and negative for k < − 1, with f = ω/(2π) ∈ (0, 1/2]. Thus

$$ \left( 1+\frac{1}{k-f}\right)^{p+2}-\frac{k+1}{k} > \left( 1+\frac{1}{k-f}\right)^{2}-\frac{k+1}{k} > 0 $$

for k > 0 and

$$ \left( 1+\frac{1}{k-f}\right)^{p+2}-\frac{k+1}{k} < \left( 1+\frac{1}{k-f}\right)^{2}-\frac{k+1}{k} < 0 $$

for k < − 1. Multiplying with k (and changing the direction of the inequality for negative k) shows that the numerator is positive and leads to the inequalities in (51).

Returning to the second sum in (49), it is positive for p even since

$$ {S^{p}_{2}}(\omega) = {\sum}_{k=-\infty}^{\infty} k{f^{p}_{k}}(\omega) ={\sum}_{k=-\infty}^{\infty} k|{f^{p}_{k}}(\omega)| ={\sum}_{k=1}^{\infty} k\left( |{f^{p}_{k}}(\omega)|-|f^{p}_{-k}(\omega)|\right) >0, $$

owing to (50). The sum is positive for p odd since

$$ \begin{array}{@{}rcl@{}} {S^{p}_{2}}(\omega) &=& {\sum}_{k=-\infty}^{\infty} k{f^{p}_{k}}(\omega) \\ &=&{\sum}_{k=1}^{\infty} \left( -\left( 2k-1\right)f^{p}_{2k-1}(\omega) + (2k)f^{p}_{2k-1}(\omega)\right) \\&&+ \left( (-2k)f^{p}_{-2k}(\omega) - \left( -2k+1\right)f^{p}_{-2k+1}(\omega)\right) \\ &>0 \end{array} $$

because of (51). This ends the proof. □

Previous work already proved the exponential decay of the dual generator coefficients of some cases. Earlier results include the dual taken with respect to the inner product (f,g) defined in (6) see Theorem 1; and, the dual with respect to integer grid, i.e., the discrete inner product 〈f,g〉₁ of which the definition is given in (39), see [31]. The exponential decay of the discrete dual generator coefficients in the oversampled case, 〈f,g〉_q with q > 1, has, to the best of our knowledge, not been described in literature.

Theorem 2

For p > 0, (44) is uniquely solvable for s_q and the coefficients s_q(k) decay exponentially as $|k|\to \infty$.

Proof Proof (Proof of 6)

The Fourier transform of (43) follows from taking the Fourier transform of both sides in (44). Noting that the latter involves the convolution of a sequence with itself, which results in squaring the corresponding Fourier transform. This is followed by downsampling by a factor of q, leading to [35, eqn (4.20)], which we repeat here:

$$ S_{q}(\omega) = \left( \frac1q\sum\limits_{k=0}^{q-1} B_{q}\left( \frac{\omega+2\pi k}{q}\right)^{2}\right)^{-1}. $$

By Theorem 5 and since the Fourier coefficients of periodic functions analytic on the real line decay exponentially, the result holds if we can show that S_q(ω) is analytic on the interval [0, 2π] and by extension the real line, i.e., 1/S_q(ω) has no zeros on the unit circle.

To obtain an expression for B_q, we use known convolution expressions for the sampled B-splines [35]. They depend on the parity of p and q. For q odd:

$$ b_{q}(k)=\frac{1}{q}\underbrace{b^{0}_{q}\star b^{0}_{q}\star{\dots} \star b^{0}_{q}}_{p+1\text{ times}}\star b_{1}(k). $$

For p odd and q even:

$$ b_{q}(k)=\frac{1}{q}\delta_{(p+1)/2}\star\underbrace{b^{0}_{q}\star b^{0}_{q}\star{\dots} \star b^{0}_{q}}_{p+1\text{ times}}\star b_{1}(k). $$

Finally, for p even and q even:

$$ b_{q}(k)=\frac{1}{q}\delta_{(p+2)/2}\star\underbrace{b^{0}_{q}\star b^{0}_{q}\star{\dots} \star b^{0}_{q}}_{p+1\text{ times}}\star c_{1}(k). $$

The z-transform of the rectangular pulse is $\hat B^{0}_{q}(z)=z^{\lfloor q/2\rfloor }\left (\tfrac {1-z^{-q}}{1-z^{-1}}\right )$, where ⌊x⌋ truncates to the smaller integer. Therefore, we arrive at the following expressions for $B_{q}(\omega )=\hat B_{q}(e^{i\omega })$, which again depend on the parity of p and q. When either p or q is odd:

$$ B_{q}(\omega) = \frac{1}{q} T_{q}(\omega) B_{1}(\omega). $$

Here, $T_{q}(\omega ) = \hat T_{q}(e^{i\omega })$ with $\hat T_{q}(z) = z^{i_{0}}\left (\frac {1-z^{-q}}{1-z^{-1}}\right )^{p+1}$ for i₀ = (q − 1)(p + 1)/2 [35, eqn (3.10)]. This leads to

$$ T_{q}(\omega) = \left( \frac{\sin(q\tfrac\omega 2)}{\sin(\tfrac\omega 2)}\right)^{p+1}. $$

When both p and q are even:

$$ B_{q}(\omega) = \frac{e^{-i\omega/2}}{q}T_{q}(\omega)C_{1}(\omega). $$

The function T_q(ω) has zeros at $\omega = 2\pi \tfrac kq + 2\pi l$ for $k=1,\dots ,q-1, l\in \mathbb {Z}$, while Lemma 2 shows that B₁(ω) is strictly positive and C₁(ω) has zeros at ω = π + 2πl with $l\in \mathbb {Z}$. Noting that the zero of C₁ coincides with one of the zeros of T_q for p even, we see that B_q(ω) has the same zeros as T_q(ω).

Since

$$K_{q,k}(\omega)\triangleq B_{q}\left( \frac{\omega+2\pi k}{q}\right)^{2}$$

is a positive 2πq-periodic function with zeros at ω = 2πl, $l\in \mathbb {Z}$, with the exception of ω = − 2πkq + 2πql, $l\in \mathbb {Z}$, $K_{q,k_{1}}$ and $K_{q,k_{2}}$ don’t share zeros if k₁≠k₂. Therefore, $L(\omega )\triangleq \tfrac 1q{\sum }_{k=0}^{q-1}K_{q,k}(\omega )$ is 2π-periodic and strictly positive. And, its inverse, S_q(e^iω), meets the previously stated requirements. □

Theorem 3

Let p = 0 and q≠ 2. Then (44) is uniquely solvable for s_q and the coefficients s_q(k) decay exponentially as $|k|\to \infty$. More specifically, for q odd we have $s^{0}_{q}(k) = \tfrac 1q\delta _{0}(k)$. For q even, but q≠ 2, we have

$$ s^{0}_{q}(k) = \left\{ \begin{array}{ll} \frac{-1}{(1-q)^{k+1}} & k \leq 0, \\ 0 & k > 0. \end{array}\right. $$

Proof Proof of Theorem 7

If q is even, then $b^{0}_{q}(k)=1$ for − q/2 ≤ k ≤ q/2 − 1 and 0 otherwise, see the comment under (35). Hence, $b^{0}_{q}\star b^{0}_{q}(k)$ is non-zero for − q ≤ k ≤ q − 2, and $[b^{0}_{q}\star b^{0}_{q}]_{\downarrow q}(k)$ is non-zero only if − 1 ≤ k ≤ 0. To be precise, $[b^{0}_{q}\star b^{0}_{q}]_{\downarrow q}(0)=q-1$ and $[b^{0}_{q}\star b^{0}_{q}]_{\downarrow q}(-1)=1$, therefore

$$ (S^{0}_{q})^{-1}(\omega) = e^{-i\omega}+q-1. $$

The modulus is bounded above and below by $q-2\leq |(S^{0}_{q})^{-1}(\omega )|\leqq$ for ω ∈ [0, 2π]. Thus, $(S^{0}_{q})^{-1}(\omega )$ vanishes in [0, 2π] only if q = 2 and ω = π at the same time. This means that ${S^{0}_{2}}(\omega )$ has a pole at ω = π and (44) is not uniquely solvable for ${s^{0}_{2}}$.

Still for q even, but q≠ 2, using the series expansion $\frac {1}{1-z}={\sum }_{k=0}^{\infty } z^{k}$, $S^{0}_{q}$ can be written as

$$ \begin{array}{@{}rcl@{}} S^{0}_{q}(\omega) = \frac{1}{e^{-i\omega}+q-1} = -\sum\limits_{k=0}^{\infty}\frac{e^{-i\omega k}}{(1-q)^{k+1}}, \end{array} $$

so $|s^{0}_{q}(k)|$ decays exponentially for $k\rightarrow -\infty$ and q ≥ 4 and it is zero if k > 0.

If q is odd, then $b^{0}_{q}(k)=1$ for − (q − 1)/2 ≤ k ≤ (q − 1)/2. Hence, $b^{0}_{q}\star b^{0}_{q}(k)$ is non-zero for − q + 1 ≤ k ≤ q − 1, and $[b^{0}_{q}\star b^{0}_{q}]_{\downarrow q}(k)$ is non-zero if k = 0. To be precise, $[b^{0}_{q}\star b^{0}_{q}]_{\downarrow q}(0)=q$ and

$$ (S^{0}_{q})^{-1}(\omega) = q. $$

Thus, $s^{0}_{q}(k)=\tfrac 1q\delta _{0}(k)$. □

1.5 B.5 Compact discrete duals to B-splines

We end this section with a description of duals to the sampled B-splines that have compact support. The q-shift biorthogonality conditions (37) admit alternative solutions outside of the shift-invariant space S_q, in particular solutions with compact support. For these discrete solutions we are not able to provide a continuous analogue as in (38).

For clarity of the presentation we switch to a new notation for these duals. Thus, we are looking for a sequence $\tilde {h}_{q}$ that satisfies

$$ \sum\limits_{k=-\infty}^{\infty} \tilde h_{q}(k) b_{q}(k-q l) = \delta_{l}, \quad l\in\mathbb{Z}. $$

(52)

If this holds, then q-shifts of $\tilde {h}_{q}$ define a dual generating sequence for a discrete shift-invariant space

$$ \tilde{U}_{q} = \text{span} \{ \delta_{q k} \star \tilde{h}_{q}\}_{k \in \mathbb{Z}}. $$

This space may in general be different from S_q. We are encouraged by the observation that if both $\tilde h_{q}$ and b_q are compact sequences, (52) reduces to a finite number of conditions.

Proof Proof (Proof of Theorem 2)

Assume that w is a sequence with support [−K,K], i.e., that w(k) = 0 for |k| > K. Recall that the support of b_q for p > 0 is given by [−Q,Q] with $Q= \lceil q \frac {p+1}{2}-1 \rceil$, see (35). Substituting w into (52) yields, with $L = \left \lfloor \frac {K+Q}{q} \right \rfloor$,

$$ \sum\limits_{k=-K}^{K} w(k) b_{q}(k-q l) = \delta_{0}(l), \qquad -L \leq l \leq L, $$

(53)

where the range of l is restricted such that |k − ql| > Q for k ∈ [−K,K]. Indeed, if $l > \frac {K+Q}{q}$, then

$$ k- q l < k - (K+Q) < -Q + (k-K) < -Q. $$

The case $l < -\frac {K+Q}{q}$ similarly leads to k − ql > Q. The conditions (23) correspond to a linear system with $2L+1 = 2\left \lfloor \frac {K+Q}{q} \right \rfloor +1$ equations for 2K + 1 unknowns.

We are interested in choosing K such that there are more unknowns than conditions,

$$ 2K + 1 \geq 2\left \lfloor \frac{K+Q}{q} \right\rfloor+1. $$

This leads to $K \geq \left \lfloor \frac {K+Q}{q} \right \rfloor$ and hence $\frac {K+Q}{q} < K+1$, leading to

$$ K > \frac{Q-q}{q-1}. $$

Substituting $Q= \lceil q \frac {p+1}{2}-1 \rceil$ results in

$$ K > \frac{q(p-1)-2}{2(q-1)} \quad\text{and}\quad K > \left\lceil\frac{q(p-1)-2}{2(q-1)}\right\rceil $$

for odd and even p, respectively. This leads, after rearrangement, to the statements of the theorem.

The system matrix of (53) can also be written as

$$ A(l,k) = \beta\left( \tfrac{k}{q}-l\right),\qquad -L\leq l\leq L ,\quad -K\leq k \leq K. $$

This is precisely the collocation matrix of a sequence of 2K + 1 spline functions $\beta (\cdot -\tfrac {k}{q})$ evaluated in the 2L + 1 integers − L ≤ l ≤ L. The result follows if this system has full rank, i.e., if A has rank L (recall that L < K). We select a subset of L splines, in such a way that each spline can be associated uniquely with one integer in its support. This is always possible because both the integers and the spline centres are regularly distributed on the same interval by construction. The corresponding L × L submatrix of A represents a spline interpolation problem that is known to be uniquely solvable by [16, Theorem 1]. Therefore, at least one exact solution of the underdetermined linear system exists. □

Remark 1

Theorem 2 guarantees the existence of a dual sequence with small support, but does not state anything about its stability, in the sense of having a bound on a discrete norm of $\tilde {h}_{q}$. In practice we observe that, owing to the norm-minimization property of least squares solutions, compact duals with smaller discrete norm can be found by solving (53) in a least squares sense for a larger value of K than the minimal one.