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Regularized Kernel-Based Reconstruction in Generalized Besov Spaces

Abstract

We present a theoretical framework for reproducing kernel-based reconstruction methods in certain generalized Besov spaces based on positive, essentially self-adjoint operators. An explicit representation of the reproducing kernel is given in terms of an infinite series. We provide stability estimates for the kernel, including inverse Bernstein-type estimates for kernel-based trial spaces, and we give condition estimates for the interpolation matrix. Then, a deterministic error analysis for regularized reconstruction schemes is presented by means of sampling inequalities. In particular, we provide error bounds for a regularized reconstruction scheme based on a numerically feasible approximation of the kernel. This allows us to derive explicit coupling relations between the series truncation, the regularization parameters and the data set.

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Notes

  1. 1.

    It follows from [10, Theorems 3.4 & 3.7] and (39) that such integral kernels exist for \({\sigma } > d/2\) with \(\varvec{K}^{({\sigma })} (x, \cdot ) \in L^p(M; d{\mu })\) for \(1 \le p \le \infty \) and \(\varvec{K}^{(\sigma )} (x, \cdot ) \in B_{2,2}^{\sigma }\) \((M; D)\).

  2. 2.

    For \(p\ne \infty \), the convergence rate is \(\delta ^{{\sigma -}d/r}\) instead of the anticipated rate \(\delta ^{{\sigma -}d(1/r-1/p)_+}\), where \((x)_{+}=\max \{x,0\}\) (cf. [61] for the case of classical Sobolev spaces). This is most likely due to the fact that we work with the global estimates (51) and (52) instead of local estimates on a cover (see also [41, 61]).

  3. 3.

    There are also lower bounds for the covering number, cf. [11, Theorem 5.21].

  4. 4.

    For practical consideration other parameter choices could be more useful. We do not give the details here, but leave those considerations to the reader since we work in a very general framework and hence do not have a model for the numerical costs for realizing \(\varepsilon _{\max }\). In many specific applications, an estimate for these costs is available and can be employed in an exhaustive cost–benefit discussion.

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Acknowledgements

We are grateful for the comments and suggestions of the anonymous referees. The authors acknowledge support of the Deutsche Forschungsgemeinschaft (DFG) through the Sonderforschungsbereich 1060: The Mathematics of Emergent Effects.

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Correspondence to Christian Rieger.

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Communicated by Pencho Petrushev.

Appendix

Appendix

In this appendix, we give an explicit bound on the constant \(\tilde{b}\) of Remark 2 for the Euclidean space \(\mathbb {R}^d\) with \(d\ge 2\), where we closely follow the lines of the proof of the statement in [10, Lemma 3.19]. We recall the general strategy first and then perform the necessary estimates in our setting. For measurable sets \(\varOmega \subset \mathbb {R}^d\), we set \(|\varOmega |:=\mu (\varOmega )\). In this case, (8) and (9) hold with \(\beta =d\), i.e.,

$$\begin{aligned}0<|B(x,2t)|=2^d|B(x,t)|\quad \text {for all }x\in \mathbb {R}^d, \quad \text {and all }t>0. \end{aligned}$$

Furthermore,

$$\begin{aligned}|B(x,\sqrt{t})|=\frac{(\pi t)^{d/2}}{\varGamma (\frac{d}{2}+1)},\quad \text {and }p_t(x,y)=\frac{1}{(4\pi t)^{d/2}}\exp \left( -\Vert x-y\Vert ^2/(4t)\right) .\end{aligned}$$

Consequently,

$$\begin{aligned}p_t(x,x)=(4\pi t)^{-d/2}=\frac{1}{2^d\varGamma (d/2+1)}|B(x,\sqrt{t})|^{-1}, \end{aligned}$$

and in particular, if \(\mathbf {1}\) denotes the characteristic function,

$$\begin{aligned} \mathbf {1}_{[0,\tau ]}(\sqrt{\mathcal {D}})(x,x)\le e \cdot p_{\tau ^{-2}}(x,x)=\frac{e}{2^d\varGamma (d/2+1)}|B(x,\tau ^{-1})|^{-1} \quad \text {for all }\tau >0.\nonumber \\ \end{aligned}$$
(118)

It is shown in [10, Lemma 3.19] that for \(\tau >0\) and \(r\in \mathbb {N}\), we can set \(\tau \sqrt{t}=2^{r}\) such that

$$\begin{aligned}\frac{2^{-rd}}{|B(x,\tau ^{-1})|}\left( c'-2^dc_4\sum _{k\ge r}\exp (-2^{2k})2^{kd}\right) \le \mathbf {1}_{[0,\tau ]}(x,x), \end{aligned}$$

holds, where the constants \(c_4:=\frac{e}{2^d\varGamma (d/2+1)}=:ec'\) can be obtained from (118). Hence, to make the lower bound positive, we need to choose \(r\in \mathbb {N}\) large enough such that

$$\begin{aligned} \exp (-1)>\sum _{k\ge r}\exp (-2^{2k})2^{kd}. \end{aligned}$$
(119)

Once we have an appropriate \(r\in \mathbb {N}\) at hand, we follow the argument in [10] and set (see [10, (3.44)])

$$\begin{aligned}c_3:=\frac{2^{-r}}{2^d\varGamma (d/2+1)}\left( 1-\sum _{k\ge r}\exp (-2^{2k})2^{(k+1)d}\right) >0, \end{aligned}$$

and choose a \(\ell >0\) large enough such that

$$\begin{aligned} 0<c_32^{d\ell }-c_4. \end{aligned}$$
(120)

Then, following the proof of [10, Lemma 3.19], we may set \(\tilde{b}:=2^{\ell }\).

Lemma 10

If we choose for \(d\ge 2\), \(r(d)\in \mathbb {N}\) as the smallest integer such that

$$\begin{aligned}r(d)\ge \frac{7}{2\ln (2)}\ln \left( \frac{d\ln (2)+1}{2\ln (2)}\right) , \end{aligned}$$

then (119) holds.

Proof

We determine \(r\in \mathbb {N}\) such that

$$\begin{aligned} \exp (-2^{2k})2^{(k+1)d}<e^{-k}\text { for all }k\ge r, \end{aligned}$$
(121)

since then

$$\begin{aligned} \sum _{k\ge r} \exp (-2^{2k})2^{kd}< & {} \sum _{k\ge r}\exp (-k)=\exp (-(r+1))/(1-e^{-1})\le 2\exp (-(r+1)) \nonumber \\ {}< & {} 2\exp (-2). \end{aligned}$$
(122)

To determine \(r:=r(d)\) such that (121) holds, we set

$$\begin{aligned}h_d(x):=-\exp (2x\ln (2))+d(x+1)\ln (2)+x. \end{aligned}$$

Then, \(h_d'(x)=-2\ln (2)\exp (2x\ln (2))+d\ln (2)+1\), and, since \(h_d(x)\rightarrow -\infty \) as \(x \rightarrow \pm \infty \), \(h_d\) has a unique global maximum at

$$\begin{aligned} \tilde{x}:=\frac{1}{2\ln (2)}\ln \left( \frac{d\ln (2)+1}{2\ln (2)}\right) . \end{aligned}$$

Note that \(h_d(\tilde{x})>0\). Therefore, we look for \(r\ge \tilde{x}\) such that \(h_d(r)<0\), and then (121) follows. We make the ansatz \(r=:s\tilde{x}\) with \(s\ge 1\) and use the abbreviation \(a_d:=d\ln (2)\) and \(b:=2\ln (2)\). Then,

$$\begin{aligned} h_d(s\tilde{x})= & {} -\left( \frac{a_d+1}{b}\right) ^s+\frac{s}{b}\ln \left( \frac{a_d+1}{b}\right) (a_d+1)+a_d. \end{aligned}$$
(123)

If \(d=2,\dots ,10\), it suffices to choose \(s=6\). If \(d\ge 10\), we set

$$\begin{aligned} \tilde{h}_d(s):= -\left( \frac{a_d+1}{b}\right) ^s+\frac{(a_d+1)}{b}\left( s\ln \left( \frac{a_d+1}{b}\right) +b\right) \ge h_d(s\tilde{x}). \end{aligned}$$

Now we set

$$\begin{aligned} s(d):=\max \left\{ 2,\ \frac{1+2\ln (d/2)}{\ln (d/2)-1}\right\} , \end{aligned}$$

and estimate very roughly as follows: Since \(\ln (\frac{a_d+1}{b})\le \frac{a_d+1}{b}\) and \(b\le 2\frac{a_d+1}{b}\), we have

$$\begin{aligned} \tilde{h}(s)\le \left( \frac{a_d+1}{b}\right) ^2\left[ -\left( \frac{a_d+1}{b}\right) ^{s-2}+s+2\right] <0, \end{aligned}$$

since by the choice of s,

$$\begin{aligned} \ln \left( \frac{a_d+1}{b}\right) \ge \ln (d/2)\ge \frac{s+1}{s-2}\ge \frac{\ln (s+2)}{s-2}. \end{aligned}$$

Note that for \(d\ge 10\), s(d) is monotonically decreasing. Therefore, \(s(d)\le s(10)\le 7\), and with \(r(d)\ge 7\tilde{x}\) the assertion follows. \(\square \)

We now turn to (121) and use r as obtained in Lemma 10. By (122), it suffices to choose \(\ell >0\) such that

$$\begin{aligned} 2^{d\ell -r(d)}(1-\frac{2}{e})>e, \end{aligned}$$

which holds for

$$\begin{aligned} \ell \ge \frac{1}{d}\left[ \frac{3-\ln (e-2)}{\ln (2)}+r(d)\right] . \end{aligned}$$

In particular, \(\ell \rightarrow 0\) as \(d\rightarrow \infty \), and thus \(\tilde{b}\rightarrow 1\).

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Griebel, M., Rieger, C. & Zwicknagl, B. Regularized Kernel-Based Reconstruction in Generalized Besov Spaces. Found Comput Math 18, 459–508 (2018). https://doi.org/10.1007/s10208-017-9346-z

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Keywords

  • Reproducing kernels
  • A priori error analysis
  • Generalized Besov spaces
  • Feasible reconstruction schemes
  • Spline smoothing

Mathematics Subject Classification

  • 41A17
  • 41A25
  • 41A58
  • 42A82
  • 62G08