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.

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 $\tilde{b}\rightarrow 1$.