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With the following Theorem we will have an explicit desrctiption for the
limes of RN as the amount of nodes is increased.
\begin{Theorem}[Ridge weight penaltiy corresponds to adapted spline]
\label{theo:main1}
For arbitrary training data \(\left(x_i^{train}, y_i^{train}\right)\) it holds
\[
\plimn \norm{\mathcal{RN^{*, \tilde{\lambda}}} - f^{*,
\tilde{\lambda}}_{g, \pm}}_{W^{1,\infty}(K)} = 0.
\]
With
\begin{align*}
\label{eq:1}
\tilde{\lambda} &\coloneqq \lambda n g(0), \\
g(x) &\coloneqq
g_{\xi}(x)\mathbb{E}\left[ v_k^2 \vert \xi_k = x \right], \forall x
\in \mathbb{R}
\end{align*}
and \(RN^{*, \tilde{\lambda}}\), \(f^{*,\tilde{\lambda}}_{g, \pm}\)
as defined in ??? and ??? respectively.
\end{Theorem}
In order to proof Theo~\ref{theo:main1} we need to proof a number of
auxilary Lemmata first.
\begin{Definition}[Sobolev Norm]
\label{def:sobonorm}
The natural norm of the sobolev space is given by
\[
\norm{f}_{W^{k,p}(K)} =
\begin{cases}
\left(\sum_{\abs{\alpha} \leq k}
\norm{f^{(\alpha)}}^p_{L^p}\right)^{\nicefrac{1}{p}},&
\text{for } 1 \leq p < \infty \\
max_{\abs{\alpha} \leq k}\left\{f^{(\alpha)}\right\},& \text{for
} p = \infty
\end{cases}
.
\]
\end{Definition}
\begin{Lemma}[Poincar\'e typed inequality]
Let \(f:\mathbb{R} \to \mathbb{R}\) differentiable with \(f' :
\mathbb{R} \to \mathbb{R}\) Lesbeque integrable. Then for \(K=[a,b]
\subset \mathbb{R}\) with \(f(a)=0\) it holds that
\begin{equation}
\label{eq:pti1}
\exists C_K^{\infty} \in \mathbb{R}_{>0} :
\norm{f}_{w^{1,\infty}(K)} \leq C_K^{\infty}
\norm{f'}_{L^{\infty}(K)}.
\end{equation}
If additionaly \(f'\) is differentiable with \(f'': \mathbb{R} \to
\mathbb{R}\) Lesbeque integrable then additionally
\begin{equation}
\label{eq:pti2}
\exists C_K^2 \in \mathbb{R}_{>0} : \norm{f}_{W^{1,\infty}(K)} \leq
C_K^2 \norm{f''}_{L^2(K)}.
\end{equation}
\proof
With the fundamental theorem of calculus, if
\(\norm{f}_{L^{\infty}(K)}<\infty\) we get
\begin{equation}
\label{eq:f_f'}
\norm{f}_{L^{\infty}(K)} = \sup_{x \in K}\abs{\int_a^x f'(s) ds} \leq
\sup_{x \in K}\abs{\int_a^x \sup_{y \in K} \abs{f'(y)} ds} \leq \abs{b-a}
\sup_{y \in K}\abs{f'(y)}.
\end{equation}
Using this we can bound \(\norm{f}_{w^{1,\infty}(K)}\) by
\[
\norm{f}_{w^{1,\infty}(K)} \stackrel{\text{Def~\ref{def:sobonorm}}}{=}
\max\left\{\norm{f}_{L^{\infty}(K)},
\norm{f'}_{L^{\infty}(K)}\right\}
\stackrel{(\ref{eq:f_f'})}{\leq} max\left\{\abs{b-a},
1\right\}\norm{f'}_{L^{\infty}(K)}.
\]
With \(C_k^{\infty} \coloneqq max\left\{\abs{b-a}, 1\right\}\) we
get (\ref{eq:pti1}).
By using the Hölder inequality, we can proof the second claim.
\begin{align*}
\norm{f'}_{L^{\infty}(K)} &= \sup_{x \in K} \abs{\int_a^bf''(y)
\mathds{1}_{[a,x]}(y)dy} \leq \sup_{x \in
K}\norm{f''\mathds{1}_{[a,x]}}_{L^1(K)}\\
&\hspace{-6pt} \stackrel{\text{Hölder}}{\leq} sup_{x
\in
K}\norm{f''}_{L^2(K)}\norm{\mathds{1}_{[a,x]}}_{L^2(K)}
= \abs{b-a}\norm{f''}_{L^2(K)}.
\end{align*}
Thus (\ref{eq:pti2}) follows with \(C_K^2 \coloneqq
\abs{b-a}C_K^{\infty}\).
\qed
\end{Lemma}