\subsection{Proof of Lemma \ref{lem:relating norms}}
\label{sec:proof of lem relating norms}
Let $\psi_1, \ldots, \psi_n \in \mathcal{H}$ denote the KPCA directions which span $\mathcal{S}_\mathcal{X}$ such that $\langle\psi_i, \phi_j\rangle=(\varphi_j)_i\in \mathbb{R}$, where $(v)_i$ denotes the $i^{th}$ entry of vector $v$. Furthermore, let $\mathbf{B}_i$, denote the $ith$ row of a matrix $\mathbf{B}$. For any $L:\mathcal{H} \rightarrow \mathcal{H}$ and $\phi_x \in \mathcal{H}$ we may write $L \mathcal{P}_{\mathcal{S}_\mathcal{X}}\phi_x = \sum_{i=1}^n\sum_{j=1}^nw_{i,j}\Psi_i\otimes \Psi_j \phi_x$, where $\Psi_i\otimes \Psi_j \phi_x = \langle \Psi_j, \phi_x\rangle_\mathcal{H}\Psi_i$. Let $\Wb$ be the matrix of $w_{ij}$ weights. Lastly, let $a^Tb$ denote the standard Euclidean inner product for $a, b \in \mathbb{R}^n$. Then, for $\phi_x, \phi_y \in \mathcal{H}$,
\begin{eqnarray}
    \|L\mathcal{P}_{\mathcal{S}_\mathcal{X}}\phi_x-L\mathcal{P}_{\mathcal{S}_\mathcal{X}}\phi_y\|_\mathcal{H}^2 
    &=&\langle L\mathcal{P}_{\mathcal{S}_\mathcal{X}}\phi_x-L\mathcal{P}_{\mathcal{S}_\mathcal{X}}\phi_y, L\mathcal{P}_{\mathcal{S}_\mathcal{X}}\phi_x-L\mathcal{P}_{\mathcal{S}_\mathcal{X}}\phi_y\rangle \label{firstline Lem1}
    \\ &=& \left\langle\sum_{i=1}^n\sum_{j=1}^nw_{i,j}\Psi_i\otimes \Psi_j (\phi_x-\phi_y), \sum_{i=1}^n\sum_{j=1}^nw_{i,j}\Psi_i\otimes \Psi_j (\phi_x-\phi_y)\right\rangle_\mathcal{H} \nonumber
    \\ &=&\left\langle\sum_{i=1}^n\sum_{j=1}^nw_{i,j}\langle\Psi_j, \phi_x-\phi_y\rangle_\mathcal{H}\Psi_i, \sum_{i=1}^n\sum_{j=1}^nw_{i,j}\langle\Psi_j, \phi_x-\phi_y\rangle_\mathcal{H}\Psi_i\right \rangle_\mathcal{H} \nonumber
    \\ &=& \left\langle\sum_{i=1}^n\sum_{j=1}^nw_{i,j}\left((\varphi_\bx)_j-(\varphi_\by)_j\right)\Psi_i, \sum_{i=1}^n\sum_{j=1}^nw_{i,j}\left((\varphi_\bx)_j-(\varphi_\by)_j\right)\Psi_i \right\rangle_\mathcal{H} \nonumber
    \\ &=& \left\langle \sum_{i=1}^n \Wb^T_i(\varphi_\bx-\varphi_\by)\Psi_i, \sum_{i=1}^n \Wb^T_i(\varphi_\bx-\varphi_\by)\Psi_i \right \rangle_\mathcal{H} \nonumber
    \\ &=& \sum_{i=1}^n \Wb^T_i(\varphi_\bx-\varphi_\by) \left\langle \Psi_i, \sum_{j=1}^n\Wb^T_j(\varphi_\bc-\varphi_\by)\Psi_j\right\rangle_\mathcal{H} \nonumber
    \\ &=& \sum_{i=1}^n \left(\Wb^T_i(\varphi_\bx-\varphi_\by)\right)^2\langle\Psi_i,\Psi_i\rangle_\mathcal{H} \nonumber
    \\ &=& \sum_{i=1}^n (\varphi_\bx-\varphi_\by)^T\Wb_i\Wb_i^T(\varphi_\bx-\varphi_\by) \nonumber
    \\ &=& (\varphi_\bx-\varphi_\by)^T\Wb\Wb^T(\varphi_\bx-\varphi_\by) \nonumber
    \\ &=& \|\varphi_\bx-\varphi_\by\|_\Mb^2 \label{lastline lem1}
\end{eqnarray}
where in the final step we have defined $\Mb := \Wb \Wb^T$. Then the eigenvalues of $\Mb$ are equal to the square of the singular values of $\Wb$. In general we note that the eigenvalues of $(L\mathcal{P}_{\mathcal{S}_\mathcal{X}})^\dagger L\mathcal{P}_{\mathcal{S}_\mathcal{X}}$ are equal to the eigenvalues of $\Mb$ where $L^\dagger$ denote the adjoint of $L$. Note that $\|L\phi_x-L\phi_y\|_\mathcal{H}^2=\|L\mathcal{P}_{\mathcal{S}_\mathcal{X}}\phi_x-L\mathcal{P}_{\mathcal{S}_\mathcal{X}}\phi_y\|_\mathcal{H}^2$ for $\phi_x,\phi_y \in \mathcal{S}_\mathcal{X}$. Hence, we have $\|L\phi_\xb\|_\mathcal{H}=\|\varphi_\xb\|_\Mb$ for any $\phi_x\in \mathcal{S}_\mathcal{X}$ from (\ref{lastline lem1}).