log.txt

\subsection{Regularity condition}
\label{subsubsec:regularity}
We first establish that $\frac{\partial}{\partial \delta}\expectation
\tilde{K}(\delta) = \expectation \frac{\partial}{\partial
\delta}\tilde{K}(\delta)$. Since $\tilde{K}$ is a finite dimensional
matrix-valued function, we verify the integrability coefficient-wise, following
\citet{sutherland2015}'s demonstration. Namely, without loss of generality we
show
\begin{dmath*}
    \left[\frac{\partial}{\partial \delta}\expectation
    \tilde{K}(\delta)\right]_{\ell m} = \expectation \frac{\partial}{\partial
    \delta} \left[\tilde{K}(\delta)\right]_{\ell m}
\end{dmath*}
\begin{proposition}[Differentiation under the integral sign]
    \label{pr:diff_under_int}
    Let $\mathcal{X}$ be an open subset of $\mathbb{R}^d$ and $\Omega$ be a
    measured space. Suppose that the function
    $f:\mathcal{X}\times\Omega\to\mathbb{R}$ verifies the following conditions
    \begin{itemize}
        \item $f(x,\omega)$ is a measurable function of $\omega$ for each $x$
        in $\mathcal{X}$.
        \item For almost all $\omega$ in $\Omega$, the derivative $\partial
        f(x, \omega)/\partial x_i$ exists for all $x$ in $\mathcal{X}$.
        \item There is an integrable function $\Theta:\Omega\to\mathbb{R}$ such
        that $\abs{\partial f(x, \omega)/\partial x_i}\le\Theta(\omega)$ for
        all $x$ in $\mathcal{X}$.
    \end{itemize}
    Then
    \begin{dmath*}
        \frac{\partial}{\partial x_i} \int_\Omega f(x,\omega)d\omega =
        \int_\Omega \frac{\partial}{\partial x_i}f(x,\omega)d\omega.
    \end{dmath*}
\end{proposition}
\begin{proof}
    Define the function
    $\tilde{G}_{x,z }^{i,\ell,m}(t,\omega):\mathbb{R}\times\Omega\to\mathbb{R}$
    by
    \begin{dmath*}
    \tilde{G}_{x, z}^{i,\ell,m}(t,\omega) = \left[\tilde{K}(x + te_i -
    y)\right]_{\ell m} \hiderel{=}
    \left[\tilde{G}_{x,y}^{i}(t,\omega)\right]_{\ell m},
    \end{dmath*}
    where $e_i$ is the $i$-th standard basis vector. Then
    $\tilde{G}_{x, z}^{i,\ell,m}$ is integrable \acs{wrt}. $\omega$ since
    \begin{dmath*}
        \int_\Omega \tilde{G}_{x, z}^{i,\ell,m}(t,\omega) d\omega =
        \expectation \left[\tilde{K}(x+te_i - z)\right]_{\ell m} = \left[K(x +
        t e_i - z)\right]_{\ell m} < \infty.
    \end{dmath*}
    Additionally for any $\omega$ in $\Omega$, $\partial/\partial t
    \tilde{G}_{x,y}^{i,\ell,m}(t,\omega)$ exists and satisfies
    \begin{dmath*}
        \expectation \abs{\frac{\partial}{\partial
        t}\tilde{G}_{x,y}^{i,\ell,m}(t,\omega)} 
        = \expectation \abs{\frac{1}{D}\sum_{j=1}^DA(\omega_j)_{\ell m}
        \left(\sin\inner{y ,\omega_j}\frac{\partial}{\partial
        t}\sin(\inner{x,\omega_j} +
        t\omega_{ij})+\cos\inner{y,\omega_j}\frac{\partial}{\partial
        t}\cos(\inner{x,\omega_j} + t\omega_{ij})\right)} 
        = \expectation \abs{\frac{1}{D}\sum_{j=1}^D A(\omega_j)_{\ell m}
        \left(\omega_{ji}\sin\inner{y, \omega_j}\sin(\inner{x, \omega_j} +
        t\omega_{ji})-\omega_{ji}\cos\inner{y, \omega_j}\cos(\inner{x
        ,\omega_j}+t\omega_{ji})\right)} \le\expectation\left[ \frac{1}{D}
        \sum_{j=1}^D\abs{A(\omega_j)_{\ell m}\omega_{ji}\sin\inner{y,
        \omega_j}\sin(\inner{x, \omega_j} + t\omega_{ji})} +
        \abs{A(\omega_j)_{\ell m}\omega_{ji}\cos\inner{y, \omega_j}\cos(\inner{x,
        \omega_j} + t\omega_{ji})}\right] 
        \le \expectation
        \left[\frac{1}{D}\sum_{j=1}^D2\abs{A(\omega_j)_{\ell
        m}\omega_{ji}}\right].
    \end{dmath*}
    Hence
    \begin{dmath*}
        \expectation \abs{\frac{\partial}{\partial
        t}\tilde{G}_{x,y}^{i,\ell,m}(t,\omega)} \le 2\expectation \left[\norm{\omega
        \otimes A(\omega)}_1\right].
    \end{dmath*}
    which is assumed to exist since in finite dimensions all norms are
    equivalent and $\expectation_{\mu}\left[
    \norm{\omega}^2_{\dual{\mathcal{X}}}
    \norm{A(\omega)}^2_{\mathcal{Y},\mathcal{Y}} \right]$ is assume to exists.
    Thus applying \cref{pr:diff_under_int} we have
    \begin{dmath*}
        \left[\frac{\partial}{\partial \delta_i}\expectation
        \tilde{K}(\delta)\right]_{\ell m} = \expectation
        \frac{\partial}{\partial \delta_i} \left[\tilde{K}(\delta)\right]_{\ell
        m}
    \end{dmath*}
    The same holds for $y$ by symmetry. Combining the results for each
    component $x_i$ and for each element $\ell m$, we get that
    $\frac{\partial}{\partial \delta}\expectation \tilde{K}(\delta) =
    \expectation \frac{\partial}{\partial \delta}\tilde{K}(\delta)$.
    \subsection{Bounding the Lipschitz constant}
    Since $F$ is differentiable, $L_{F}=\norm{\frac{\partial F}{\partial
    \delta}(\delta^*)}_{\mathcal{Y},\mathcal{Y}}$ where
    $\delta^*=\argmax_{\delta\in\mathcal{D}_{\mathcal{C}}}\norm{\frac{\partial
    F}{\partial \delta}(\delta)}_{\mathcal{Y},\mathcal{Y}}$.
    \begin{dmath*}
        \expectation_{\mu,\delta^*}\left[ L_F^2 \right] 
        = \expectation_{\mu,\delta^*} \norm{\frac{\partial \tilde{K}}{\partial
        \delta}(\delta^*)-\frac{\partial K_0}{\partial
        \delta}(\delta^*)}^2_{\mathcal{Y},\mathcal{Y}}
        \le \expectation_{\delta^*}\left[ \expectation_{\mu}
        \norm{\frac{\partial \tilde{K}}{\partial
        \delta}(\delta^*)}^2_{\mathcal{Y},\mathcal{Y}} - 2\norm{\frac{\partial
        K_0}{\partial
        \delta}(\delta^*)}_{\mathcal{Y},\mathcal{Y}}\expectation_{\mu}
        \norm{\frac{\partial \tilde{K}}{\partial
        \delta}(\delta^*)}_{\mathcal{Y},\mathcal{Y}} + \norm{\frac{\partial
        K_0}{\partial \delta}(\delta^*)}^2_{\mathcal{Y},\mathcal{Y}} \right]
    \end{dmath*}
    Using Jensen's inequality $\norm{\expectation_\mu\frac{\partial
    \tilde{K}}{\partial
    \delta}(\delta^*)}_2\le\expectation_\mu\norm{\frac{\partial
    \tilde{K}}{\partial \delta}(\delta^*)}_2$ and $\frac{\partial}{\partial
    \delta}\expectation \tilde{K}(\delta) = \expectation
    \frac{\partial}{\partial \delta}\tilde{K}(\delta)$. (see
    \cref{subsubsec:regularity}), $\expectation_\mu\frac{\partial
    \tilde{K}}{\partial \delta}(\delta^*)=\frac{\partial}{\partial
    \delta}\expectation_\mu\tilde{K}(\delta^*)=\frac{\partial K_0}{\partial
    \delta}(\delta^*)$ thus
    \begin{dmath*}
        \expectation_{\mu,\delta^*}\left[ L_F^2 \right] \le
        \expectation_{\delta^*}\left[ \expectation_\mu \norm{\frac{\partial
        \tilde{K}}{\partial \delta}(\delta^*)}_2^2 - 2\norm{\frac{\partial
        K_0}{\partial \delta}(\delta^*)}_2^2 + \norm{\frac{\partial
        K_0}{\partial \delta}(\delta^*)}_2^2 \right] 
        = \expectation_{\mu,\delta^*}\norm{\frac{\partial \tilde{K}}{\partial
        \delta}(\delta^*)}_2^2-\expectation_{\delta^*}\norm{\frac{\partial
        K_0}{\partial \delta}(\delta^*)}_2^2 \le
        \expectation_{\mu,\delta^*}\norm{\frac{\partial \tilde{K}}{\partial
        \delta}(\delta^*)}_2^2
        = \expectation_{\mu,\delta^*}\norm{\frac{\partial }{\partial
        \delta^*}\cos h_{\omega}(x) A(\omega) }_2^2
        = \expectation_{\mu,\delta^*}\norm{-h'_{\omega}(x)
        \sin\left(h_{\omega}(x)\right) \otimes A(\omega)}_2^2 
        \le
        \expectation_{\mu}\left[H_\omega_2^2 \norm{A(\omega)}_2^2\right]
        \colonequals \sigma_p^2
    \end{dmath*}
\end{proof}