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Suppose $h:\mathcal{X \times Y \times W}\rightarrow\mathcal{X}$, with $\mathcal{X,Y,W} \subseteq \mathbb R^d$, $d \in \mathbb N$ is Lipschitz in $x,y$, i.e., $$ \| h(x,y,w) - h(x',y',w) \|_2 \le L_h (\|x - x' \|_2 + \| y -y'\|_2) \quad \forall \, x,x' \in \mathcal X, \ \forall \, y,y'\in \mathcal Y. $$ We also have $f: \mathcal X \rightarrow \mathbb R$ is Lipschitz in $x$, i.e., $$ | f(x) - f(x') | \le L_f \|x - x' \|_2 \quad \forall \, x,x' \in \mathcal X.$$ Can we say that $$ \max_y \left|\int_{\mathcal W} p(w) f(h(x,y,w)) dw - \int_{\mathcal W} p(w) f(h(x',y,w)) dw \right| \le K \|x-x' \|_2 \quad \forall \, x,x' \in \mathcal X,$$ (where $p(w)$ is a probability density function), i.e., $\int_{\mathcal W} p(w) f(h(x,y,w)) dw$ is Lipschitz in $x$? If yes what is the Lipschitz constant $K$?

We can assume for example that $h(x,y,w)= g(x,y) + w$ where $g:\mathcal{X\times Y}\rightarrow\mathcal{X}$ is a deterministic function of $x,y$ and $w \sim \mathcal N(\boldsymbol{\mu}, \Sigma)$. Thanks

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Assume that $$|f(x)-f(x')|\le L_f\|x-x'\|_2\quad\forall \ x,x' \in \mathcal X,$$ rather than $$|f(x)-h(x')|\le L_f\|x-x'\|_2\quad\forall \ x,x' \in \mathcal X.$$ Then $$|f(h(x,y,w))-f(h(x',y,w))|\le L_f\|h(x,y,w)-h(x',y,w)\|_2 \le L_fL_h\|x-x'\|_2,$$ whence $$\left|\int_{\mathcal W} p(w) f(h(x,y,w))\, dw - \int_{\mathcal W} p(w) f(h(x',y,w)) \,dw \right| \\ \le\int_{\mathcal W} p(w) |f(h(x,y,w))-f(h(x',y,w))| dw \\ \le\int_{\mathcal W} p(w) L_fL_h\|x-x'\|_2 dw = L_fL_h\|x-x'\|_2.$$ So, your desired inequality holds with $K=L_fL_h$.

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  • $\begingroup$ Thanks a lot! Yes, it should be f(x') rather than h(x'), I edited the question to correct this mistake. $\endgroup$
    – PJORR
    Commented May 9, 2020 at 16:06

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