Let $f(x,y)$ be a Lipschitz continuous density function on $\mathbb{R}^2$. And let $f(x) = \int\limits_\mathbb{R} f(x,y)dy$ be marginal density function. Is $f(x)$ Lipschitz continuous?
More generally let $$g(u,v)=\int\limits_{xu +by=1} f(x,y) ds$$ (path integral along the line). Given $(u_0,v_0)\neq (0,0)$. Can we prove that $$g(u_1,v_1)-g(u_0,v_0) = O(\|(u_1,v_1)-(u_0,v_0)\|)$$ when $\|(u_1,v_1)-(u_0,v_0)\|$ is sufficiently small?
The "more generally" part is not more general than the first question. The answer to your question is "yes" if the density is compactly supported and "no" in general.
To see the counterexample, let $R_n=[2^n,2^n+4^{-n}]\times [0,8^n]$ and let $f_n(x,y)=(4^{-n}-d((x,y),R_n))^+$, so that $\int f_n(x,y)\,dx\,dy\approx 2^{-n}$. Now if $f=\sum f_n$, the marginal of $f$ jumps by $2^{-n}$ between $2^n$ and $2^n+4^{-n}$, so that the marginal is not Lipschitz.
