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Suppose I have a random variable $X$ with a density on $\mathbb{R}^n$ and a family of random variables $Y_\delta$ on $\mathbb{R}^n$ with the property that $\|Y_\delta\|_2 \leq \delta$ with probability $1$. Is it true that $X + Y_\delta$ (adding independent copies) tends to $X$ in total variation distance as $\delta \to 0$? If the $Y_\delta$'s had a densities this is well known, but I can't find a reference in case $Y_\delta$ does not have a density. It still feels true to me, however.

Cole_Franks
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I only consider the case $X+\delta Y$, where $X$ and $Y$ are independent, and $\delta>0$.

Suppose $\nu$ any Borel probability measure on $\mathbb{R}^d$, and let $f\in L^+_1(\lambda_d)$ (where $\lambda_d$ is Lebesgue's measure on $\mathbb{R}^d$) be such that $\int f=1$. Suppose $X$ and $Y$ are independent random variables with laws $\mu:=d\lambda_d$ and $\nu$, respectively. Then $X+ \delta Y$ has law with density $$h_\delta(x)=\int_{\mathbb{R}^n}f(x-\delta y)\nu(dy)$$ By Fubini's theorem $$\begin{align} \int_{\mathbb{R}^d}|h_\delta(x)-f(x)|\,dx&=\int_{\mathbb{R}^d}\Big|\int_{\mathbb{R}^d}f(x-\delta y)-f(x)\,\nu(dy)\Big|\,dx\\ &\leq\int_{\mathbb{R}^d}\Big(\int_{\mathbb{R}^d}|f(x-\delta y)-f(x)|\,dx\Big)\,\nu(dy) \end{align}$$ Since $\int_{\mathbb{R}^d}|f(x-\delta y)-f(x)|\,dx\leq 2\|f\|_1=2$ and $\lim_{\delta\rightarrow0}\int_{\mathbb{R}^d}|f(x-\delta y)-f(x)|\,dx=0$, dominated convergence yields $$\|h_\delta - f\|_{L_1(\lambda_d)}\xrightarrow{\delta\rightarrow0}0.$$

Mittens
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    How do you know that $\int_{\mathbb R^d}\vert f(x-\delta y)-f(x)\vert,dx\underset{\delta\to0}{\longrightarrow}0$? – Will Jul 29 '21 at 07:28
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    The translation operator is continuous in $L_p(\lambda_d)$ for any $p\geq1$. This can be seen in many textbooks in real analysis. There are a few postings in MSE that discuss this too, for example this one – Mittens Jul 29 '21 at 12:56
  • Alright, thanks – Will Jul 29 '21 at 13:16
  • Great answer, thanks! Should $\mu$ be $f$ (or whatever the appropriate notation is for the measure with density $f$ w.r.t. Lebesgue)? And do you happen to know if this fact appears in any textbook? – Cole_Franks Jul 29 '21 at 16:16
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    With regards to $\mu$, yes, I just made an edit. I forgot to write $\mu=fd\lambda_d$, where $\lambda_d$ is Lebesgue's measure. As for references, almost any textbook an real analysis (Follland's, Rudin's, Widden-Sygmund's) will dedicate some part to the continuity of the translation operator, Fubini's theorem, as well as the Radon-Nikodym theorem. There is also a concept, which I did not comment about, known as good kernels and approximations to identity. I think one can generalize the, carefully with kernels that are singular with respect Lebesgue measure. – Mittens Jul 29 '21 at 16:21