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Let $f(x)$ be a probability density function, i.e $f(.) \geq 0$ and $\int_{\mathbb{R}}f(x) \ dx=1$. Let $I_{n,k}= \big[\frac{k}{n}, \frac{k+1}{n}\big), \ k \in \mathbb{Z}$ and $p_{n,k}$ denote the area of $f(.)$ in the interval $I_{n,k}$, i.e, $$ p_{n,k}=\int_{\frac{k}{n}}^{\frac{k+1}{n}} f(x) \ dx $$

We know that $\displaystyle \sum_{k \in \mathbb{Z}} p_{n,k}=1=\displaystyle \sum_{k \in \mathbb{Z}} p_{n,k+1}$. Define $\{a_{n,m}\}$ as the convolution of $\{p_{n,k}\}$ with itself, $$ a_{n,m}=\displaystyle \sum_{k \in \mathbb{Z}} \ p_{n,m-k} \ p_{n,k} $$

I am trying to prove that $\sum_{m \in \mathbb{Z}}|a_{n,m+1}-a_{n,m}| \rightarrow 0$ as $n \rightarrow \infty$

Is this true ? Atleast intuitively I feel that as our partitions become smaller and smaller difference should converge because $\{a_m\}$ is also a probability distribution on integers.

pikachuchameleon
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1 Answers1

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This is indeed true. We compute \begin{eqnarray*} \sum_{m}\left|a_{n,m+1}-a_{n,m}\right| & = & \sum_{m}\left|\sum_{k}\left[p_{n,\left(m+1\right)-k}-p_{n,m-k}\right]p_{n,k}\right|\\ & \leq & \sum_{k}\left[p_{n,k}\cdot\sum_{m}\left|p_{n,\left(m+1\right)-k}-p_{n,m-k}\right|\right]\\ & \overset{\ell=m-k}{=} & \sum_{k}\left[p_{n,k}\sum_{\ell}\left|p_{n,\ell+1}-p_{n,\ell}\right|\right]\\ & \overset{\sum_{k}p_{n,k}=1}{=} & \sum_{\ell}\left|\int_{\left(\ell+1\right)/n}^{\left(\ell+2\right)/n}f\left(y\right)\,{\rm d}y-\int_{\ell/n}^{\left(\ell+1\right)/n}f\left(x\right)\,{\rm d}x\right|\\ & \overset{x=y-\frac{1}{n}}{=} & \sum_{\ell}\left|\int_{\ell/n}^{\left(\ell+1\right)/n}\left[f\left(x+\frac{1}{n}\right)-f\left(x\right)\right]\,{\rm d}x\right|\\ & \leq & \sum_{\ell}\int_{\ell/n}^{\left(\ell+1\right)/n}\left|f\left(x+\frac{1}{n}\right)-f\left(x\right)\right|\,{\rm d}x\\ & = & \int_{\mathbb{R}}\left|f\left(x+\frac{1}{n}\right)-f\left(x\right)\right|\,{\rm d}x\\ & \xrightarrow[n\to\infty]{} & 0. \end{eqnarray*} Here, the last step used continuity of translation with respect to the $L^{1}$ norm, see e.g. Continuity of $L^1$ functions with respect to translation.

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PhoemueX
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