Let $f \in L^p(\mathbb R^n)$ be given. Consider a partition of rectangles $I_{ij}:=[x_i,x_{i+1}]\times [x_j,x_{j+1}]$ of $\mathbb R^2.$

Then, we may define the coefficients

$$\alpha_{ij}= \frac{1}{\left\lvert I_{ij} \right\rvert} \int_{I_{ij}} f(s) \ ds$$

and consider the function $$g(x):=\sum_{ij} \alpha_{ij}\chi_{I_{ij}}(x).$$

I ask: If the distance between $x_i$ and $x_{i+1}$ goes uniformly in $i$ to zero. Does this imply that $g$ converges in $L^p$ to $f$? The functions $\chi$ are just indicator functions.

  • 1
    $\begingroup$ Since you have accepted @IosefPinelis's answer, it is probably polite to upvote it. $\endgroup$ – LSpice May 1 '18 at 19:19

$\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \renewcommand{\bar}{\overline}$

The answer is yes, at least for $p\ge1$. This follows because compactly supported continuous functions are dense in $L^p$ and such functions are uniformly continuous.

Here are details. For any $f\in L^p(\R^n)$, let $\bar f$ be defined as your function $g$: \begin{equation*} \bar f:=\sum_{ij} \al_{f;ij}\chi_{I_{ij}}, \end{equation*} where \begin{equation*} \al_{f;ij}:=\frac1{|I_{ij}|} \int_{I_{ij}} f(s)\,ds. \end{equation*}

Take any real $\ep\in(0,1)$ and take any continuous function $f_\ep$ with a support $K_\ep\subseteq[-M_\ep,M_\ep]^n$ for some real $M_\ep>0$ such that \begin{equation*} \|f-f_\ep\|_p<\ep. \tag{1} \end{equation*} Let $$\ep_1:=\ep/(2M_\ep+2)^{n/p}.$$ By the uniform continuity of $f_\ep$, there is some real $\de_\ep>0$ such that \begin{equation*} |f_\ep-\bar{f_\ep}|=\sum_{ij} |f_\ep-\bar{f_\ep}|\,\chi_{I_{ij}}\le \sum_{ij} \ep_1\chi_{I_{ij}}=\ep_1 \end{equation*} as soon as \begin{equation*} \max_i(x_{i+1}-x_i)<\de_\ep. \tag{2} \end{equation*} Hence, \begin{multline*} \|f_\ep-\bar{f_\ep}\|_p^p =\int_{\R^n} |f_\ep-\bar{f_\ep}|^p =\int_{[-M_\ep-1,M_\ep+1]^n} |f_\ep-\bar{f_\ep}|^p \\ \le\ep_1^p(2M_\ep+2)^n=\ep^p. \tag{3} \end{multline*} Also, for $h:=f-f_\ep$, by Jensen's inequality and (1), \begin{multline*} \|\bar f-\bar{f_\ep}\|_p^p=\|\,\bar h\,\|_p^p=\int_{\R^n}|\,\bar h\,|^p =\sum_{ij}\int_{I_{ij}}|\,\bar h\,|^p =\sum_{ij}|\,\bar h\,|^p\,|I_{ij}| \\ \le\sum_{ij}\bar{|h|^p}\,|I_{ij}| =\sum_{ij}\int_{I_{ij}}|h|^p =\|h\|_p^p=\|f-f_\ep\|_p^p<\ep^p. \tag{4} \end{multline*} Thus, by (1), (3), (4), and Minkowski's inequality, \begin{equation*} \|f-\bar f\|_p\le3\ep \end{equation*} as soon as (2) holds.

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