1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Since you have accepted @IosefPinelis's answer, it is probably polite to upvote it. $\endgroup$ – LSpice May 1 '18 at 19:19
2
$\begingroup$

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

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.