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Iosif Pinelis
Iosif Pinelis
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$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\J{\mathcal J}$No. Consider first the case $d=1$. Let $\J$ denote the set of all subsets of $\Z$. For $J\in\J$, let $$f_J:=\sum_{j\in J}1_{[j,j+1)},$$ so that $f_J\in\tilde L^p(\R^d)$. For any two distinct $J$ and $K$ in $\J$, we have $\|f_J-f_K\|_{\tilde L^p(\R^d)}\ge1$. Since the set $\J$ is uncountable, your space $\tilde L^p(\R^d)$ is not separable. (Indeed, if $S$ is a dense subset of $\tilde L^p(\R^d)$, then for each $J\in\J$ there is some $s_J\in S$ such that $\|f_J-s_J\|_{\tilde L^p(\R^d)}<1/2$. So, by the norm inequality, the $s_J$'s must be pairwise distinct, so that the set $S$ is uncountable.)

The case $d\ge2$ is similar: then partition $\R^d$ into cubes rather than intervals.

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\J{\mathcal J}$No. Consider first the case $d=1$. Let $\J$ denote the set of all subsets of $\Z$. For $J\in\J$, let $$f_J:=\sum_{j\in J}1_{[j,j+1)},$$ so that $f_J\in\tilde L^p(\R^d)$. For any two distinct $J$ and $K$ in $\J$, we have $\|f_J-f_K\|_{\tilde L^p(\R^d)}\ge1$. Since the set $\J$ is uncountable, your space $\tilde L^p(\R^d)$ is not separable.

The case $d\ge2$ is similar: then partition $\R^d$ into cubes rather than intervals.

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\J{\mathcal J}$No. Consider first the case $d=1$. Let $\J$ denote the set of all subsets of $\Z$. For $J\in\J$, let $$f_J:=\sum_{j\in J}1_{[j,j+1)},$$ so that $f_J\in\tilde L^p(\R^d)$. For any two distinct $J$ and $K$ in $\J$, we have $\|f_J-f_K\|_{\tilde L^p(\R^d)}\ge1$. Since the set $\J$ is uncountable, your space $\tilde L^p(\R^d)$ is not separable. (Indeed, if $S$ is a dense subset of $\tilde L^p(\R^d)$, then for each $J\in\J$ there is some $s_J\in S$ such that $\|f_J-s_J\|_{\tilde L^p(\R^d)}<1/2$. So, by the norm inequality, the $s_J$'s must be pairwise distinct, so that the set $S$ is uncountable.)

The case $d\ge2$ is similar: then partition $\R^d$ into cubes rather than intervals.

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Iosif Pinelis
Iosif Pinelis
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$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\J{\mathcal J}$No. Consider first the case $d=1$. Let $\J$ denote the set of all subsets of $\Z$. For $J\in\J$, let $$f_J:=\sum_{j\in J}1_{[j,j+1)}.$$$$f_J:=\sum_{j\in J}1_{[j,j+1)},$$ Then, forso that $f_J\in\tilde L^p(\R^d)$. For any two distinct $J$ and $K$ in $\J$, we have $\|f_J-f_K\|_{{\tilde L}^p (\mathbb R^d)}\ge1$$\|f_J-f_K\|_{\tilde L^p(\R^d)}\ge1$. Since the set $\J$ is uncountable, your space ${\tilde L}^p (\mathbb R^d)$$\tilde L^p(\R^d)$ is not separable.

The case $d\ge2$ is similar: then partition $\R^d$ into cubes rather than intervals.

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\J{\mathcal J}$No. Consider first the case $d=1$. Let $\J$ denote the set of all subsets of $\Z$. For $J\in\J$, let $$f_J:=\sum_{j\in J}1_{[j,j+1)}.$$ Then, for any two distinct $J$ and $K$ in $\J$, we have $\|f_J-f_K\|_{{\tilde L}^p (\mathbb R^d)}\ge1$. Since the set $\J$ is uncountable, your space ${\tilde L}^p (\mathbb R^d)$ is not separable.

The case $d\ge2$ is similar: then partition $\R^d$ into cubes rather than intervals.

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\J{\mathcal J}$No. Consider first the case $d=1$. Let $\J$ denote the set of all subsets of $\Z$. For $J\in\J$, let $$f_J:=\sum_{j\in J}1_{[j,j+1)},$$ so that $f_J\in\tilde L^p(\R^d)$. For any two distinct $J$ and $K$ in $\J$, we have $\|f_J-f_K\|_{\tilde L^p(\R^d)}\ge1$. Since the set $\J$ is uncountable, your space $\tilde L^p(\R^d)$ is not separable.

The case $d\ge2$ is similar: then partition $\R^d$ into cubes rather than intervals.

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Iosif Pinelis
Iosif Pinelis
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$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}$$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\J{\mathcal J}$No: Your space. Consider first the case ${\tilde L}^p (\mathbb R^d)$ contains$d=1$. Let $\J$ denote the spaceset of all subsets of $L^\infty (\mathbb R^d)$$\Z$. For $J\in\J$, whichlet $$f_J:=\sum_{j\in J}1_{[j,j+1)}.$$ Then, for any two distinct $J$ and $K$ in $\J$, we have $\|f_J-f_K\|_{{\tilde L}^p (\mathbb R^d)}\ge1$. Since the set $\J$ is uncountable, your space ${\tilde L}^p (\mathbb R^d)$ is not separable.

The case $d\ge2$ is similar: then partition $\R^d$ into cubes rather than intervals.

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}$No: Your space ${\tilde L}^p (\mathbb R^d)$ contains the space $L^\infty (\mathbb R^d)$, which is not separable.

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\J{\mathcal J}$No. Consider first the case $d=1$. Let $\J$ denote the set of all subsets of $\Z$. For $J\in\J$, let $$f_J:=\sum_{j\in J}1_{[j,j+1)}.$$ Then, for any two distinct $J$ and $K$ in $\J$, we have $\|f_J-f_K\|_{{\tilde L}^p (\mathbb R^d)}\ge1$. Since the set $\J$ is uncountable, your space ${\tilde L}^p (\mathbb R^d)$ is not separable.

The case $d\ge2$ is similar: then partition $\R^d$ into cubes rather than intervals.

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Iosif Pinelis
Iosif Pinelis
  • 128k
  • 8
  • 107
  • 229