Is the space $L^p_{\text{loc}} (\mathbb R^d)$ separable w.r.t. the norm $\|f\|_{\tilde L^p} := \sup_{x \in \mathbb R^d} \|1_{B(x, 1)} f\|_{L^p}$?

Question

Fix $p \in [1, \infty)$. Let $(L^p (\mathbb R^d), \|\cdot\|_{L^p})$ be the Lesbesgue space of $p$-integrable real-valued functions on $\mathbb R^d$. Let ${\tilde L}^p (\mathbb R^d)$ be the space of Lebesgue measurable functions $f:\mathbb R^d \to \mathbb R$ such that $$ \|f\|_{\tilde L^p} := \sup_{x \in \mathbb R^d} \|1_{B(x, 1)} f\|_{L^p} < \infty, $$ where $B(x, 1)$ is the open unit ball centered at $x$.

Is $(\tilde L^p (\mathbb R^d), \|\cdot\|_{\tilde L^p})$ separable?

Thank you so much for your elaboration!

Answer 1

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\J{\mathcal J}$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.