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 $x$.



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


Thank you so much for your elaboration!



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


