Given a compact set $\Omega\subset \mathbb{R}^N$, I am wondering if $L^\infty(\Omega)$ is dense in the weak $L^p$ space $L^{p,\infty}(\Omega)$ with $1< p<\infty$ (see here the definition).
I would like to know if the proof is as simple as using that $L^\infty(\Omega)$ is dense in $L^p(\Omega)$ along with the fact that $L^p(\Omega)\subset L^{p,\infty}(\Omega)\subset L^q(\Omega)$ for any $q \leq p$.
Actually, the closure of $L^\infty$ in $L^{p,\infty}$ is $$ L^{p,\infty}_0:=\left\{ f:\lim_{t\to\infty}t^p\mu\{x:\lvert f(x)\rvert>t\}=0 \right\}. $$ Indeed, if $f$ belongs to the closure of $L^\infty$ in $L^{p,\infty}$, for each $N$, there exists a function $f_N\in L^\infty$ such that $\sup_{t>0}t^p\mu\{x:\lvert f(x)-f_N(x)\rvert>t\}<N^{-1}$. Then $$ t^p\mu\{x:\lvert f(x)\rvert>t\}\leqslant t^p\mu\{x:\lvert f(x)-f_N(x)\rvert>t/2\}+t^p\mu\{x:\lvert f_N(x)\rvert>t/2\}\leqslant \frac{2^p}N+t^p\mu\{x:\lvert f_N(x)\rvert>t/2\} $$ and since $f_N$ is bounded, $$ \limsup_{t\to\infty}t^p\mu\{x:\lvert f(x)\rvert>t\}\leqslant \frac{2^p}N $$ hence $f$ belongs to $L_0^{p,\infty}$.
Conversely, if $f$ belongs to $L_0^{p,\infty}$, then we can show that $f_N\colon x\mapsto f(x)\mathbf{1}_{\lvert f(x)\rvert\leqslant N}$ approximates $f$ with respect to the semi-norm $$ \lVert g\rVert_{p,\infty}=\left(\sup_{t>0}t^p\mu\{x:\lvert g(x)\rvert>t\}\right)^{1/p}, $$ since $$ \lVert f-f_N\rVert_{p,\infty}^p=\sup_{t>0}t^p\mu\{x:\lvert f(x)\rvert>\max\{t,N\}\}\leqslant \sup_{t\geqslant N}t^p\mu\{x:\lvert f(x)\rvert>t\}. $$
This density is not true.
Let $\Omega=(0,+1)$. Let $p\in (0,\infty)$ and define $$ f(x) = x^{-\frac1p}. $$ Then $$ \mu(\{x:\ |f(x)| > t\} ) = t^{-p} $$ for $t\ge1$, so that $\|f\|_{L^{p,\infty}} = 1$. Let now $g\in L^\infty(\Omega)$ with $N:=\|g\|_{L^\infty}$. Let $t>0$. Then $$ \mu(\{x:\ |f(x)-g(x)| > t\} ) \ge \mu(\{x:\ |f(x)| > t+N \} ) = (t+N)^{-p}, $$ so that $$ t^p \mu(\{x:\ |f(x)-g(x)| > t\} ) \ge t^p (t-N)^{-p} \to 1 $$ for $t\to\infty$.
