How to show that $C^\infty_0$ is not dense in $L^p_{weak} (\mathbb{R}^n)$?

Question

Let $C^\infty_0$ denote the smooth, compactly supported functions on $\mathbb{R}^n$. Let $L^p_{\mathrm{weak}}(\mathbb{R}^n)$ denote the space of all functions $f:\mathbb{R}^n \rightarrow \mathbb{R}$ which satisfy

$$ \Big| \big\{x \in \mathbb{R}^n : |f(x)| > \lambda \big\} \Big| \lesssim \lambda^{-p} $$

for all $\lambda > 0$. This space when endowed with the quasi-norm

$$ \Vert f\Vert_{L^p_{\mathrm{weak}}}^* : = \sup_{\lambda > 0} \lambda \Big| \big\{x \in \mathbb{R}^n : |f(x)| > \lambda \big\} \Big|^{1/p} $$

is a quasi-Banach space. How do you show that $C^\infty_0$ is not dense in $L^p_{\mathrm{weak}} (\mathbb{R}^n)$ for $1< p < \infty$? This question was inspired by a real analysis qualifying problem, and I know that considering the function $f(x) = |x|^{-n/p}$ is important.

Answer 1

EDITED:

$|x|^{-n/p} > \lambda \Rightarrow |x| <\lambda^{-p/n}$, so $$ x \in (-\lambda^{-p/n} , \lambda^{-p/n} ) , $$ so $$ m( \{ x : |f(x)| >\lambda \} ) \leq 2 \lambda^{-p/n} \leq 2 \lambda^{-p} $$

We know that $|x|^{-n/p}$ is not in $L^p$ (see Stein/Shakarchi Book IV Ch. 1 problem 1). Which is good, since otherwise density of smooth functions with compact support in $L^p(\mathbb{R}^n)$ prevents the counterexample.

Then, you need to show that $|x|^{-n/p}$ is not a limit of $C^\infty_0$ functions (hint, whats the support of $|x|^{-n/p}$?).

Answer 2

The point is that the function $f\colon x\mapsto \left\lvert x\right\rvert^{-n/p}$ belongs to $\mathbb L^p_{\operatorname{weak}}\left(\mathbb R^n\right)$ and for each fixed $R$, $$ \left\lVert f-f\cdot\mathbf 1_{[-R,R]^n}\right\rVert_{\mathbb L^p_{\operatorname{weak}}}\geqslant c $$ where $c$ depends only on $n$. Therefore, for any function $g$ with compact support, $\left\lVert f-g\right\rVert_{\mathbb L^p_{\operatorname{weak}}}\geqslant c$.