Let $p > 0$, and denote by $L^{p,\infty}(\mathbb{R})$ the space of all measurable functions $f : \mathbb{R} \rightarrow \mathbb{R}$ for which $$ \|f\|_{p,\infty}:=\sup_{\alpha > 0} \alpha^p |\{x \in \mathbb{R} : |f(x)| > \alpha\}|<\infty $$ Prove that the simple functions are not dense in $ L^{p,\infty}(\mathbb{R})$, in the sense that there exists a function $ f \in L^{p,\infty}(\mathbb{R})$ such that $ \|f - h_k\|_{p,\infty} \rightarrow 0 $ fails to hold for every sequence of simple functions $h_1, h_2,\ldots$.
An other related problem:
Let $1<p<\infty$ and $f(x)=|x|^{−n/p}$ for $ x\in R^n$. Prove that $f$ is not the limit of a sequence $f_k \in C_0^\infty (R^n)$ in the sense of convergence in $L^{p,\infty}(R^n)$. That is,
$$\lim \sup_{k>0} \sup \lambda >0 \lambda^p|{x\in R^n ||f(x)−f_k(x)>\lambda}|>0 ,$$
for any such sequence.
Q2: Does $C_0^\infty$ mean the auction that converges to zero at the tails? or the function with compact support?
My attempt for Q1:
Consider the function $f:[0,1] \to [0,\infty]$ where $f(x) = \lambda^{-1/p}$ on $\{|f| > \lambda\}$ and $0$ otherwise. Then $\|f\|_{L^{p,\infty}_w}=1$. Consider any sequence of simple functions $\{h_k\}$ such that $0 \leq h_k \leq f$ for all k. Then
$$\|f-h_k\|_{p,\infty} \leq 2.$$
Let $m$ denote Lebesgue measure on $\mathbb{R}$. First note that for any function $s\in L^{p,\infty}(\mathbb{R})$ and any $\alpha >0$, $$m(\{x\in \mathbb{R}:|s(x)|>\alpha\}) \leqslant \|s_{p,\infty}\|/\alpha^p<\infty.$$ In particular, if $s\in L^{p,\infty}(\mathbb{R})$ is simple and measurable, we can write $s=\sum_{i=1}^n c_i1_{A_i}$, where $A_i$ are pairwise disjoint, measurable and where $\min_i |c_i|>0$. Fix $0<\alpha<\min_i |c_i|$ and note that $$m(\{x\in \mathbb{R}:s(x)\neq 0\})=m(\{x\in \mathbb{R}:|s(x)|>\alpha\})<\infty.$$ Thus $\sum_{i=1}^n m(A_i)<\infty$.
Define $$f(x)=\left\{\begin{array}{ll} x^{-1/p} & : x > 0 \\ 0 & : x\leqslant 0.\end{array}\right.$$ Of course, $f$ is measurable. Moreover, for any $\alpha>0$, $$\{x\in \mathbb{R}:|f(x)|>\alpha\}=(0,\alpha^{-p}),$$ and $$\alpha^p m(\{x\in \mathbb{R}:|f(x)|>\alpha\})=1$$ for all $\alpha>0$. Thus $f\in L^{p,\infty}(\mathbb{R})$. The idea is that the supremum used in the definition of $\|\cdot\|_{p,\infty}$ is obtained for lots of different $\alpha$ (in fact, all of them).
Let $s$ be simple. Let $A=\{x\in \mathbb{R}:s(x)\neq 0\}$ and recall that $m(A)<\infty$. Then for $\alpha>0$, $$\{x\in \mathbb{R}:|f(x)-s(x)|>\alpha\}\supset \{x\in \mathbb{R}:|f(x)-s(x)|>\alpha\}\setminus A=(0,\alpha^{-p})\setminus A,$$ and $$\alpha^p m(\{x\in \mathbb{R}:|f(x)-s(x)|>\alpha\}) \geqslant \alpha^p[\alpha^{-p}-m(A)]=1-m(A)\alpha^p.$$ Taking $\alpha\to 0^+$ yields that the supremum is at least $1$.
For the second question about $\mathbb{R}^n$, Let $$B_n=\{x\in\mathbb{R}^n:|x|< 1\}$$ and note that $|x|^{-n/p}>\alpha$ if and only if $|x|< \alpha^{-p/n}$. This is equivalent to saying that $$\{x\in \mathbb{R}^n:|f(x)|>\alpha\}= \alpha^{-p/n}B_n.$$ The measure of this set is $(\alpha^{-p/n})^n m(B_n)=\alpha^{-p}m(B_n)$. Therefore for any $\alpha>0$, $$\alpha^pm(\{x\in \mathbb{R}^n:|f(x)|>\alpha\})=m(B_n)<\infty.$$ Fix any continuous function $g:\mathbb{R}^n\to \mathbb{R}$. Note that since $g$ is continuous on the compact set $\overline{B_n}$, it is bounded there. Fix $b\geqslant 1$ such that $|g(x)|\leqslant b$ for all $x\in B_n$. For $\alpha>0$, note that $ (\alpha+b)^{-p/n}B_n\subset B_n$ because we chose $b\geqslant 1$, so $(\alpha+b)^{-p/n}<1$. For $x\in (\alpha+b)^{-p/n}B_n$, $$|f(x)-g(x)|\geqslant f(x)-|g(x)|> (\alpha+b)-b=\alpha.$$ Therefore $$\{x\in\mathbb{R}^n:|f(x)-g(x)|>\alpha\}\supset (\alpha+b)^{-p/n}B_n,$$ which has measure $(\alpha+b)^{-p}m(B_n)$. Multiplying by $\alpha^p$ yields that $$\|f-g\|_{p,\infty}\geqslant \lim_{\alpha\to\infty}\alpha^p (\alpha+b)^{-p}m(B_n)=m(B_n).$$