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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.

Feng
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JMill.
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  • What exactly does $\lesssim$ mean? Is it an inequality up to a multiplicative constant depending on $f$? – Bart Michels Aug 12 '18 at 20:34
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    $C_0^{\infty}$ is not even contained in $L^p_{weak}$. Maybe you meant $C_0^{\infty} \cap L^p_{weak}$. – shalin Aug 12 '18 at 20:46
  • Another possiblity is that you instead meant to consider the space of smooth compactly supported functions which is sometimes also denoted by $C_0^\infty$ and is contained in $L_{\text{weak}}^p$? – Rhys Steele Aug 12 '18 at 20:47
  • Barto: The constant depends of $f$. – JMill. Aug 12 '18 at 20:51
  • Yes, I do think that if $C_0^\infty$ means smooth, compactly supported functions the question makes more sense. – JMill. Aug 12 '18 at 20:52

2 Answers2

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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, $\left\lVert f-g\right\rVert_{\mathbb L^p_{\operatorname{weak}}}\geqslant c$.

Davide Giraudo
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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 $|x|^{-n/p}$?).

djkat
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  • You don't get to choose $n$, it's the dimension of the space. Thankfully it doesn't matter since $|x|^{-n/p}$ actually isn't in $L^q(\mathbb{R}^n)$ for any $q>0$. Also I'm a bit confused by the fact that you seem to want to split in to cases and show there's no approximation for the separate cases, but $x \mapsto |x|^{-n/p}$ is one function so this doesn't seem okay. Could you clarify? – Rhys Steele Aug 12 '18 at 21:02
  • IT revolved around the fact that I wanted to choose $n$. If you cannot choose $n$ there's no point in splitting to cases. – djkat Aug 12 '18 at 21:08
  • @Rhys Steele Edited it after referencing the exercise I based the argument out of. – djkat Aug 12 '18 at 23:50
  • I'm also quite confused by your hint to be honest. Note that for $f \in L^p(\mathbb{R}^d)$ we have $|f|{L{\text{weak}}^p}^* \leq |f|{L^p}$. In particular, the closure of $C_0^\infty$ in $L{\text{weak}}^p$ contains $L^p$. This means that there are functions with full support in the closure of $C_0^\infty$ so I'm not sure how your hint helps (although I may well be missing your point). – Rhys Steele Aug 13 '18 at 09:13
  • I was thinking about an argument similar to what Davide Giraudo just left in another response (didn't want to give away the full story). It's important to consider the $L^p_{weak}$ norm, but it sounds like you're already considering this and still find a problem with the argument? Do you have an argument showing that the closure of $C_0^\infty$ in $L^p_{weak}$ contains $L^p$? The inequality you have shows that $L^p_{weak}$ contains $L^p$. – djkat Aug 14 '18 at 04:57
  • If you were thinking of that argument then there's no problem. I thought that you intended something else by the hint since really that argument uses something about the way $|x|^{-n/p}$ decays and not just its support (As can be seen by the fact that the closure of $C_0^\infty$ contains $L^p$, see my next comment). – Rhys Steele Aug 14 '18 at 08:43
  • In fact, the inequality I gave trivially shows both facts. I didn't give any of these details since they are straightforward but long for a comment. The inequality says that the obvious embedding $e: L^p \hookrightarrow L_{\text{weak}}^p$ is continuous. As a result, by a standard topological fact about continuous maps, $e(L^p) = e(\overline{C_0^\infty}) \subseteq \overline{e(C_0^\infty)}$. Alternatively, you can see this by hand. Take $f \in L^p$ and $f_n \in C_0^\infty$ with $f_n \to f$ in $L^p$. The inequality then gives you that $f_n \to f$ in $L_{\text{weak}}^p$ which is what you want. – Rhys Steele Aug 14 '18 at 08:45