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Let $\chi$ a non negative continuous function whose support is included in $]-1,1[$ with $\int\chi=1$. Let $\varphi$ a function in $L^{p'}$ non negative with compact support in $\mathbb{R}\setminus\{0\}$. Let $q,q'\in]1,+\infty[$ such that $\frac1q+\frac1{q'}=1$.

  1. Show that $$\lim\limits_{n\to\infty}n\int_{\mathbb{R}^2}\chi(nt')\varphi(t)\frac{1}{\vert t-t'\vert^{\frac{1}{q}}}dt'dt=\int_{\mathbb{R}}\varphi(t)\frac{1}{\vert t\vert^{\frac{1}{q}}}dt$$
  1. Deduce that there does not exist a positive constant $C$ such that $$\forall(f,g)\in L^1\times L_\omega^q,~\Vert f\star g\Vert_{L^q}\leqslant C\Vert f\Vert_{L^1}\Vert g\Vert_{L_\omega^q}$$ where we denote $L_\omega^p$ the weak $L^q$ space endowed with the norm $\Vert g\Vert_{L_\omega^q}=\sup\limits_{\lambda>0}\lambda^q\mu([\vert g\vert >\lambda])$.

  2. Deduce that there does not exist a positive constant $C$ such that $$\forall(f,g)\in L^{q'}\times L_\omega^q,~\Vert f\star g\Vert_{L^q}\leqslant C\Vert f\Vert_{L^{q'}}\Vert g\Vert_{L_\omega^q}$$

My work

  1. Suppose $Supp(\varphi)\subset[-K,-\varepsilon]\cup[\varepsilon,K]$ for $\varepsilon,K>0$. Let $\chi_n$ defined as $\chi_n(x)=n\chi(nx)$. Then $Supp(\chi_n)\subset]-\frac1n,\frac1n[$. $\chi_n$ is an approximation of unity derived from $\chi$. Let $n>\frac2\varepsilon$ so that for any $(t,t')\in Supp(\phi)\times Supp(\chi_n)$, $\vert t-t'\vert>\frac\varepsilon2$. Note that if $m\geqslant n$, this is still true for all $\chi_m$. Define $g:t\mapsto\frac{1}{\vert t\vert^{\frac1q}}$ for $t\in[-K-\frac\varepsilon2,-\frac\varepsilon2]\cup[\frac\varepsilon2,K+\frac\varepsilon2]$ and $0$ everywhere else. Then $g$ is in every $L^p(\mathbb{R})$, $1\leqslant p\leqslant +\infty$.

In this way, the integral has a meaning and we have: $$n\int_{\mathbb{R}^2}\chi(nt')\varphi(t)\frac{1}{\vert t-t'\vert^{\frac{1}{q}}}dt'dt=\int_{\mathbb{R}}\varphi(t)\mathbf{1}_{[-K,-\varepsilon]\cup[\varepsilon,K]}(t)\int_{\mathbb{R}}\chi_n(t)\mathbf{1}_{[-\frac\varepsilon2,\frac\varepsilon2]}(t')\frac1{\vert t-t'\vert^{\frac1q}}dt'dt=\int_{\mathbb{R}}\varphi(t)\int_{\mathbb{R}}\chi_n(t)g(t-t')dt'dt=\int_{\mathbb{R}}\varphi(t)[\chi_n\star g](t)dt$$

Because $g\in L^1(\mathbb{R})$ and $\chi_n$ is an approximation of unity, $\chi_n\star g\xrightarrow[n\to+\infty]{L^1}g$. This means that, up to an extraction, $\chi_n\star g\xrightarrow[n\to+\infty]{} g$ almost everywhere.

In this way, we can apply the dominated convergence theorem: $\varphi(t)[\chi_n\star g](t)$ converges almost everywhere to $\varphi(t) g(t)$. And for almost all $t$, we have thanks to the Young inequality: $$\vert\varphi(t)[\chi_n\star g](t)\vert\leqslant\vert\varphi(t)\vert\Vert\chi_n\star g\Vert_{\infty}\leqslant\vert\varphi(t)\vert\underbrace{\Vert\chi_n\Vert_{L^1}}_{=1}\Vert g\Vert_{L^\infty}$$ and $\vert\varphi\vert\Vert g\Vert_{L^\infty}$ is integrable thanks to the Hölder inequality: $$\int_{R}\vert\varphi(t)\Vert g\Vert_{L^\infty}\leqslant\Vert g\Vert_{L^\infty}\Vert\varphi\Vert_{L^{q'}}\Bigl(\mu(Supp(\varphi)\Bigr)^{\frac1q}$$

So we get the wanted convergence.

  1. First we show that $\vert\cdot\vert^{\frac{-1}q}\in L_\omega^q(\mathbb{R})$. Indeed, let $\lambda>0$. $$\lambda^q\mu([\vert\cdot\vert>\lambda])=\lambda^q\mu([\vert\cdot\vert<\frac{1}{\lambda^q}])=2$$

Then, suppose this constant $C$ exists. Then from the first question we have $\chi_n\star g\in L^q$ as $\chi_n\in L^1$ and $$\int_{\mathbb{R}}\varphi(t)[g\star\chi_n](t)dt\leqslant C\Vert\varphi \Vert_{L^{q'}}\Vert g\Vert_{L_\omega^q}$$ which means that, when $n$ goes to infinity, $$\int_\mathbb{R}\varphi(t)g(t)dt\leqslant C\Vert\varphi \Vert_{L^{q'}}\Vert g\Vert_{L_\omega^q}$$

I tried using some well thought $\varphi$ to find a contradiction but could not find one. For example, I tried characteristic functions, but it does not lead to a contradiction. Any hints ?

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    $\begingroup$ If such a constant exists then, by reflexivity of $L^q$ for $1<q<\infty$, we see that $\chi_n*|\cdot|^{-1/q}$ converges weakly to some element in $L^q$ as $n\to \infty$. But the computations in the first part show that, as $n\to \infty$, the same quantity tends, in the sense of distributions, to $|\cdot|^{-1/q}$ outside the origin. This is a contradiction. $\endgroup$ – Jose27 Sep 24 at 6:36
  • $\begingroup$ How is this a contradiction ? Can not both limits be the same ? $\endgroup$ – Flewer47 Sep 24 at 9:08
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    $\begingroup$ No, because that function is not in $L^q$. $\endgroup$ – Jose27 Sep 24 at 15:03

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