# Convolution and weak $L^p$ spaces

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 ?

• 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. – Jose27 Sep 24 at 6:36
• How is this a contradiction ? Can not both limits be the same ? – Flewer47 Sep 24 at 9:08
• No, because that function is not in $L^q$. – Jose27 Sep 24 at 15:03