I want to show the following asymtotic estimate in Hardy space over $\mathbb{R}^n$: Let $a\in \mathbb{R}^n$. I want to show the function $$ f(x)=\mathbb{1}_{B(0,1)}-\mathbb{1}_{B(a,1)} $$ is asymtotic to $\ln|a|$; that is, $$ ||f||_{\mathcal{H}^1}\sim \ln|a| $$ The definitions we use here is the following: If $\varphi\in C_c^\infty(B(0,1))$, then there is a maximal operator $$ M_\varphi(f)(x)=\sup_t |\varphi_t*f| = \sup \frac{1}{t^n}\int_{\mathbb{R}^n}\varphi\left(\frac{x-y}{t}\right) f(y)dy $$ where $\varphi_t:=\varphi(x/t)/t^n$ and $\varphi$ satisfies $$ \int_{B(0,1)}\varphi = 1, \varphi\geq 0,\mbox{ and }\varphi = 1/\mathcal{L}(B(0,1))\;\forall x\in B(0,1/2) $$ The Hardy norm of $f$ is defined to be the $L^1$-norm of this maximal operator evaluated at $f$.

It can be shown that $||f||_{\mathcal{H}^1}/\ln|a|$ is certainly bounded by showing $M_\varphi(f)(x)\geq c|x|^{-n}$, however the upper bound is the difficulty.

  • $\begingroup$ Do you mean $a\in\mathbb{R}^n$? $\endgroup$ – Piotr Hajlasz Jan 5 at 16:46
  • $\begingroup$ Yes, I'll correct this. $\endgroup$ – user495490 Jan 6 at 1:07

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