# Local Cancellation in Real Hardy Space

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.

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