Given the following class of normalised test functions: \begin{equation} \mathcal{T}=\{\phi\in C_c^{\infty}(\mathbb{R}^n)\ |\ \text{supp }\phi \subseteq B(0, 1),\ \|D\phi\|_{\infty}\leq 1\} \end{equation} and a locally integrable function $f$ on $\mathbb{R}^n$, we define the grand maximal function of $f$ as \begin{equation} f^{\ast}(x)\equiv \sup_{t>0}\sup_{\phi\in\mathcal{T}}|\phi_t\ast f(x)|\equiv\left|\int_{\mathbb{R}^n}\frac{1}{t^n}\phi\left(\frac{x-y}{t}\right)f(y)\ \mathrm{d}y\right|\quad(x\in\mathbb{R}^n). \end{equation}

The Hardy-Littlewood maximal function is defined as

\begin{equation} Mf(x)\equiv\sup_{t>0}\frac{1}{|B(x, t)|}\int_{B(x, t)}|f(y)|\ \mathrm{d}y\quad (x\in\mathbb{R}^n). \end{equation}In the following I will assume $f\geq 0$.

I wish to show that $Cf^{\ast}\geq Mf$ for some constant $C$ that depends solely on $n$ and my reasoning is as follows:

Let $\phi\in\mathcal{T}$ such that $0\leq\phi\leq 1$ and $\phi\equiv 1$ on $B(0, 1/2)$. Given $t>0$ and $x\in\mathbb{R}^n$ we have \begin{align} f^{\ast}(x)&\geq \int_{B(x, t)}\phi_t(x-y)f(y)\ \mathrm{d}y\\ &=\int_{B(0, t)}\frac{1}{t^n}\phi\left(\frac{z}{t}\right)f(x-z)\ \mathrm{d}z\\ &\geq \frac{\alpha(n)/2^n}{\alpha(n)(t/2)^n }\int_{B(0, t/2)}f(x-z)\ \mathrm{d}z\\ &=C_1(n)\frac{1}{|B(x, t/2)|}\int_{B(x, t/2)}f(y)\ \mathrm{d}z. \end{align} Since the above inequality holds for all $t>0$ and $x\in\mathbb{R}^n$ we have \begin{equation} C(n)f^{\ast}\geq Mf \end{equation}for $C(n)\equiv1/C_1(n)$.

However, I am not sure if there exists such a $\phi$. I tried to show there exists one by taking any non-negative $\eta\in\mathcal{T}$ and defining $\phi(x)=\eta_{1/2}\ast\chi_{B(0, 1)}(x)$ for $x\in\mathbb{R}^n$. While this function is smooth with compact support in the unit ball and identically one on $B(0, 1/2)$, I'm not sure if $\int_{\mathbb{R}^n}\phi(x)\mathrm{d}x=1$ and I don't think that $\|D\phi\|_{\infty}\leq 1$ or rather I don't know how to show if it is.

I would like to know if the overall idea of my proof is correct and if so how to show that the test function I have used exists.

  • $\begingroup$ Where the assumption $||\phi||_1 = 1$ is used? $\endgroup$ – Voliar Jul 4 '14 at 13:23
  • $\begingroup$ @ Voliar - Sorry, my mistake... I meant $\int \phi(x)\ \mathrm{d}x=1$...I've fixed this in the question. $\endgroup$ – Nirav Jul 4 '14 at 20:44
  • $\begingroup$ @Voliar I was using a positive test function $\phi$ in my work. $\endgroup$ – Nirav Jul 4 '14 at 20:53
  • $\begingroup$ Probably I miss something, but it seems to me that in your approach you don't need $||\phi||_1 = 1$. (at least it is not clear where this assumption is used) $\endgroup$ – Voliar Jul 4 '14 at 21:01
  • $\begingroup$ @Voliar- as in normalised test functions must integrate over $\mathbb{R^n}$ to equal 1. $\endgroup$ – Nirav Jul 4 '14 at 22:56

It is not an answer, but just some logical reasoning about existence of an admissible function.

Consider instead of $\phi(x)$, which you define above, the function $\psi(x) = K \phi(x)$, where $K > 0$ is some constant such that $||D \psi||_\infty = K ||D \phi||_\infty \leq 1$.

Hence, $\psi \in \mathcal{T}$, and you can use $\psi(x)$ to obtain exactly the same estimations for $f^*(x)$ that you have, but with the additional constant $K > 0$ (which, of course, doesn't play any role).

  • $\begingroup$ Yes, I thought of rescaling $\phi$ but then while $\|D\psi\|_{\infty}\leq 1$ I don't have that $\psi$ is a normalised test function. $\endgroup$ – Nirav Jul 5 '14 at 1:11

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