# Inequality for two Hardy-Littlewood Maximal Type Operators

$$\textbf{The Problem:}$$ Let $$Mf(x)=\sup\limits_{x\in B}\frac{1}{m(B)}\int_{B}\vert f\vert\quad\text{for }x\in\mathbb R^d$$ denote the Hardy-Littlewood maximal function, where the supremum is taken over all Euclidean balls $$B\in\mathbb R^d$$. Let $$\|\cdot\|_{\ast}$$ be a norm on $$\mathbb R^d$$ and define $$\tilde{M}f(x)=\sup\limits_{r>0}r^{-d}\int_{\mathbb R^d}\vert f(x+y)\vert\mathbf1_{\|y\|_{\ast}\leq r}dy.$$ Prove that there exists a constant $$c\in(0,\infty)$$ depending only on $$d$$ and $$\|\cdot\|_{\ast}$$ such that $$Mf(x)\leq c\tilde{M}f(x)$$ for all $$x\in\mathbb R^d$$ and $$f\in L^{1}_{\text{loc}}(\mathbb R^d).$$

$$\textbf{My Thoughts:}$$ Let $$B\subset\mathbb R^d$$ be an arbitrary but fixed ball containing $$x$$. Then we can choose a ball $$B_{r}'(x)$$ centered at $$x$$ with twice the radius of $$B$$. Then we have with $$m(B_{1}(0))=v,$$ \begin{align*}\frac{1}{m(B)}\int_{B}\vert f\vert\leq\frac{1}{m(B)}\int_{B'}\vert f\vert\leq\frac{2^d}{m(B')}\int_{B'}\vert f\vert=\frac{2^d r^{-d}}{v}\int_{B'}\vert f\vert. \end{align*} Since $$\mathbb R^d$$ is a finite dimensional normed vector space, all norms on it are equivalent, that is there are positive constants such that $$c_{1}\|y\|_{\ast}\leq\|y\|_{2}$$. It follows that $$\large\frac{2^d r^{-d}}{v}\int_{B'}\vert f\vert\leq \frac{2^{d}r^{-d}}{v}\int_{\|y\|_{\ast}\leq\frac{r}{c}}\vert f(x+y)\vert dy,$$ so putting $$c=\frac{2^d}{v}$$ and taking supremums we have what we need.

Is my proof above correct? Any comments are welcomed, be it about the style, lack of details, and most important, the correctness.

Thank you for your time.