# 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.