Define the uncentered Hardy-Littlewood maximal operator $M$ by $$Mf(x):=\sup_{x\in B}\dfrac{1}{\left|B\right|}\int_{B}\left|f\right|,$$ where we the supremum is taken over all (open) balls $B$ containing the point $x\in\mathbb{R}^{n}$. Let $f_{0}$ denote the characteristic function of the unit ball. I am trying to show that for $1<p<\infty$,
$$\lim_{n\rightarrow\infty}\dfrac{\left\|Mf_{0}\right\|_{L^{p}}}{\left\|f_{0}\right\|_{L^{p}}}=\infty > \tag{1}$$
by following the suggestion in Exercise 2.1.8 of L. Grafakos, Classical Fourier Analysis (Third Edition). Actually, I would like to show that the growth of the $L^{p}\rightarrow L^{p}$ is exponential in the dimension $n$, but I'll settle for the above.
Let $B_{0}$ denote the unit ball, and for $\left|x\right|>1$, let $B_{x}$ be the ball with center $c$ and radius $r$ given by $$c=\dfrac{(\left|x\right|-\left|x\right|^{-1})}{2}\dfrac{x}{\left|x\right|},\quad r=\dfrac{(\left|x\right|+\left|x\right|^{-1})}{2}$$
Observe that $x\in B_{x}$, so $$Mf(x)\geq\dfrac{1}{\left|B_{x}\right|}\int_{B_{x}}\left|f_{0}\right|=\dfrac{2^{n}}{(\left|x\right|+\left|x\right|^{-1})^{n}\left|B_{0}\right|}\left|B_{x}\cap B_{0}\right|$$ Since \begin{align*} \dfrac{1}{4}\left(\left|x\right|-\left|x\right|^{-1}\right)^{2}+1&=\dfrac{\left|x\right|^{2}}{4}+\dfrac{\left|x\right|^{-2}}{4}+\dfrac{1}{2}=\left(\dfrac{\left|x\right|+\left|x\right|^{-1}}{2}\right)^{2}, \end{align*} we see that $B_{x}\cap B_{0}$ contains $B_{x}\cap B_{0}$ contains a "half-ball" of $B_{1}(0)$and a hyperspherical cap of height $\left|x\right|^{-1}$ and radius $(\left|x\right|+\left|x\right|^{-1})/2$, which we denote by $C_{x}$. In particular, $$\left|B_{x}\cap B_{0}\right|\geq\dfrac{1}{2}\left|B_{0}\right|+\left|C_{x}\right|$$ My initial thought was to simplify things and ignore the term $\left|C_{x}\right|$ to get the lower bound $$Mf(x)\geq\dfrac{2^{n-1}}{(\left|x\right|+\left|x\right|^{-1})^{n}},\quad\forall \left|x\right|>1$$ Thus, \begin{align*} \dfrac{\left\|Mf\right\|_{L^{p}}}{\left\|f_{0}\right\|_{L^{p}}}&\geq\dfrac{2^{n-1}}{(n^{-1}\omega_{n-1})^{1/p}}\left(\int_{\left|x\right|>1}\dfrac{1}{(\left|x\right|+\left|x\right|^{-1})^{np}}dx\right)^{1/p}\\ &=\dfrac{2^{n-1}}{(n^{-1}\omega_{n-1})^{1/p}}\left(\int_{1}^{\infty}\dfrac{r^{n-1}}{(r+r^{-1})^{np}}dr\int_{S^{n-1}}d\theta dr\right)^{1/p}\\ &=2^{n-1}n^{1/p}\left(\int_{1}^{\infty}\dfrac{r^{n-1}}{(r+r^{-1})^{np}}dr\right)^{1/p} \end{align*} I'm having trouble giving a useful asymptotic for the integral factor above. Perhaps, I errored in ignoring $\left|C_{x}\right|$. I would greatly appreciate any suggestions in how to proceed.
Edit 1: Plugging in some special values of $p$ into Wolfram Alpha, it looks like hypergeometric functions might be relevant here.
I've met this question recently. Yet I solved it by the same means: $$||\tilde M||_p^p\geq \frac{\int(\tilde{M}f_0(x))^pdx}{\int f_0(x)^pdx}\geq\frac{2^{p(n-1)}}{V_n}\int_{x\geq 1}(|x|+|x|^{-1})^{-pn}dx=I$$ The only question is to estimate the integral. To this end, we need it to take a "suitable" form: $$I=n2^{p(n-1)}\int_{r\geq1}\frac{r^{pn+n-1}}{(1+r^2)^{pn}}dr\\=n2^{p(n-1)}\int_{\pi/4}^{\pi/2}(\cos\theta)^{pn-n-1}(\sin\theta)^{pn+n-1} d\theta\\=n2^{1-p}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2\sin \theta \cos \theta)^{pn-1}(\tan \theta)^nd\theta.$$ Observing that $\tan\theta\geq1$ and $\sin2\theta=2\sin \theta\cos\theta$, we obtain $$I\geq n2^{-p}\int_{0}^{\frac{\pi}{2}} (\sin u)^{pn-1} du.$$ So $||\tilde{M}||_p\rightarrow \infty$ as $n\rightarrow \infty$ due to Wallis formula, and the growth is much slower than "exponential".