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We fix $p \in [1, \infty)$. We have for every $f \in L^p (\mathbb R^d)$ that $\lim_{|z| \to \infty} \|1_{B(z, 1)} f\|_{L^p} =0$. I wonder if there is an estimate of above decay, i.e.,

Is there a measurable function $\varphi : \mathbb R_+ \to \mathbb R_+$ and constants $c, \delta \ge 0$ (possibly depending on $d, p$) such that $\lim_{r \to \infty} \varphi (r) = \infty$ and that $$ \|1_{B(z, 1)} f\|_{L^p} \le \frac{c(1+\|f\|_{L^{p+\delta}})}{\varphi (|z|)}, \quad \forall f \in L^{p+\delta} (\mathbb R^d), z \in \mathbb R^d? $$

Thank you so much for your elaboration!

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Such a bound does not exist: Consider $f=f_z:=1_{B(z,1)}$.

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  • $\begingroup$ I am sorry for being sloppy. I meant the inequality to be satisfied for all $f \in L^{p+\delta} (\mathbb R^d)$ and $z \in \mathbb R^d$... $\endgroup$
    – Akira
    Commented Mar 4 at 21:58
  • $\begingroup$ Yes, this is how I understood it. The answer shows that such an inequality cannot hold. $\endgroup$
    – Iosif Pinelis
    Commented Mar 4 at 21:59

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