Muckenhoupt Weights and Weighted Estimates — Examples

If $\alpha \le -n$, then ${\int }_{B(0,1)}\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }dx=+\mathrm{\infty }$ by [F2], so $w$ is not locally integrable and hence not a weight. If $\alpha >-n$, then $w$ is locally integrable (apply [F2] on each ball, using [F3] to compare ${\int }_{B(x_0,R)}\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }$ with the radial integral), it is positive and finite off the origin, and it is assigned the value $0$ at the single point $0$ of measure zero; hence $w$ is a weight.

Let $\alpha >-n$ and let $B=B(x_0,R)$ be a ball with $\mathrm{\mid }{x}_0\mathrm{\mid }\ge 3R$. On $B$ one has $(2\mathrm{/}3)\mathrm{\mid }{x}_0\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }\le (4\mathrm{/}3)\mathrm{\mid }{x}_0\mathrm{\mid }$, so both ${\int }_B\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }dx$ and ${\int }_B\mathrm{\mid }x{\mathrm{\mid }}^{-\alpha \mathrm{/}(p-1)}dx$ are comparable to $\mathrm{\mid }{x}_0{\mathrm{\mid }}^{\alpha }{R}^n$ and $\mathrm{\mid }{x}_0{\mathrm{\mid }}^{-\alpha \mathrm{/}(p-1)}{R}^n$ respectively, and the defining product is bounded by a constant depending only on $n,\alpha ,p$.

Let $\alpha >-n$ and let $B=B(x_0,R)$ with $\mathrm{\mid }{x}_0\mathrm{\mid }<3R$. By [F3] the two integrals ${\int }_B\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }dx$ and ${\int }_B\mathrm{\mid }x{\mathrm{\mid }}^{-\alpha \mathrm{/}(p-1)}dx$ are comparable to $R^{n+\alpha }$ and $R^{n-\alpha \mathrm{/}(p-1)}$ provided both exponents exceed $-n$; the normalized product $⟨w{⟩}_B⟨w^{-1\mathrm{/}(p-1)}{⟩}_B^{p-1}$ is then comparable to $R^{\alpha }{R}^{-\alpha \mathrm{/}(p-1)\cdot (p-1)}=1$, uniformly in $B$. If $-\alpha \mathrm{/}(p-1)\le -n$, that is $\alpha \ge n(p-1)$, the second exponent does not exceed $-n$ and the corresponding integral over $B(0,R)$ diverges, so the product is $+\mathrm{\infty }$ on the ball $B(0,R)$.

Steps 1.1–2.1 give the stated weight and $A_p$ ranges for ball averages; the ball/cube comparison [F3] gives the same result for the defined cube characteristic. For doubling, if $\mathrm{\mid }{x}_0\mathrm{\mid }<3R$, the integral on $B(x_0,2R)$ is bounded above by the radial integral on $B(0,5R)$, of order $R^{n+\alpha }$, while [F3] bounds the integral on $B(x_0,R)$ below by a positive multiple of that order. If $\mathrm{\mid }{x}_0\mathrm{\mid }\ge 3R$, both balls have $\mathrm{\mid }x\mathrm{\mid }$ comparable to $\mathrm{\mid }{x}_0\mathrm{\mid }$ (on the larger ball, $\mathrm{\mid }{x}_0\mathrm{\mid }\mathrm{/}3\le \mathrm{\mid }x\mathrm{\mid }\le 5\mathrm{\mid }{x}_0\mathrm{\mid }\mathrm{/}3$); their integrals are therefore comparable up to a fixed constant. Hence $w(B(x_0,2R))\le C_{n,\alpha }w(B(x_0,R))$ for every $\alpha >-n$.

At the upper endpoint: for $\alpha =n(p-1)$ one has $-\alpha \mathrm{/}(p-1)=-n$, so polar coordinates give ${\int }_{B(0,R)}\mathrm{\mid }x{\mathrm{\mid }}^{-\alpha \mathrm{/}(p-1)}dx={\int }_{B(0,R)}\mathrm{\mid }x{\mathrm{\mid }}^{-n}dx=\mathrm{\mid }{\mathbb{S}}^{n-1}\mathrm{\mid }{\int }_0^R{r}^{-1}dr=+\mathrm{\infty }$ for every $R>0$ by the logarithmic divergence of ${\int }_0^1{r}^{-1}dr$. The second factor of the defining product $⟨\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }{⟩}_Q⟨\mathrm{\mid }x{\mathrm{\mid }}^{-\alpha \mathrm{/}(p-1)}{⟩}_Q^{p-1}$ is therefore $+\mathrm{\infty }$ on the cube $Q=Q(0,R)$ containing $B(0,R)$, so the defining supremum is $+\mathrm{\infty }$ and $\mathrm{\mid }x{\mathrm{\mid }}^{n(p-1)}$ is not in $A_p$, while it is locally integrable and hence a weight.

At the lower endpoint: for $\alpha =-n$ polar coordinates give ${\int }_{B(0,R)}\mathrm{\mid }x{\mathrm{\mid }}^{-n}dx=\mathrm{\mid }{\mathbb{S}}^{n-1}\mathrm{\mid }{\int }_0^R{r}^{-1}dr=+\mathrm{\infty }$, so $\mathrm{\mid }x{\mathrm{\mid }}^{-n}\mathrm{\notin }{L}_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$ and no $A_p$ membership is defined; this is the same logarithmic divergence of the radial integral ${\int }_0^R{r}^{-1}dr$.

Steps 1.1 and 1.2 show the failure at both endpoints, so the range $-n<\alpha <n(p-1)$ determined in The A_p range of a power weight is exactly the open admissible interval and cannot be enlarged.

Let $-n<\alpha \le 0$ and let $B(y,r)\ni x\mathrm{\ne }0$. If $\mathrm{\mid }x\mathrm{\mid }\ge 4r$, then $\mathrm{\mid }z-x\mathrm{\mid }<2r$ for $z\in B(y,r)$, so $\mathrm{\mid }z\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$ and $\mathrm{\mid }z{\mathrm{\mid }}^{\alpha }\le 2^{-\alpha }\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }$. If $\mathrm{\mid }x\mathrm{\mid }<4r$, then $B(y,r)\subseteq B(0,\mathrm{\mid }x\mathrm{\mid }+2r)\subseteq B(0,6r)$; [F2] bounds its integral by $C_{n,\alpha }{r}^{n+\alpha }$ and $r^{\alpha }\le 4^{-\alpha }\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }$. Dividing by the ball volume gives an average bounded by $C_{n,\alpha }^{\mathrm{\prime }}\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }$ in both cases. Taking the supremum gives the uncentred bound, hence also the centred bound and $A_1$ membership.

For $\alpha >0$ and a cube $Q$ centred at the origin, ${\mathrm{ess\hspace{0.17em}inf}}_Q\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }=0$ because every positive threshold has a subball around the origin on which $\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }$ lies below it; that subball has positive Lebesgue measure, while $⟨\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }{⟩}_Q>0$; hence the cube-average/essential-infimum form of $A_1$ fails on $Q$, and by [F1] the pointwise form fails as well.

For $\alpha \le -n$ the function is not locally integrable, hence not a weight, by The A_p range of a power weight. Together with steps 1.1 and 1.2 this shows that $A_1$ membership holds exactly for $-n<\alpha \le 0$, while the doubling range for the measure $\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }dx$ is the strictly larger interval $\alpha >-n$ (the direct doubling computation in The A_p range of a power weight, the final verification step).

Direct computation: $\mathrm{\parallel }{\mathbf{1}}_{(0,r)}{\mathrm{\parallel }}_{L^p(w)}^p={\int }_0^r\mathrm{\mid }x{\mathrm{\mid }}^{\alpha }dx={\int }_0^r{x}^{\alpha }dx=r^{\alpha +1}\mathrm{/}(\alpha +1)$ for $\alpha >-1$ by [F2], since $\mathrm{\mid }x\mathrm{\mid }=x$ on $(0,r)$; taking $p$-th roots gives the displayed formula.

As $r\to 0^{+}$ the expression $(r^{\alpha +1}\mathrm{/}(\alpha +1){)}^{1\mathrm{/}p}$ tends to $0$ because $\alpha +1>0$; as $r\to \mathrm{\infty }$ it grows like the constant multiple $r^{(\alpha +1)\mathrm{/}p}$ of the power function. At $\alpha =-1$ one has ${\int }_0^r{x}^{-1}dx=+\mathrm{\infty }$ by [F2], so for fixed $r>0$ the full expression $(r^{\alpha +1}\mathrm{/}(\alpha +1){)}^{1\mathrm{/}p}$ diverges as $\alpha \downarrow -1$, since $r^{\alpha +1}\to 1$. The density $\mathrm{\mid }x{\mathrm{\mid }}^{-1}$ does not satisfy this page's local-integrability definition of a weight; its integral of the indicator is nevertheless well defined and infinite.