Singular integrals of stable subordinator

Abstract

It is well known that ${\int }_0^1{t}^{-\theta }dt<\infty$ for $\theta \in (0,1)$ and ${\int }_0^1{t}^{-\theta }dt=\infty$ for $\theta \in [1,\infty )$. Since $t$ can be taken as an $\alpha$-stable subordinator with $\alpha =1$, it is natural to ask whether ${\int }_0^1{t}^{-\theta }d{S}_t$ has a similar property when $S_t$ is an $\alpha$-stable subordinator with $\alpha \in (0,1)$. We show that $\theta =\frac{1}{\alpha }$ is the border line such that ${\int }_0^1{t}^{-\theta }d{S}_t$ is finite a.s. for $\theta \in (0,\frac{1}{\alpha })$ and blows up a.s. for $\theta \in [\frac{1}{\alpha },\infty )$. When $\alpha =1$, our result recovers that of ${\int }_0^1{t}^{-\theta }dt$. Moreover, we give a $p$-th moment estimate for the integral when $\theta \in (0,\frac{1}{\alpha })$.

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