稳定从属子的奇异积分
Singular integrals of stable subordinator
摘要：
众所周知，$\int_{0}^{1} t^{-θ} d t<\infty$用于$θ\in (0,1)$，$\int_{0}^{1} t^{-θ} d t=\infty$用于$θ\in [1,\infty)$。由于$t$可以被看作是$α=1$的$α$稳定从属子，所以很自然地要问，当$S_{t}$是$α\in (0,1)$的$α$稳定从属子时，$\int_{0}^{1} t^{-θ} d S_{t}$是否具有类似的性质.我们证明了$θ=\frac 1α$是使得$\int_{0}^{1} t^{-θ} d S_{t}$是有限a.s.的边界线.为$θ\in (0, \frac 1α)$和炸毁了为$θ\in [\frac1α,\infty)$。当$α=1$时，我们的结果恢复了$\int_{0}^{1} t^{-θ} d t$的结果。当$θ\in (0,\frac 1α)$满足时，给出了积分的$p$阶矩估计.
英文摘要：
It is well known that $\int_{0}^{1} t^{-θ} d t<\infty$ for $θ\in (0,1)$ and $\int_{0}^{1} t^{-θ} d t=\infty$ for $θ\in [1,\infty)$. Since $t$ can be taken as an $α$-stable subordinator with $α=1$, it is natural to ask whether $\int_{0}^{1} t^{-θ} d S_{t}$ has a similar property when $S_{t}$ is an $α$-stable subordinator with $α\in (0,1)$. We show that $θ=\frac 1α$ is the border line such that $\int_{0}^{1} t^{-θ} d S_{t}$ is finite a.s. for $θ\in (0, \frac 1α)$ and blows up a.s. for $θ\in [\frac1α,\infty)$. When $α=1$, our result recovers that of $\int_{0}^{1} t^{-θ} d t$. Moreover, we give a $p$-th moment estimate for the integral when $θ\in (0,\frac 1α)$.
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