On convergence of random multiple Dirichlet series |
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Author |
andriykuryliak@gmail.com1, olskask@gmail.com2, n-stas@ukr.net3
1) Ivan Franko National University of L’viv, Lviv, Ukraine; 2) Ivan Franko National University of L’viv, Lviv, Ukraine; 3) Ivan Franko National University of L’viv, Lviv, Ukraine
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Abstract |
Let $F_{\omega}(s)=\sum_{\|n\|=0}^\infty
f_{(n)}(\omega)\exp\{(\lambda_{(n)},s)\},$\ where the exponents $\lambda_{(n)}=(\lambda_{n_1}^{(1)},\ldots,\lambda_{n_p}^{(p)})\in\mathbb{R}^p_+$,
$(n)=(n_1,\ldots,n_p)\in\mathbb{Z}^p_+$, $p\in\mathbb{N}$, $\|n\|=n_1+\ldots+n_p$, and the coefficients $f_{(n)}(\omega)$ are pairwise independent random complex variables. In the paper, in particular, we prove the following statements: 1) If $\tau(\lambda)=\varlimsup_{\|n\|\to +\infty}\ln \|n\|/\|\lambda_{(n)}\|=0$, then in order that a Dirichlet series be convergent a.s. in the whole space $\mathbb{C}^p$, it is necessary and sufficient that
$$
(\forall \Delta>0)\colon\ \sum\nolimits_{\|n\|=0}^{+\infty}\big(1-F_{(n)}\big(\exp(-\Delta\|\lambda_{(n)}\|)\big)\big)\le+\infty.
$$
2)
If $\tau(\lambda)=0$, then in order that $\sigma\in \partial G_{a}\cap\big(\mathbb{R}_+\setminus\{0\}\big)^p$ a.s., it is necessary and sufficient that
$$(\forall \varepsilon>0)\colon\
\sum_{\|n\|=0}^{+\infty}\Big(1-F_{(n)}\big(e^{(-1+\varepsilon)(\sigma,\lambda_{(n)})}\big)\Big)\le+\infty\ \wedge\ \sum_{\|n\|=0}^{+\infty}\Big(1-F_{(n)}\big(e^{(-1-\varepsilon)(\sigma,\lambda_{(n)})}\big)\Big)=+\infty,$$
where $ F_{(n)}(x):=P\{ \omega : |f_{(n)}(\omega)|\le x\},\ x\in \mathbb R,\ (n)\in\mathbb{Z}^p_+$ is the distribution function of $|f_{(n)}(\omega)|$,
$\partial G_a$ is the set of conjugate abscissas of absolute convergence of the random Dirichlet series $F_{\omega}$.
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Keywords |
multiple Dirichlet series; conjugate abscissas of convergence; random exponents
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DOI |
doi:10.15330/ms.49.2.122-137
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Reference |
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Pages |
122-137
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Volume |
49
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Issue |
2
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Year |
2018
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Journal |
Matematychni Studii
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Full text of paper | |
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