I have a Borel non-decreasing measure $\mu$ such that

$$ \int_{-\infty}^{+\infty}\left|\sum_{i=1}^n \xi_i e^{-y_i t}\right|^2 d\mu(t)\geq 0 $$ and finite for every $n\in\mathbb{N}$, every $\{\xi_i\}_{i=1\ldots n}$ complex sequence and every $\{y_i\}_{i=1\ldots n}$ real sequence different from (0,...0).

I can conclude that $\mu(\mathbb{R})>0$, but can I say that $\mu(\mathbb{R})<+\infty$? I would like to have finite $\mu$ in order to prove that $\mu$ is Borel-finite and non-negative.

  • $\begingroup$ What about $n=1$, $\xi_1 = 1$, $y_1 = 0$, then by assumption $\int_{\mathbb R} 1\, d\mu = \mu(\mathbb R)$ is positive and finite?! $\endgroup$ – martini Jun 18 '13 at 15:52
  • $\begingroup$ And if my assumption is true for every $\{y_i\}_{i=1\ldots n}$ different from the (0,...,0) vector? can you still prove that $\mu$ is finite? $\endgroup$ – alemou Jun 18 '13 at 16:03
  • $\begingroup$ Let $n = 1$, $\xi_1 = 1$, $y_1 = \pm 1$. Then $|\xi_1 e^{-ty_1}|^2 = e^{\pm 2t}$. As $1 \le e^{2t} + e^{-2t}$, we have $\int_{\mathbb R} 1 \le \int e^{2t} + e^{-2t} < \infty$. $\endgroup$ – martini Jun 18 '13 at 21:20

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