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Suppose that

  1. $\mu_k$ is an increasing sequence of numbers such that $0 < \mu_1 \leq \mu_2 \leq ..$ with $\mu_k \to \infty$ as $k \to \infty$

  2. $\sum_{k=1}^\infty |u_k|^2 < \infty$ and $\sum_{k=1}^\infty \sqrt{\mu_k}|u_k|^2 < \infty$ where $u_k$ is a given sequence of real numbers

I want to show that (if true) the sum $$\sum_{k=1}^\infty |u_k|^2 \mu_k \frac{\cosh(2\sqrt{\mu_k}(T-t))}{\sinh^2(\sqrt{\mu_k}T)}$$ is uniformly convergent in the variable $t \in [\epsilon, T]$ for $\epsilon > 0$? (in the previous version of this thread I forgot to exclude $t=0$.)

The problem is that the numerator contains an exponential term with the "wrong" sign.

The motivation is, I want to integrate this sum term by term because it gives me a bound on a norm (the sum comes from a solution to an differential equation). Thanks for help.

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The answer is negative.

Take $\mu_{k}=k$, $u_{k}=\frac{1}{k}$ and $t=0$. Conditions 1 and 2 hold. Notice that $\frac{\cosh(2\sqrt{\mu_{k}}T)}{\sinh^{2}(\sqrt{\mu_{k}}T)}=1+o(1)$ and the sum diverges.

If you want it to be uniformly convergent you need either $\sum |u_{k}|^{2}\mu_{k}<\infty$ or $t\in [\varepsilon,T]$ for some $\varepsilon >0$.

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  • $\begingroup$ Thank you. I stupidly forgot to exclude $t=0$. Can you explain your last sentence a little bit, why is it we get uniform convergences for $[\epsilon, T]$? This is what I was having trouble with. $\endgroup$
    – Eson
    Jun 26, 2015 at 13:46
  • $\begingroup$ $\cosh$ attains its maximum when $t=\varepsilon$ therefore $\frac{\cosh(2\sqrt{\mu_{k}}(T-\varepsilon))}{\sinh^{2}(\sqrt{\nu_{k}}T)}=e^{-2 \sqrt{\mu_{k}} }\varepsilon}+o(1)$ and since $\mu_{k}e^{-2\sqrt{\mu_{k}}\varepsilon}<1$ for sufficiently large $k$ the series is dominated by $\sum |u_{k}|^{2}$ $\endgroup$
    – Paata Ivanishvili
    Jun 26, 2015 at 13:52
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    $\begingroup$ I am sorry I had some problems with typing tex. Here is why: $\cosh$ attains its maximum when $t=\varepsilon$ therefore $\frac{\cosh(2\sqrt{\mu_{k}}(T-\varepsilon))}{\sinh^{2}(\sqrt{\mu_{k}}T)}=e^{- 2\sqrt{\mu_{k}}\varepsilon}+o(1)$ and since $\mu_{k}e^{-2\sqrt{\mu_{k}}\varepsilon}<1$ for sufficiently large $k$ the series is dominated by $\sum |u_{k}|^{2}$ $\endgroup$
    – Paata Ivanishvili
    Jun 26, 2015 at 13:59
  • $\begingroup$ In my last comment it should be $o(e^{-2\sqrt{\mu_{k}}\varepsilon})$ instead of $o(1)$. $\endgroup$
    – Paata Ivanishvili
    Jun 26, 2015 at 14:14

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