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Let $V$ be the set of sequences $a \in\mathbb{R}^\mathbb{N}$ such that $\lim_{n\to\infty} a_n = 0$. The set $V$ can be seen as a real vector space, with pointwise addition and scalar multiplication.

For $a\in V$ we define the "associated $\zeta$-function" to be \begin{eqnarray} \zeta(a) := \sum_{n=1}^\infty \frac{1}{n^{1+|a_n|}} \end{eqnarray} (similar to post https://mathoverflow.net/questions/187610/almost-zeta-function)

Let $D = \{a\in V: \zeta(a) = \infty\}$. Is $D\subseteq V$ a subspace?

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  • $\begingroup$ What do you want to do about $D$ not containing $\vec 0$? $\endgroup$
    – Allen Knutson
    Commented Nov 22, 2014 at 15:43
  • $\begingroup$ $D$ does contain the constant 0 sequence because the harmonic series diverges. $\endgroup$
    – Dominic van der Zypen
    Commented Nov 22, 2014 at 16:15
  • $\begingroup$ Oops, of course. $\endgroup$
    – Allen Knutson
    Commented Nov 22, 2014 at 19:19

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Define $$ a_n=\begin{cases}\frac{2\log\log n}{\log n}&:&\text{$n$ is odd,}\\0&:&\text{$n$ is even,}\end{cases}\hspace{1cm}b_n=\begin{cases}0&:&\text{$n$ is odd,}\\\frac{2\log\log n}{\log n}&:&\text{$n$ is even.}\end{cases} $$ It's not hard to check that $\zeta(a)=\zeta(b)=\infty$, but $\zeta(a+b)<\infty$, so $D$ is not closed under addition.

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