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Given a sequence of complex numbers $\{a_n\}_n$, one says that this sequence admits $a$ as a sequential density if $$\underset{N_s\to\infty}{\lim}\frac{1}{N_s}\sum_{n=1}^{N_s} a_n = a$$ where $N_s = 2^{2^s}$ for instance. The sequence admits $a$ as a logarithmic density if $$\underset{N\to\infty}{\lim}\frac{1}{\log N}\sum_{n=1}^N \frac{1}{n} a_n = a.$$ If the the sequence has a sequential density, does it also have a logarithmic one, with the same limit ?

Motivation: In a paper called Problems of Almost Everywhere convergence related to harmonic analysis and number theory, Jean Bourgain states that for any function $f$ in $L^2(\mathbf{T})$ the sequence of its Riemann sums $\{R_nf(x)\}$ admits $\int_0^1f$ as a logarithmic density. In the proof, it turns out that Bourgain reduces this convergence to an $L^2$-maximal inequality

$$\left\Vert\underset{N_s=2^{2^s}}{\sup}\frac{1}{N}\left\vert\sum_{n=1}^NR_nf\right\vert\right\Vert_2\leq C \left\Vert f \right\Vert_2 $$

which actually proves the sequential density of the sequence $\{R_nf(x)\}$ for almost every $x$ in $\mathbf{T}$.

In number theory, the Davenport-Erdös theorem states the equivalence of this two notions of densities for sets. See this post for instance. I don't know where to find a proof of this theorem in order to adapt it for sequences.

Thank you in advance

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  • $\begingroup$ Also posted on math.stackexchange $\endgroup$ – Wulfenite Aug 10 at 14:30
  • $\begingroup$ So it is considered impolite to simultaneously post on both sites. If you're not sure, try MSE first, and if there is no answer in 2 or 3 days, consider posting on MO. The issue is that this leads to wasted effort. $\endgroup$ – Anthony Quas Aug 10 at 16:17
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    $\begingroup$ So I think the answer to your question is "no" unless you have some kind of growth conditions on the $a_n$. For instance if $a_{2^{2^n}}=2^{2^n}$ and $a_{2^{2^n}+1}=-2^{2^n}$ and all other terms are 0, the first sequence has limit 1, whereas the second sequence has limit 0. $\endgroup$ – Anthony Quas Aug 10 at 16:30
  • $\begingroup$ As usually I will link to the discussion of cross-posting on MathOverflow Meta. (And a similar post on Meta Mathematics Stack Exchange. This is meant as a kind of follow up on Anthony Quas' comment.) $\endgroup$ – Martin Sleziak Aug 10 at 16:39
  • $\begingroup$ Sorry for publishing on the two forums. I understand and won't do that again, I'm not familiar with all of this. I was about to delate my post from MathStackExange when @Martin Sleziak answered me. Thanks a lot for your help. $\endgroup$ – Wulfenite Aug 10 at 22:19
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The answer is "no" even if we assume that the $a_n$ are bounded. For example, take $a_n=1$ if the fractional part of $\log_2\log_2 n$ is between $0$ and $\frac12$ (equivalently, if $n$ is between $2^{2^k}$ and $2^{2^{k+1/2}}$ for some integer $k$), and $a_n=0$ otherwise. Then the sequential limit will equal $0$, but the lim sup of the logarithmic-density expression equals $1$ while its lim inf equals $0$.

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