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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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I think that for positive real numbers you could do basically the same proof as for densities of sets.

Let $s_n=\sum\limits_{k=1}^n a_k$. Then we have $a_n=s_n-s_{n-1}$ and \begin{align*} \frac{a_n}n &= \frac{s_n-s_{n-1}}n\\ D_N:=\sum_{n=1}^N \frac{a_n}n &= \frac{s_N}N + \sum_{k=1}^{N-1} \frac{s_k}{k(k+1)} \end{align*} We know that $\lim\limits_{N\to\infty} s_N/N=a$. For a given $\varepsilon>0$ there is $N_0$ such that $N\ge N_0$ we have $a-\varepsilon<s_N/N < a+\varepsilon$ and thus $$D_N\le (a+\varepsilon) \left(1+\sum_{k=N_0}^{N-1} \frac1{k+1}\right) + C$$ where $C=\sum\limits_{k=1}^{N_0-1} \frac{s_k}{k(k+1)}$.

Taking $N\to\infty$, we get $$\limsup_{N\to\infty} \frac{D_N}{\log N} \le a+\varepsilon$$ using $1+\sum\limits_{k=N_0}^{N-1} \frac1{k+1} \sim \log N$.

In a similar way we can get $\liminf_{N\to\infty} \frac{D_N}{\log N}\ge a-\varepsilon$ and, since this is true for every $\varepsilon>0$, we get $$\lim_{N\to\infty} \frac{D_N}{\log N} = a.$$

Remark. Basically in the same way we can show $$ \liminf_{N\to\infty} \frac1N \sum_{k=1}^N a_k \le \liminf_{N\to\infty} \frac{1}{\log N} \sum_{k=1}^N \frac{a_k}k \le \limsup_{N\to\infty} \frac{1}{\log N} \sum_{k=1}^N \frac{a_k}k \le \limsup_{N\to\infty} \frac1N \sum_{k=1}^N a_k.$$


EDIT: This could work for complex numbers.

Let $s_n=\sum\limits_{k=1}^n a_k$. Then we have $a_n=s_n-s_{n-1}$ and \begin{align*} \frac{a_n}n &= \frac{s_n-s_{n-1}}n\\ D_N:=\sum_{n=1}^N \frac{a_n}n &= \frac{s_N}N + \sum_{k=1}^{N-1} \frac{s_k}{k(k+1)} \end{align*} We know that $\lim\limits_{N\to\infty} \frac{s_N}N=a$. For a given $\varepsilon>0$ there is $N_0$ such that $k\ge N_0$ we have $\left|\frac{s_k}k-a\right|<\varepsilon$ and thus$\newcommand{\abs}[1]{|#1|}\newcommand{\absl}[1]{\left|#1\right|}$ \begin{align*} \abs{D_N-a \log N}&\le \absl{\frac{s_N}N -a} + \sum_{k=1}^{N-1} \frac1{k+1} \absl{\frac{s_k}{k}-a} + \absl{a(1+\sum_{k=1}^{N-1}\frac1{k+1}-\log N)}\\ \absl{\frac{D_N}{\log N}-a} &\le \varepsilon \cdot \frac{1+\sum_{k=1}^{N-1} \frac1{k+1}}{\log N} + \abs{a} \cdot \left(1+\sum_{k=1}^{N-1}\frac1{k+1}-\log N\right)/\log N \end{align*} Since the last expression tends to zero for $N\to\infty$ we get $$\lim\limits_{N\to\infty} \absl{\frac{D_N}{\log N}-a} \le \varepsilon.$$ As this is true for every $\varepsilon>0$, we get $$\lim\limits_{N\to\infty} \frac{D_N}{\log N}=a.$$

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  • $\begingroup$ Thanks ! Working a bit more your proof allowed me to solve my initial problem related to Bourgain's paper, by using maximal inequalities. But I didn't solved the general problem when one only assumes that ${\frac{s_N}{N}}$ converges up to a subsequence. $\endgroup$ – Wulfenite Aug 10 at 16:09

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