# Logarithmic vs Sequential density of a sequence

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.

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 $$a-\varepsilon and thus $$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$$\newcommand{\abs}{|#1|}\newcommand{\absl}{\left|#1\right|}$$ \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 $$\lim\limits_{N\to\infty} \frac{D_N}{\log N}=a.$$

• 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. – Wulfenite Aug 10 at 16:09