15.11

Altheman

Suppose that $f$ is a real, continuous, and periodical
function such that the sequence $\left(\displaystyle\sum_{k=1}^n \displaystyle\frac{|f(k)|}{k}\right)_{n\ge 1}$ is bounded. Prove
that $f(k)=0$ for all positive integers $k$ . Give a necessary and
sufficient condition ensuring the existence of a constant $c>0$
such that $\displaystyle\sum_{k=1}^n \displaystyle\frac{|f(k)|}{k}>c\ln n$ for all $n$ .

**Posted:** Wed Jul 22, 2009 6:44 pm

_________________
-Alex
Altheman's Problem Column

fedja

Let $T>0$ be a period of $f$ . Note that for every $\delta>0$ , there exist positive integer $m_+$ and $m_-$ such that $n_+T\le m_+\le n_+T+\delta$ , $n_-T-\delta \le m_-\le n_-T$ with some integer $n_-,n_+$ (indeed, just use Dirichlet lemma to choose integer $M>0$ for which $M/T$ is almost an integer $N$ . If it is an exact integer, put $m_+=m_-=M$ . If it is only an approximate integer, then it works for either $m_-$ or $m_+$ and to get the other one, just multiply $M$ by $\left\lfloor \frac 1{|(M/T)-N|}\right\rfloor$ ).

Now, if $|f(k)|>0$ for some integer $k$ , then there is an open interval $I$ of length $2\gdelta>0$ such that $|f|\ge c>0$ on $I$ and $k\in I\mod T$ . Now, put $k_0=k$ and define $k_j$ as follows. If $k_{j-1}\in I_+\mod T$ , put $k_j=k_{j-1}+m_-$ , otherwise put $k_{j}=k_{j-1}+m_+$ . Here, as usual, $I_+$ is the right half of $I$ . We get a sequence $A$ of integers with bounded differences on which $|f|\ge c$ . Hence, $\sum_{k\in A,k<n}\frac{|f(k)|}{k}\ge c'\log n$ , which takes care of both parts of the problem simultaneously if the condition " $f$ doesn't vanish identically on $\mathbb Z$ " is good enough for the second part.

**Posted:** Sun Jul 26, 2009 3:07 am

harazi

Of course, Fedja's solution is impecable. My approach was just a bit different: I'll exclude the trivial case when the period of $f$ is rational. Suppose $f$ does not vanish identically (in case it has an irrational period), then $s_k=|f(1)|+...+|f(k)|$ is equivalent to $kl$ for some $l>0$ (by Weyl's criterion). Now just use Abel summation to get that for large $n$ one has $\sum_{k=1}^{n}{\frac{|f(k)|}{k}}>(l/3)\ln n$ .

**Posted:** Wed Nov 04, 2009 8:45 am