If $f$ is continuous with irrational period $\xi$ and $\int_0^\xi f(x)\,dx = 0$, is it possible that $f(1)+f(2)+f(3)+\cdots=\infty$?

Question

If $f:\mathbb R\to\mathbb R$ is continuous with irrational period $\xi$ and $\int_0^\xi f(x)\,dx = 0$, is it possible that the series $f(1)+f(2)+f(3)+\cdots$ diverges to $\infty$? (This does not happen for, say, $f(x) = \sin x$, as the partial sums of $\sin 1+\sin 2+\sin 3+\cdots$ are bounded.)

Answer 1

It depends, since $\xi $ is irrational , according to Kronecker's Approximation Theorem, $\left\{m+\xi n \mid m,n \in \mathbb{Z}\right\}$ is dense in $\mathbb{R}$. The the fact that $f(x)$ is continuous and $f(x)=f(x+k\xi), \forall k \in\mathbb{Z}$ makes $f\left(m+\xi n\right)=f(m) \Rightarrow \{f(m) \mid m \in \mathbb{Z}\}$ also dense in $f$'s range.

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Now, one counter example to your statement (since you already mentioned $\sin{x}$), $f(x)=\cos{x}$, from this $$\left|\sum\limits_{k=1}^{n} \cos{k}\right|\leq \left|\frac{1}{2}\frac{\sin\left(\left(n+\frac{1}{2}\right)\right)+\sin{\frac{1}{2}}}{\sin{\frac{1}{2}}}\right| \leq \frac{1}{2}\left|\frac{1}{\sin{\frac{1}{2}}}\right|+\frac{1}{2}$$ taking the limit, we conclude that $\sum\limits_{k=1}^{\infty} \cos{k}$ is bounded and never reaches $\infty$.

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However, if we assume $f(x)\geq0$ and $f(x) \not\equiv 0$, then $f(x)$ (which is continuous) attains its maximum on $\left[0,\xi\right]$, $f(x_M)=M$. And because its periodic $f(x_M)=f(x_M+n\xi)=M$. Using the fact that $\left\{m+\xi n \mid m,n \in \mathbb{Z}\right\}$ is dense (it can be shown that $m$ can be positive integer, since theorem mentions positive $k$ in the link I provided), we have that $\exists m_M\in \mathbb{Z^+}$: $$\left|x_M-m_M+\xi n_M\right|<\delta \Rightarrow \left|f(x_M)-f(m_M+\xi n_M)\right| < \varepsilon \Rightarrow \\ f(m_M+\xi n_M)=f(m_M)>M-\varepsilon > 0$$ and this happens for all $x_M+n\xi$ thus we have a entire sequence $$f(m_{M,n}+\xi n_{M,n})=f(m_{M,n})>M-\varepsilon > 0$$ Thus $$\sum\limits_{k=1}f(k) > \sum\limits_{n=1}f(m_{M,n}) \geq \lim\limits_{n\rightarrow\infty} n(M-\varepsilon)=\infty$$

NOTE: While typing the last section I realised it doesn't consider $\int_0^\xi f(x)\,dx = 0$, but it's so technical it makes me sad to delete it.

Answer 2

This is not completely formal, but I think it could be made formal.

The idea is that, for any interval $[a,b]$, there is some positive integer $k$ such that the sequence $n\pmod{\xi}$ is in that interval every $k$ or so terms, where $k$ depends only on the length of the interval.

Define the sequences $t_n=10^{2n}$, $h_n=\frac{1}{10^n}$, and $g_n=10^3*10^{2n}$.

The idea is that we will add pairs of opposite spikes in groups of $2t_n$, $t_n$ of which are positive and $t_n$ of which are negative, where each spike in the $n$th group has height $h_n$ and has width such that an element of the sequence $n\pmod{\xi}$ is in the interval containing the spike every $g_n$ terms.

We make the positive spikes appear as early as possible and the negative spikes appear as late as possible without allowing the spikes to intersect (so the positive spikes of height $h_1$ appear for the first $t_1$ terms, and the negative spikes of height $h_1$ appear for the first time at about the $g_1-t_1$th term, then the positive spikes of height $h_2$ appear at the $t_1+1$th term until the $t_1+t_2$th term).

The reason we do this is so that the positive spikes from the $n+1$th group appear before the negative spikes of the $n$th group, but the sum of the terms in the $n+1$th group (i.e. $t_{n+1}h_{n+1}$) is greater than the sum of the terms in the $n$th group, so by the time we reach the negative spikes for the $n$th group we will have already added exponentially more than the $n$th group added, so the sums just keep increasing and by the time we reach the negative spikes they can't decrease by enough. For the same reason, it doesn't matter that the negative sequence will encounter those negative spikes again, or that when we encounter the positive spikes again we won't attain their full height.

Finally, to check that the heuristic that this must grow slower than linearly, note that the ratio of the partial sum to the number of terms is greatest immediately after adding a group of positive spikes. After the $n$th group of positive spikes we have gone about $g_{n-1}=10^{2n+1}$ terms and the to estimate the total we can basically ignore everything except for the $n$th group, whose sum gives $t_nh_n=10^n$ so the ratio is $\frac{10^n}{10^{2n+1}}=\frac{1}{10^{n+1}}\to0$.

Answer 3

This is not a full answer, more like a (very) long comment. We will give a direct proof that this is not possible if $f \in C^{1,\alpha}$ with $\alpha>1/2$ (i.e, if $f'$ exists and is Hölder-$\alpha$ for some $\alpha>1/2$). This is weaker than the result in Jack D'Aurizio's comment, but I feel as though there is room for improvement in places. Moreover, our approach suggests that Fourier series might provide a good tool.

I will use a slightly different convention than you do, and I will assume that $f:S^1 \to \Bbb R$. We will prove that if $\xi$ is an irrational multiple of $\pi$ then $\sum_{k=1}^N f(e^{i \xi k})$ must remain bounded as $N \to \infty$.

If $f \in C^{1,\alpha}$ with $\alpha>1/2$, this implies (via an imbedding theorem) that $f \in H^{1+\alpha}(S^1)$, and hence there exists $g \in L^2(S^1)$ such that $\hat g(n) = |n|^{1+\alpha} \hat f(n)$. Namely, $g=-(-\partial_x^2)^{(1+\alpha)/2}f$.

Note that $\hat f(0)=\int f=0$. Writing $$\sum_{k=1}^N f(e^{i \xi k}) = \sum_{n \in \Bbb Z \backslash 0} \hat f(n) \sum_{k=1}^N e^{i \xi kn} = \sum_{n \in \Bbb Z\backslash 0} \frac{\hat g(n)}{n^{1+\alpha}} \frac{1-e^{i\xi(N+1)n}}{1-e^{i\xi n}}$$ Now using the triangle inequality and $|1-e^{i \xi (N+1)n}| \leq 2$ to bound the last sum, we get \begin{equation}\sup_N\bigg| \sum_{k=1}^N f(e^{i \xi k}) \bigg| \leq \sum_{n \in \Bbb Z\backslash 0} \frac{|\hat g(n)|}{n^{1+\alpha}} \frac{2}{|1-e^{i \xi n}|}.\tag{1}\end{equation} Now I claim that for each $\epsilon>0$, there exists a constant $C$ such that for all $n \in \Bbb Z \backslash 0$ we have that $$\frac{1}{|1-e^{i \xi n}|} \leq C|n|^{1+\epsilon}\tag{2}$$ To prove this, note by the ergodic theorem (equivalently Weyl's equidistribution theorem) that for any $\delta>0$,$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{1}{|1-e^{i \xi k}|^{1-\delta}} = \int_{S^1} \frac{1}{|1-z|^{1-\delta}}d|z|<+\infty$$ Convergent sequences are bounded, so there exists some $C=C(\delta)$ such that for all $n$ we have $$\frac{1}{|1-e^{i \xi n}|^{1-\delta}} \leq \sum_{k=1}^n \frac{1}{|1-e^{i \xi k}|^{1-\delta}} \leq Cn$$ This implies that $$\frac{1}{|1-e^{i\xi n}|} \leq C'n^{\frac{1}{1-\delta}}$$ which proves the claim in equation (2), after perhaps renaming $\frac{1}{1-\delta}=:1+\epsilon$. Now using equation (2) with $\epsilon=(\alpha-1/2)/2$ and plugging into equation (1), then using Cauchy-Schwarz, we see that $$\sup_N\bigg| \sum_{k=1}^N f(e^{i \xi k}) \bigg| \leq C \sum_{n \neq 0} \frac{|\hat g(n)|}{|n|^{\frac{\alpha}{2}+\frac{1}{4}}} \leq C \bigg( \sum_{n \neq 0} |\hat g(n)|^2 \bigg)^{1/2} \bigg( \sum_{n \neq 0} \frac{1}{|n|^{\alpha+ \frac{1}{2}}} \bigg)^{1/2} < \infty.$$ where we used Plancherel and $\alpha>1/2$ in the final inequality.

I suspect that the result can be optimally improved to the case when $f$ is itself Hölder-$\alpha$ for some $\alpha>1/2$. A proof of this might involve getting a better bound in equation (2), but it eludes me.