# Is $H(\theta) = \sum \limits_{k=1}^{\infty} \frac{1}{k} \cos (2\pi n_k \theta)$ for a given sequence $n_k$ equal a.e. to a continuous function?

I am studying Furstenberg's article Strict ergodicty and transformation of the torus and I'm stuck with the following construction. Define sequence $(v_k)_{k \in \mathbb{N}}$ as $v_1 =1, v_{k+1}=2^{v_k}+v_k+1$ and sequence $(n_k)_{k \in \mathbb{Z} \backslash \{0\}}$ as $n_k=2^{v_k}$ for $k>0$ and $n_k = - n_{-k}$ for $k<0$. Consider function $H(\theta) : [0,1) \rightarrow \mathbb{R}$ given by the series $H(\theta) = \sum \limits_{k \neq 0} \frac{1}{|k|} e^{2\pi i n_k \theta} = 2 \sum \limits_{k=1}^{\infty} \frac{1}{k} \cos (2\pi n_k \theta)$. Of course there is $H(0) = + \infty$. Since $n_k$ tends monotonically to zero, we know that the series converges for all $\theta \in (0,1)$ and even converges unformly in every interval $[\varepsilon, 1-\varepsilon]$ for $\varepsilon \in (0, \frac{1}{2})$ (this can be seen by Theorem 2.7 in Trignonometric series by A. Zygmund). This implies that $H$ is continuous on $(0,1)$. Furstenberg says that $H$ is not equal almost everywhere (w.r.t. Lebesgue measure) to a continuous function. Why is that? It would be true if limit $\lim \limits_{\theta \rightarrow 0} H(\theta)$ would be $+ \infty$ or $- \infty$ or didn't exists, but is it the case? Furstenberg gives argument that $H \in L^2(0,1)$ and $\sum \limits_{k=1}^{\infty} \frac{1}{k} = +\infty$, so the series is not Ceasaro summable in $0$, but I don't know how is that connected to continuity.

• I don't think it's true that the series converges for all $\theta \in (0,1)$. For instance at $\theta=1/2$ it should diverge to infinity (because $n_k$ is an even integer for all $k$). The reason Theorem 2.7 in Zygmund doesn't apply is that the coefficients of this trigonometric series are not $(...,1,1/2,1/3,1/4,...)$ but rather $(...,0,...,0,1,0,...,0,1/2,0,...,0,1/3,0,...),$where 1/k occurs at the $n_k$-th ordinate, and everything else is $0$. This needn't be a sequence of bounded variation, necessary to apply that result. – Brad Rodgers Apr 16 '15 at 13:41
• That's right. Sequence $(...,0,...,0,1,0,...,0,1/2,0,...,0,1/3,0,...)$ tends to zero, but not monotonically and is not of bounded variation (its variation is the sum of series $\sum \limits_{k=1}^{\infty}$). Thanks for pointing that out! But H is in $L^2 (0,1)$ since sequence $(...,0,...,0,1,0,...,0,1/2,0,...,0,1/3,0,...)$ is in $l^2$, right? – Adam Apr 16 '15 at 21:53
• Hi! Sorry for the delayed response. You're absolutely right that $H$ is still in $L^2(0,1)$. – Brad Rodgers Apr 21 '15 at 12:00

Let $a_j$ be the Fourier coefficients of $H$, so that $a_j = 1/|k|$ if $j=n_k$ and $0$ otherwise. If $H$ were almost everywhere equal to a continuous function, say $h$, then the Cesaro partial sums $$\sigma_n(h,\theta) := \sum_{j=-n}^n a_j e^{i2\pi j \theta} (1-|j|/(n+1))$$ would tend everywhere to the finite number $h(\theta)$. (This is Fejer's theorem. See, for instance, Katznelson Theorem 3.1; this result is in Zygmund's book too, probably Chapter III, but I haven't looked up the location.)
But it is easy to see, as you point out, that $\sigma_n(h,0) \rightarrow \infty$. This contradiction shows that $H$ is not almost everywhere equal to a continuous function.