The series $\sum_{k=2}^\infty \frac{ \cos(kx)}{k \ln k}$ is bounded below by $c \log\log x$ near 0, thus fails to be uniformly convergent The convergent series
$$ \sum_{k=2}^\infty \frac{ \cos(kx)}{k \ln k} $$
defines a function in $H^{1/2}([0,2\pi])$. This is an example of such a series for which convergence on is not uniform.
I want to show that the convergence is not uniform on $(0,2\pi)$ by showing that the series is $\geq c\log \log \frac{1}{x}$ as $x\to 0$.
For a fixed positive integer $N$, we can split the series into
$$ \sum_{k=2}^N \frac{ \cos(kx)}{k \ln k} + \sum_{k=N+1}^\infty \frac{ \cos(kx)}{k \ln k} = \sum_{k=2}^N \frac{ \cos(kx)}{k \ln k} + R_N(x),$$
where $R_N(x)$ denotes the tail series evaluated at $x$.
We use continuity of cosine at $x=0$ in an essential way: the first $N$ factors of cosine stay away from 0. If $x < \frac{\pi}{6N}$, then $\cos(kx) > \frac{\sqrt 3}{2}$ for $2\leq k \leq N$. Thus we can bound below the first $N$ terms in the sum by
$$  \frac{\sqrt 3}{2} \sum_{k=2}^N \frac{1}{k\ln k} \geq \frac{\sqrt 3}{2} \int_2^{N+1} \frac{1}{t \ln t}dt = \frac{\sqrt 3}{2}( \ln \ln (N+1) - \ln \ln 2),$$
where we use monotonicity and convexity of $t\to \frac{1}{t\ln t}$ on $[2,\infty)$. Since $x < \frac{1}{N+1}$, $\ln \ln (N+1) \geq \ln \ln \frac{1}{x}.$
This looks almost like what I need to show:
$$ \sum_{k=2}^\infty \frac{ \cos(kx)}{k \ln k} \geq \frac{\sqrt 3}{2} \ln \ln \frac{1}{x} - \frac{\sqrt 3}{2} \ln \ln 2 + R_N(x),$$
for all positive integers $N$ and all $x < \frac{\pi}{6N}$.
My question is: how can I get rid of the tail series $R_N(x)$?
If I can show that it is positive or at least negative but very small, then we can simply drop it.
I also see that I might be able to go back in my argument and use that for every $\epsilon >0$, the convergence of the series is uniform on $[\epsilon, 2\pi-\epsilon]$. Then I can pick some $N$ so that $R_N(x)$ is very small, at least on $[\epsilon, 2\pi - \epsilon]$. But the problem is that for a choice of $\epsilon$, the interval for which $\cos(kx) > 0$ for $2\leq k \leq N$ guaranteed by continuity may be contained in $[0,\epsilon)$.
There appear to be two competing forces: continuity is useful close to 0, but the uniform convergence can only be used away from 0.
How can I proceed? Or am I completely off the mark? I would very much appreciate any advice or criticism for this problem.
 A: The easiest - in some sense - way to show that the series doesn't converge uniformly on $(0,2\pi)$ is to note that all terms of the sequence are continuous functions with common period $2\pi$, and if the series were uniformly convergent on $(0,2\pi)$, it would by continuity be uniformly convergent on $[0,2\pi]$ (and by periodicity on all of $\mathbb{R}$), so the limit function would be a continuous (and $2\pi$-periodic) real-valued function. But the series diverges for $x = 0$, hence the convergence cannot be uniform.
This argument doesn't use any deep theory, the only mildly uncommon thing is to notice that if a sequence/series of continuous functions converges uniformly on $S$, then it also converges uniformly on $\overline{S}$ - since for continuous $f,g$ we have
$$\sup \{ \lvert f(x) - g(x)\rvert : x \in \overline{S}\} = \sup \{ \lvert f(x) - g(x)\rvert : x \in S\}$$
by continuity of $f-g$ on $\overline{S}$.
But we can also continue on the way you started. That has the advantage that we do not only show that the series doesn't converge uniformly, we also get information about the behaviour of the sum function near $0$ (and near $2\pi$). To estimate the size of $R_N(x)$, summation by parts is a useful technique. We start with
\begin{align}
\Biggl\lvert \sum_{k = m+1}^n \cos (kx)\Biggr\rvert
&\leqslant \Biggl\lvert \sum_{k = m+1}^n e^{ikx}\Biggr\rvert \tag{$\lvert \operatorname{Re} z\rvert \leqslant \lvert z\rvert$}\\
&= \Biggl\lvert \sum_{k = 0}^{n-m-1} e^{ikx}\Biggr\rvert\\
&= \biggl\lvert \frac{e^{i(n-m)x}-1}{e^{ix}-1}\biggr\rvert\\
&\leqslant \frac{2}{\lvert e^{ix}-1\rvert}\\
&= \frac{1}{\bigl\lvert \sin \frac{x}{2}\bigr\rvert}.
\end{align}
Now for fixed $N$ and $n \geqslant N$ we define $S_n(x) := \sum\limits_{k = N+1}^n \cos (kx)$ and find (using $S_N(x) = 0$)
\begin{align}
\sum_{k = N+1}^n \frac{\cos (kx)}{k\log k}
&= \sum_{k = N+1}^n \frac{S_k(x) - S_{k-1}(x)}{k\log k}\\
&= \frac{S_n(x)}{(n+1)\log (n+1)} + \sum_{k = N+1}^n S_k(x)\biggl(\frac{1}{k\log k} - \frac{1}{(k+1)\log (k+1)}\biggr).
\end{align}
Taking absolute values, the inequality $\lvert S_k(x)\rvert \leqslant \frac{1}{\bigl\lvert\sin \frac{x}{2}\bigr\rvert}$ yields
\begin{align}
\Biggl\lvert \sum_{k = N+1}^n \frac{\cos (kx)}{k\log k} \Biggr\rvert &\leqslant
\frac{1}{\bigl\lvert \sin \frac{x}{2}\bigr\rvert (n+1)\log (n+1)} + \frac{1}{\bigl\lvert  \sin \frac{x}{2}\bigr\rvert} \sum_{k = N+1}^n \biggl\lvert \frac{1}{k\log k} - \frac{1}{(k+1)\log (k+1)}\biggr\rvert\\
&= \frac{1}{\bigl\lvert \sin \frac{x}{2}\bigr\rvert (N+1)\log (N+1)}.
\end{align}
Letting $n \to \infty$, we obtain
$$\lvert R_N(x)\rvert \leqslant \frac{1}{\bigl\lvert \sin \frac{x}{2}\bigr\rvert (N+1)\log(N+1)}.$$
The $\sin \frac{x}{2}$ in the denominator of that bound means that we cannot use this estimate to reach our desired conclusion if we work with a fixed $N$ independent of $x$. But we can choose $N$ depending on $x$ so that the singularity of $\frac{1}{\sin (x/2)}$ is compensated by the vanishing of $\frac{1}{N(x)}$. If for some $a > 0$ we choose $N(x) > \frac{a}{x} - 1$, then for $0 < x \leqslant \pi$ we have
$$(N(x)+1)\sin \frac{x}{2} \geqslant (N(x)+1)\frac{x}{\pi} > \frac{a}{x}\cdot\frac{x}{\pi} = \frac{a}{\pi}$$
and hence
$$\frac{1}{\bigl(\sin \frac{x}{2}\bigr) (N(x)+1)\log (N(x)+1)} < \frac{\pi}{a\log (N(x)+1)} < \frac{\pi}{a\log \frac{a}{x}}$$
for $0 < x < \min \{a,\pi\}$.
Now we have a bound for $\lvert R_{N(x)}(x)\rvert$ that we can use.
If we have $N(x) \leqslant \frac{b}{x}$ for some $b < \frac{\pi}{2}$, your computation shows
$$\sum_{k = 2}^{N(x)} \frac{\cos (kx)}{k\log k} > \cos b \log \log (N(x)+1)$$
(since $\log \log 2 < 0$).
If we choose $N(x) = \bigl\lfloor \frac{b}{x}\bigr\rfloor$ with $0 < b < \frac{\pi}{2}$, then we have $N(x) \leqslant \frac{b}{x} < N(x)+1$, and by the above
\begin{align}
\sum_{k = 2}^{\infty} \frac{\cos (kx)}{k\log k}
&\geqslant \sum_{k = 2}^{N(x)} \frac{\cos (kx)}{k\log k} - \Biggl\lvert \sum_{k = N(x)+1}^{\infty} \frac{\cos (kx)}{k\log k}\Biggr\rvert\\
&> \cos b\cdot \log \log \frac{b}{x} - \frac{\pi}{b \log \frac{b}{x}}
\end{align}
for $0 < x < b$. On the other hand we have
$$\sum_{k = 2}^{N(x)} \frac{\cos (kx)}{k\log k} \leqslant \sum_{k = 2}^{N(x)} \frac{1}{k\log k} \leqslant \log \log N(x) - \log \log 2 + \frac{1}{2\log 2}$$
and so
$$\sum_{k = 2}^{\infty} \frac{\cos (kx)}{k\log k} \leqslant \log \log \frac{b}{x} - \log \log 2 + \frac{1}{2\log 2} + \frac{\pi}{b\log \frac{b}{x}}$$
for $0 < b < x$. Since
$$\log \log \frac{b}{x} = \log \log \frac{1}{x} + \log \biggl(1 + \frac{\log b}{\log \frac{1}{x}}\biggr) = \log \log \frac{1}{x} + O\bigl(\lvert\log x\rvert^{-1}\bigr),$$
it follows that
$$\sum_{k = 2}^{\infty} \frac{\cos (kx)}{k\log k} \sim \log \log \frac{1}{x}$$
as $x\searrow 0$.
