A Primer in Harmonic Analysis

I picked these problems from Modern Fourier Analysis Vol I - I think that they serve as a good primer for the basic techniques and theorems in harmonic analysis (a subject that I have recently started looking back into in order to deal with some of the techniques used when Levy processes in mathematical finance).

Problem I.  Fix $d\geq1$ and suppose $\psi:(0,\infty)\mapsto[0,\infty)$ is $C^{1}$, non-increasing, and $\int_{\mathbb{R}^{d}}\psi(|x|)\;dx\leq A<\infty.$  Define
$$[M_{\psi}f](x):=\sup_{0<r<\infty}\frac{1}{r^{d}}\int_{\mathbb{R}^{d}}|f(x-y)|\psi\left(\frac{|y|}{r}\right)\;dy$$
and show that $$[M_{\psi}f](x)\leq A[Mf](x)$$
where $M$ is the usual Hardy-Littelwood maximal function.

Solution.  We first observe that the translation invariance of the indicated estimate implies that it is sufficient to prove the case $x=0$ (this can be seen more explicitly by replacing $f$ by $\tau_{x}f$, where $\tau_{x}$ is the translation by $x$ operator, and applying the present case to be proven to see then that the estimate holds for all $x$).   For convenience let us define $\psi_{r}(|y|)=r^{-d}\psi(|y|/r)$.  The radial properties of the terms in the estimate suggest polar coordinates will be useful in dealing with the resultant integrals.  Let us recall that the polar coordinate formula implies as a consequence of itself that
$$\frac{d}{ds}\int_{B(0,s)}f(y)\;dy=\frac{d}{ds}\int_{0}^{s}dt\int_{\partial B(0,t)}f(\omega)\;dS(\omega)=\int_{\partial B(0,s)}f(\omega)\;dS(\omega)=s^{d-1}\int_{S^{d-1}}f(s\omega)\;dS(\omega).$$

Divergence of Harmonic Series on a Sequence of Decreasing Sub-Domains of $\mathbb{N}$

The series $\sum_{n\in\mathbb{N}}n^{-p}$ diverges if $p\leq1$ and converges if $p>1$, and so it may seem plausible that (being a "bifurcation point" of this condition) the harmonic series $\sum_{n\in\mathbb{N}}n^{-1}$ could converge on some proper subset of $A\subset\mathbb{N}$. This is obvious if $A$ is finite. If $A$ is infinite, then a moment's thought reveals that there are many subsets on which the harmonic series converges since its terms contain any other series with terms in $\mathbb{N}^{-1}$. So for instance $$\sum_{n\in\mathbb{N}}n^{-2}=\frac{\pi^{2}}{6},$$ $$\sum_{n\in\mathbb{N}}\frac{1}{n!}=e,$$ $$\sum_{n\in\mathbb{N}}\frac{1}{2^{n}}=2,$$ and so on. Given that rather "large" subsets of $\mathbb{N}$ lead to convergence of the harmonic series, the following result was somewhat surprising to me when I was first asked to prove it.

Claim. Let $$A_\epsilon := \{a \in \mathbb{N} : 1 - cos(a) < \epsilon\}.$$ Then $$\sum_{n\in A_\epsilon } \frac{1}{n}$$ diverges for all $0<\epsilon<1.$

Proof.  For $0<\epsilon< 1$, the inequality $1-\cos(a)< \epsilon$ has solutions for $$a\in(2k\pi-\theta,2k\pi+\theta)$$ where $\theta=\cos^{-1}(1-\epsilon)$ (note that $\theta\in(0,\frac{\pi}{2})$ and by using a Taylor expansion, it is easy to see $\theta=O(\epsilon^{\frac{1}{2}})$, although all that is important is $\theta\to0$ as $\epsilon\to0$). For there to be any positive integers $a:=a_{k}$ in such an interval, it is necessary and sufficient that
$$\frac{[2k\pi-\theta]}{[2k\pi+\theta]}<1,$$
where $[\cdot]$ is the "floor" function (round down, e.g. truncate the decimals). Intuitively, this condition just says there is an integer in the $k$th solution interval (note that there could be multiple integral solutions in a $k$th interval, though this is not very important since we are mostly interested in the case for small $\epsilon$; furthermore, since $\theta=O(\epsilon^{\frac{1}{2}})$, then once $\epsilon$ is sufficiently small (say $\epsilon<0.1,$ so that $2\theta$).

From the above observations and the fact that $2\pi<6.3$ (circumference of the unit circle), it is not difficult to ascertain that $\#A=\infty$ (the cardinality of the set $A$).  Therefore $A$ is countable with its elements forming an "approximate" arithmetic sequence of integers in that sense that for
$$D:=\max_{a_{i}\in A}|a_{i+1}-a_{i}|<\infty,$$
on "average" the difference of two successive integers is approximately $D$ (having analytical results that are sharp is unnecessary in the present situation as we are only after qualitative facts like convergence).

We can now determine whether or not the sum converges. Define sequences $a_{j}:=\frac{1}{j}$ for $j\in A $, and $0$ otherwise, and $b_{j}:=\frac{1}{j}$ for all $j=1,2,\ldots.$ Then $c_{j}:=\frac{a_{j}}{b_{j}}=1$ for $j\in A$, and $0$ otherwise. Therefore, $c_{j}$ has a sum which looks like $$1+0+\ldots+0+1+0+\ldots+0+1+\ldots$$ Define one more sequence $d_{j}:=1$ if $j=a_{1}D$ ($a_{1}$ being the first integral solution to the original inequality) and $0$ otherwise (in other words, $d_{j}$ really is an arithmetic sequence with common difference $D$). Recall from the theory of Cesaro summation that for zero-spacing $D$, $$\frac{1+0+\ldots+0+1+0+\ldots+0+\ldots+0+1_{n}}{n}\to\frac{1}{D+1}\;as\;n\to\infty$$ (note because Cesaro summation is an averaging process, the limit holds even if there is a finite number of instances of improper spacing for a finite number of terms). Consequently, $$\frac{d_{1}+\ldots+d_{j}}{n}=\frac{1}{D+1}\;as\;n\to\infty$$ (see previous parenthetical remark). Consequently, \begin{align*} \lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{j=1}^{n}\frac{a_{j}}{b_{j}} &=\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{j=1}^{n}c_{j}\\ &\geq\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{j=1}^{n}d_{j}\\ &=\frac{1}{D+1}\\ &>0 \end{align*} for all $\epsilon>0$, no matter how small (note that $D$ behaves something like $O(\theta^{-1})$, and by extension something like $O(\epsilon^{-\frac{1}{2}}).$ It follows that $$\sum\limits_{j=1}^{\infty}a_{j}=\infty,$$ e.g. diverges for every $\epsilon>0$ (if you don't see why or don't recognize the convergence theorem used, just apply the summation by parts formula to $\sum a_{j}$ together with the established bound).

Does the Trigonometric Harmonic Series Converge?

It is well known that the harmonic series $H(x)=\sum_{n=1}^{\infty} xn^{-1}=+\infty$ for every $x\neq0$, but what about the trigonometric harmonic series $T(x)=\sum_{n=1}^{\infty}e^{inx}n^{-1}$?  Obviously for $k=1,2,\ldots$ we have $T(2k\pi)=H(1)=+\infty$.  It is an interesting fact that the cancellation properties inherent in $T$ imply convergence.  This is relatively straight-forward to prove this using a modificaton of Leibniz's alternating series test.  More remarkable is that the convergence is actually absolute.

In order to investigate the convergence of
$$(1)\;\;\;\;\;T(x)=\sum_{n=1}^{\infty}\frac{e^{inx}}{n}<\infty,$$
first note that
$$\lim_{n\to\infty}|z^{n}|\to0$$
for every $z\in\mathbb{C}$ with $|z|<1$.  Since
$$1>\frac{1}{n}>\frac{1}{n+1}>0$$ for all $n>1$, we find $\frac{1}{n}\searrow0$ (monotonically decreases to zero) and so Dirichlet's test implies
$$(2)\;\;\;\;\;\sum\limits_{n=1}^{\infty}\frac{z^{n}}{n}<\infty,$$
the convergence taking place and being absolute for every $z$ with $|z|<1$.  To deal with the boundary $|z|=1$, note that if $|z|=1$ and $z\neq1$ (i.e. $z\neq1+0i$), then we have
$$\left|\sum_{n=1}^{N}z^{n}\right|=\left|\frac{1-z^{N+1}}{1-z}\right|\leq\frac{2}{1-z}<\infty.$$
The upper bound $M=\frac{2}{1-z}$ is independent of $N$ and so (2) holds for all $|z|\leq1$, except when $z=1$.  Putting $z\mapsto e^{inx}$ shows that (1) converges absolutely for every $x\neq 2k\pi$ ($k=1, 2, \ldots$).

To carry out the actual summation for $T(x)$ is a tedious exercise in complex analytic methods, and the resulting formulas are unworkable (although again rather remarkably, they contain only elementary functions).  Another approach is to recognize that $T(x)$ is the Fourier transform (series) of some periodic function with Fourier coefficients $\hat{f}(0)=0$ and for $n>1$
$$\hat{f}(n)=\frac{1}{n}.$$
Despite this, the computation is relatively straight-forward for certain values of $x$.  For example, take $x=1$ and note that
$$T(1)=\sum_{n=1}^{\infty}\frac{e^{in}}{n}.$$
Writing
$$\int\left(\underbrace{(e^{iz})^{1}+(e^{iz})^{2}+\ldots}_{\text{geomtric series with ratio }r=e^{iz}}\right)dz=\int\frac{e^{iz}}{1-e^{iz}}\;dz,$$
we find that (with $u=1-e^{iz}$)
$$\sum_{n=1}^{\infty}\frac{1}{in}(e^{iz})^{n}=i\int\frac{du}{u}=i\ln(1-e^{iz}).$$
Combining all of this together, we obtain
$$\begin{align*}
T(1)
&=\left(i\ln(1-e^{iz})\right)\Big|_{z=1}\\
&=i\ln\left(e^{i/2}\left((e^{-i/2}-e^{i/2}\right)\right)\\
&=i\ln\left(e^{i/2}\right)+i\ln\left(2i\sin\left(-\frac{1}{2}\right)\right)\\
&=-\frac{1}{2}+i\left(\ln(-i)+\ln\left(\sin\left(\frac{1}{2}\right)\right)\right)\\
&=-\frac{1}{2}+\frac{\pi}{2}+i\ln\left(2\sin\left(\frac{1}{2}\right)\right)
\end{align*}$$
Since
$$T(1)=\sum_{n=1}^{\infty}\left(\frac{\cos n}{n}+i\frac{\sin n}{n}\right),$$
taking real and imaginary parts yields
$$T(1)=\frac{-\ln(2-2\cos(1))}{2}+i\frac{\pi-1}{2}.$$

The graphic at the beginning of the post shows the graph of $\sin n/n$ on the $(n,x)$ plane.