Krdytkyu

Let $f(x)=chi_{[a,b]}(x)$ be the characteristic function of the interval $[a,b]subset [-pi,pi]$.

Show that if $aneq -pi$, or $bneq pi$ and $aneq b$, then the Fourier series does not converge absolutely for any $x$. [Hint: It suffices to prove that for many values of $n$ one has $|sin ntheta_0|ge c gt 0$ where $theta_0=(b-a)/2.$]

However, prove that the Fourier series converges at every point $x$.

I've computed the Fourier series and got $frac{b-a}{2pi}+sum_{nneq 0}frac{e^{-ina}-e^{-inb}}{2pi in}e^{inx}.$

Also, $|e^{-ina}-e^{-inb}|=2|sin ntheta_0|$, and $theta_0in (0,pi)$, so I can see that for infinitely many values of $n$, we have $|sin ntheta_0|ge c gt 0$. But this does not guarantee $sum_{nneq 0}|frac{e^{-ina}-e^{-inb}}{2pi in}e^{inx}|ge sum frac{c}{n}$, and in fact we might have this inequality only for the squares of integers, in which case the right hand side converges. So how does the hint solve the problem?

Moreover, for the second problem, to show that the Fourier series converges at every point, I think I need to use Dirichlet's test, using $1/n$ as the decreasing sequence to $0$, but how can I show that $frac{e^{-ina}-e^{-inb}}{2pi in}e^{inx}$ has bounded partial sums?

I would greatly appreciate any help.

Suppose $a ne -pi$ or $b ne pi$ and $ane b$. Then the function you are talking about must be discontinuous.

Suppose the series did converge absolutely for some $x$. That would mean
$$
sum_n left|frac{e^{-ina}-e^{-inb}}{2pi in}right| < infty.
$$
But that would force the uniform convergence of the Fourier series everywhere by the Weierstrass M-test. But uniform convergence would imply that the periodic extension of the limit function $chi_{[a,b]}$ must be continuous everywhere, which only happens in the case that $a=-pi$ and $b=pi$, or $a=b$.

I'm not familiar with you text, but you should have some pointwise convergence theorem that shows the Fourier series converges to the mean of the left and right hand limits for your function.

First of all, I like to write the Fourier coefficients of this function as
$$widehat{chi_{left[a,bright]}}left(nright)=frac{b-a}{2pi}expleft(-ifrac{a+b}{2}nright)mathrm{sinc}left(frac{b-a}{2pi}nright)$$ where $$mathrm{sinc},eta
= frac{sin pieta}{pieta},qquad mathrm{sinc},0=1.$$

It's quite easy to show that $sumwidehat{chi_{left[a,bright]}}left(nright)$ converges (although not absolutely), using the following lemma:

(Dirichlet Test; exercise 7b in chapter 2 of Stein & Shakarchi) If the partial sums of $sum b_n$ are bounded, and $a_0 geqslant a_1 geqslantcdotsgeqslant a_n geqslantcdotsto 0$, then $sum a_n b_n$ converges.

The proof of Dirichlet's Test is based on summation by parts.

To use this test, it's actually easier to write $chi$ the way you did:
$$chi_{left[a,bright]}left(xright) sim frac{b-a}{2pi}+frac{1}{
2pi i}sum_{nneq 0}frac{e^{inleft(x-aright)}
- e^{inleft(x-bright)}}{n}.$$

This is a series of differences. It's enough to show that each of the series $sum e^{inleft(x-aright)}/n$ and $sum e^{inleft(x-bright)}/n$ converge, because then the Fourier series of $chi$ can be constructed by just taking the difference between these two series (if two series converge, their sum-series and difference-series also converge). If you've done exercise 8 in this chapter, you'll know that you have to use Dirichlet's test by making $a_n=1/n$ and $b_n=e^{intheta}$, with $theta=x-a$ or $x-b$. It's easy to show that the partials of $sum e^{intheta}$ are bounded. Just use the fact that it's a power series:
$$left|sum_{1}^{N-1}e^{inx}right|=left|frac{1-e^{iNx}}{1-e^{ix}}-1right|=frac{left|1-e^{ileft(N-1right)x}right|}{left|1-e^{ix}right|}
leqslantfrac{2}{left|1-e^{ix}right|}$$
(This is an acceptable bound because $x$ is fixed.) (I did it just for $n>0$, but you can just do the same for $n<0$.)

Now, there's actually two senses in which the Fourier series of $chi$ could converge absolutely. First, you look at the complex Fourier series, and ask whether the following series converges:
$$sum_{ninmathbb{Z}}left|widehat{chi_{left[a,bright]}}left(nright)e^{inx}right|to,?$$
Showing that it does not converge is reasonably doable. Now, what's interesting is to ask whether the real Fourier series (that is, the series of sinusoids, instead of the series of complex exponentials) converges absolutely. We can re-write it (yes, again) like this:
$$frac{b-a}{2pi}+frac{2}{pi}sum_{ngeqslant1}frac{1}{n}left|sinleft(frac{b-a}{2}nright)right|cdotleft|cosleft(left(x-frac{a+b}{2}right)nright)right|
to ?$$
You can show show that it does not converge (with the exception that, if $b-a=pi$, then it converges trivially on $x=b$ and $x=a$). I won't go into all of the details, because I can't seem to find an easy, quick proof (it would actually be quite tedious if I detailed all of it), but I'll do my best.

Let $theta_{0}=left(b-aright)/2$ e $omega=x-left(a+bright)/2$. We want to show that $sum_{ngeqslant1}frac{1}{n}sin ntheta_{0}cos nomega$ is not absolutely convergent. The strategy is the obvious one: show that $left|sin ntheta_0right|geqslant c>0$ more than 50% of the time, and show the same for $left|cos nomegaright|$. Then, by the pigeonhole principle, we'll have
$$left|frac{1}{n}sin ntheta_{0}cos nomegaright|geqslantfrac{c^2}{n}$$ at least once per period, which is enough to compare $sum_{ngeqslant 1}left|sin ntheta_{0}right|cdotleft|cos nomegaright|/n$ to the harmonic series.]

Let's start by getting out of the way the case in which both $theta_0$ and $omega$ are rational multiples pf $pi$. Let $theta_0=left(p/qright)2pi$, with $p/q$ already in lowest terms. Then $ntheta_{0}$ is periodic with period $q$ and assumes all values $$frac{0}{q},frac{1}{q},frac{2}{q},cdots,frac{q-1}{q}$$ in each period. We're interested in the sequence $sin ntheta_{0}$. We know that $qgeqslant 3$, since $theta_{0}inleft]0,piright[$. Also, let $omega=left(r/sright)2pi$ with $r/s$ in lowest terms, so that $nomega$ has period $s$.

If $q=3$, then $left|sin ntheta_{0}right|$ spends $2/3$ of the time being equal to $sqrt{3}/2$. If $q=4$, the sequence $left|sin ntheta_{0}right|$ is $1,0,1,0,ldots$. If $qgeqslant 5$, then $left|sin ntheta_{0}right|$ has period $5$ and is zero at most twice per period, and therefore is greater than its minimum more than 50% of the time. As to $omega$, it's also easy to see that the sequence $left|cos nomegaright|$ is also $geqslant c>0$ more than 50% of the time, unless $s=4$.

That's enough to show that, unless $q=s=4$, the sequence $left|sin ntheta_{0}cos nomegaright|$ is periodic with period $qs$, and by the pigeonhole principle there's at least one non-zero element per period. Let's say that this non-zero element is the $k$-th:
$$forall dgeqslant0text{,} left|sinleft(dqs+kright)theta_{0}cosleft(dqs+kright)omegaright|
=left|sin ktheta_{0}cos komegaright|>0text{.}$$
Then:
$$sum_{ngeqslant1}frac{1}{n}left|sin ntheta_{0}cos nomegaright|geqslantleft|sin ktheta_{0}cos komegaright|sum_{dgeqslant1}frac{1}{dqs+k}=inftytext{,}$$
and we're good, unless $q=s=4$.

When $q=s=4$, the series is actually absolutely convergent in a trivial fashion: all of its terms are zero, because $left|sinright|$ will be $0,1,0,1,ldots$ and $left|cosright|$ will be $1,0,1,0,ldots$. Well, $q=4$ means $theta_{0}=pi/2$, which means $b-a=pi$; and $s=4$ means $omega=pmpi/2$, meaning $x=a$ or $x=b$.

Now, the case in which $theta_0$ or $omega$ are irrational multiples of $pi$ is the tedious part. (Actually, we could use the equidistribution theorem here, but that'll only show up in chapter 3 of the book! We'll try to do without it.) Let $H$ be the set of angles $theta$ for which $left|sinthetaright|<sinleft(pi/50right)$. Also, let $H'$ be the set of $theta$ such that $left|costhetaright|<sinleft(pi/50right)$. $H$ and $H'$ together make up 8% of the whole circle, so we'd expect that any reasonable periodic-ish sequence in the the circle ends up outside of $Hcup H'$ around 92% of the time.

Let's say that a finite sequence $Theta_{p}left(N_{0}right)=left{ N_{0}theta,left(N_{0}+1right)theta,ldots,left(N_{0}+pright)thetaright}$ is reasonably well-distributed in the circle (rwd for short) when $Theta$ has more than 50% of its elements outside $H$, and also more than 50% outside $H'$. I'll start by showing the following result:

Let $thetainleft]0,pi/3right[setminuspimathbb{Q}$. There exists $pgeqslant 1$ such that, given any $x_{0}notinpimathbb{Q}$, the finite set
$$Theta=left{ x_{0},x_{0}+theta,ldots,x_{0}+pthetaright} $$
is rwd.

Well, since $thetainleft]0,pi/3right[setminuspimathbb{Q}$, there is $pinmathbb{Z},pgeqslant6$ such that
$$frac{2pi}{p+1}<theta<frac{2pi}{p},qquadtext{i.e.,}qquad ptheta<2pi<left(p+1right)thetatext{.}$$
Let $x_0notinpimathbb Q$. We want to show that, out of all of the $p+1$ elements in $Theta$, more than $left(p+1right)/2$ are outside $H$. Well, it's not very hard to show that, since the elements in $Theta$ are equally spaced (except for the pair $x_0$ and $x_0+ptheta$, which could be quite close to each other by roing around the circle), the number of elements in $Thetacap H$ is:
$$#left(Thetacap Hright)leqslant 2leftlceil frac{2pi/50}{theta}rightrceil +1text{.}$$
Now, we plug in $theta>2pi/left(p+1right)$ and $pgeqslant6$ to find that
$$#left(Thetacap Hright)<frac{p+1}{2}=frac{#Theta}{2}text{.}$$
(Yes, that "50" in there was important; I didn't just "make up a huge number" when designing $H$ and $H'$. We need that 50 because it makes the inequality $2leftlceil frac{p+1}{50}rightrceil +1<frac{p+1}{2}$ valid for $pgeqslant 6$.)

Anyway, ok, we showed that for $H$, and the proof is exactly the same for $H'$, so $Theta$ is rwd, as promised.

Moving on, we'd like to show the same proposition above without the restriction $thetainleft]0,pi/3right[$. However, we already know that $theta=pi/2$ is a problem, and that will make $thetaapproxpi/2$ a problem too. We'll write $theta=pi/2-vartheta$, with $varthetainleft]0,pi/6right[setminuspimathbb{Q}$ and abuse the formulas: $$left|sin nthetaright| =begin{cases}
left|sin nvarthetaright|, & ntext{ even}\
left|cos nvarthetaright|, & ntext{ odd}
end{cases}text{; and}$$
$$left|cos nthetaright| =begin{cases}
left|cos nvarthetaright|, & ntext{ even}\
left|sin nvarthetaright|, & ntext{ odd}
end{cases}$$

Let $thetainleft]pi/3,pi/2right[setminuspimathbb{Q}$. There exists $pgeqslant1$ such that, given any $N_{0}inmathbb{Z}$, the finite set $$Theta=left{ N_{0}theta,left(N_{0}+2right)theta,ldots,left(N_{0}+2pright)thetaright}$$ is rwd.

Ok, let $vartheta=pi/2-theta$ so that $2varthetainleft]0,pi/3right[$ and using the first proposition above, $left{ x_{0},x_{0}+2vartheta,ldots,x_{0}+2pvarthetaright}$ is rwd for all $x_0$. Letting $x_{0}=N_{0}vartheta,N_0inmathbb Z$, we get that $left{ N_{0}vartheta,left(N_{0}+2right)vartheta,ldots,left(N_{0}+2pright)varthetaright} $ is rwd. By considering the cases $N_0$ even and $N_0$ odd separately, we show that we can change $vartheta$ for $theta$ in this set, and the elements inside $H$ remain inside $H$ (or, if $N_0$ is odd, become elements of $H'$ and vice-versa). Anyway, it's proved.

Even though that second proposition above looks like it doesn't help much, it does:

Let $thetainleft]0,pi/2right[setminuspimathbb{Q}$. There exists $pgeqslant1$ such that, given any $N_{0}inmathbb{Z}$, the finite set $Theta_{p}left(N_{0}right)$ is rwd.

We just have to show for $thetainleft]pi/3,pi/2right[setminuspimathbb{Q}$, because the other case is already contemplated in the first proposition above. Applying the second proposition, we get $p'geqslant 1$ such that $Theta=left{ N_{0}theta,left(N_{0}+2right)theta,ldots,left(N_{0}+p'right)thetaright}$ is rwd for any $N_0$. Now, let $p=2p'+1$. Given any $N_0$, the set $Theta_{p}left(N_{0}right)=left{ N_{0}theta,left(N_{0}+1right)theta,ldots,left(N_{0}+pright)thetaright}$ can be partitioned into two sets of size $p'+1$: one with the even multiples of $N_0$, and the other with the odd multiples. Both partitions are rwd, and therefore their union also is.

Now, we extend this result to $thetainleft]0,piright[$ using the relevant trigonometric identities again, and then extend it, finally, to the whole circle:

Let $thetanotinpimathbb{Q}$. There exists $pgeqslant1$ such that given any $N_{0}$, $Theta_{p}left(N_{0}right)$ is rwd.

I'm translating this from my notes in portuguese from when I was taking this course (all of that LaTeX was already typed!), and I'm getting tired of writing, so I hope you can see that this last result is enough to tackle the cases in which each of $theta_0$ or $omega$ or both are irrational multiples of $pi$. By the way, those notes were typed using the notation $tau=2pi$, and in the process of translating from portuguese I also translated from tau, and I might have forgotten to double or halve some of the integers above, so double-check everything if you're going to use it for real.