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Most proofs I found for the pointwise convergence of the Fourier series finishes by applying the Riemann-Lebesgue Lemma which states that if $f$ is a Riemann integrable function over $[a,\ b]$, then: $$ \lim_{\mu \to \infty} \int_{a}^{b} f(u)\cos(\mu u)du=0 $$ Now since the Dirichlet kernel is even and $2\pi$ periodic given ANY interval of this length it holds: $$ \frac{1}{\pi} \int_{x-\pi}^{x+\pi}D_N(u)du=1 $$ So multiplying by arbitrary $f(x)$: $$ f(x) = \frac{1}{\pi} \int_{x-\pi}^{x+\pi} f(x)D_N(u)du $$ Then we know the following form of the $N$th Fourier partial sum: $$ S_N(x) = \frac{1}{\pi}\int_{-\pi}^{\pi}f(u)D_N(x-u)du $$ Then we make the substitution $u = x - t$ and we obtain: $$ S_N(x)=-\frac{1}{\pi}\int_{x+\pi}^{x-\pi}f(x - t)D_N(t)dt=\frac{1}{\pi}\int_{x-\pi}^{x+\pi}f(x-t)D_N(t)dt $$ Then usally we are told that since we assume $f$ to be $2 \pi$ periodic this latter integral is: $$ \frac{1}{\pi}\int_{-\pi}^{\pi}f(x-t)D_N(t)dt $$ But I thought, what if I did not require $f$ to be periodic? Then for the difference $f(x) - S_N(x)$ we obtain:

$$ \frac{1}{\pi}\int_{x-\pi}^{x+\pi}(f(x) - f(x - t))D_N(t)dt $$ Then we expand the Dirichlet kernel to its closed form so that we can apply the Riemann Lebesgue Lemma: $$ f(x)-S_N(x)=\frac{1}{\pi}\int_{x-\pi}^{x+\pi}(f(x) - f(x - t))\frac{\sin((N + \frac{1}{2})t)}{2\sin(\frac{t}{2})}dt $$ Some little rearrangement: $$ \frac{1}{\pi}\int_{x-\pi}^{x+\pi}\frac{f(x) - f(x - t)}{2\sin(\frac{t}{2})}(\sin(Nt)\cos(\frac{t}{2}) + \cos(Nt)\sin(\frac{t}{2}))dt $$ And finally we split the integral into 2: $$ \frac{1}{\pi}\int_{x-\pi}^{x+\pi} \frac{f(x) - f(x - t)}{2}\cos(Nt)dt+ \frac{1}{\pi}\int_{x-\pi}^{x+\pi} \frac{(f(x) - f(x - t))\cos(\frac{t}{2})}{2\sin(\frac{t}{2})}\sin(Nt)dt $$ And principally, this formula is indeed true for $f(x) - S_N(x)$. To obtain it, I never ever had to use such information that $f$ should be $2\pi$ periodic. That being said, in this form, I can not state that, by the Riemann-Lebesgue Lemma, the two integrals tend to zero? That would be false since the Fourier series is periodic by nature and is said to converge to a periodic extension of $f$. The original proof eliminates the $x$ in the bounds of the integral. As I have mentioned, that is ONLY possible if $f$ is $2\pi$ periodic. But in no other step of the proof it seems evident that $f$ must be periodic. So we get in the original proof: $$ \frac{1}{\pi}\int_{-\pi}^{\pi} \frac{f(x) - f(x - t)}{2}\cos(Nt)dt + \frac{1}{\pi}\int_{-\pi}^{\pi} \frac{(f(x) - f(x - t))\cos(\frac{t}{2})}{2\sin(\frac{t}{2})}\sin(Nt)dt $$ Then the proof shows that in both of the integrals the integrand is differentiable and due to the Riemann-Lebesgue Lemma the two integrals tend to zero and finishes the proof. So why is it that for a fixed $x$ I cannot apply the lemma in former, but I can apply it in the latter case?

Gary
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Patrik Nusszer
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    You can always redefine the function at $-\pi$ (more generally fourier series assume the function is periodic in the sense that only the values on the period interval matter - so if you want for any function on the line, only its restriction to the period interval matters and the end values are where there is any subtlety in the continuous case say ) to be periodic and apply the above at any point inside the interval; the only difference comes at the end point if the function has different left/right limits where then one gets the semisum of that instead – Conrad Feb 25 '22 at 20:42
  • @Conrad Thanks for the input. I know that the function is not required to be periodic, but its series will converge to its periodic extension. That being said, I still have no clue in which part of the proof this fact is evident, otherwise the result would not be true. The usual argument that "the Fourier series is periodic by nature" explains it well but still what I am looking for is in WHICH part of the proof this fact is evident? You see I can show with the Riemann Lebesgue lemma that the difference tends to zero without noticing that $f$ can be anything and it bothers me. – Patrik Nusszer Feb 25 '22 at 20:57
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    Not sure what are you talking about as in all computations only the values on the period interval of $f$ appear so it doesn't matter what $f$ is elsewhere, which is why for convenience $f$ is just taken periodic; try doing the computations with a single simple function like $x^2$ and convince yourself – Conrad Feb 25 '22 at 21:38
  • @Conrad I am sure the computations are in favour of the original theorem. I did not mean to state that I can use the Riemann-Lebesgue Lemma to conclude the same result under the wrong conditions. You take the portion of the function on $[-\pi,\ \pi]$, and the coefficients you obtain over this interval will cause the series to approach this portion of the function, and of course, if you were to plot the series you would see this portion repeating with a period of $2\pi$ since the series is $2\pi$ periodic. I just do not know how to obtain a precise proof for this. – Patrik Nusszer Feb 25 '22 at 21:55
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    Again, you do not use anywhere the values of the function outside of the period so this means that what you have is a function defined on $(-\pi,\pi)$ say or whatever period interval you use, so it doesn't really matter what the function is anywhere else; of course if you want the Fourier series to hold outside the period interval, you need to impose periodicity, but that is not necessary as ultimately the Fourier series is a representation on the period interval first and foremost – Conrad Feb 26 '22 at 01:14
  • @Conrad is the only problem that I use function values outside the interval in the bad version of the proof? – Patrik Nusszer Feb 26 '22 at 13:51
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    Again not sure what is unclear; use a specific function to do the exact computations; when you say $S_N(x) = \frac{1}{\pi}\int_{-\pi}^{\pi}f(u)D_N(x-u)du$, why do you think that holds true for $x$ outside $[-\pi, \pi]$? Take the function $x$; on $(-\pi,\pi)$ it has a sine series (more or less $\sum -2\sin nx/n$) but on $(0, 2\pi)$ it has a very different Fourier series with both $\sin, \cos$ terms, so for $x \in (0, \pi)$ you get two very different formally series but taking same value there and check the formulas above for $x=\pi/2, 3\pi/2$ and see how they work or fail to work in each case – Conrad Feb 26 '22 at 16:33
  • @Conrad $S_N(x)$ is just a definition. This is how the Fourier series of $f$ is defined on the interval $[-\pi,\ \pi]$. A definition can not be wrong. A better question would be if $x$ is outside the interval, then why we can not say that the difference of $S_N(x)$ and $f$ tend to zero. But I actually can not answer it. – Patrik Nusszer Feb 26 '22 at 20:20
  • There are periodic, continuous, functions whose Fourier series are not pointwise convergent. An example was constructed by du Bois-Raymond, and I wrote about it here https://math.stackexchange.com/a/287854/448 . The key to having a non-pointwise convergent Fourier series is a high level of non-differentiability, not being non-periodic. – David E Speyer Mar 04 '22 at 14:15
  • @DavidESpeyer Are the conditions for pointwise concedgence stated by Dirichlet in 1829 wrong? What is the correct way to prove it, where can I find a good source for it? – Patrik Nusszer Mar 04 '22 at 14:49
  • I'm not sure what Dirichlet wrote, but the Wikipedia article seems reasonably clear and I don't see any errors in it https://en.wikipedia.org/wiki/Convergence_of_Fourier_series – David E Speyer Mar 04 '22 at 14:54

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Ignoring the typos where you lost a factor of $\frac{1}{\pi}$ at a few stages, everything is correct until the very end, where you try to apply Riemann-Lebesgue Lemma. The left hand integral does indeed vanish due to this lemma, but the lemma cannot be applied to the right hand integral because it fails the Riemann integrable condition.

Label the function $F(t)=\frac{(f(x)-f(x-t))\cos(\frac{t}{2})}{2\sin(\frac{t}{2})}$, then what we want to say is that: $$\lim_{N\to\infty}\int_{x-\pi}^{x+\pi}F(t)\sin(Nt)dt = 0$$ This cannot be guaranteed, however. Due to that $\sin(\frac{t}{2})$ in the denominator, $F$ has a singularity whenever $\sin(\frac{t}{2})=0$, which happens whenever $t$ is a multiple of $2\pi$. Since the range of $t$ in the integral is of length $2\pi$, this is guaranteed to occur at least once, no matter what $x$ is. Without imposing other conditions on $f$, we should expect $F$ explode in value when $t$ is near one of these singularities, so $F$ isn't integrable and Riemann-Lebesgue Lemma cannot apply.

This isn't an issue when $x$ is in the interval $(-\pi,\pi)$, since then the only singularity is at $t=0$. At this point, the $f(x)-f(x-t)$ in the numerator is zero, and since $f$ is differentiable, this can cancel out the previously mentioned singularity. This explains why the Fourier series converges to $f$ on the interval $(-\pi,\pi)$, even if $f$ is not periodic. If in addition to the above, $f$ is also $2\pi$ periodic, then the periodicity lets the above resolution extend to all the other singularities, so that $F$ is integrable and Riemann-Lebesgue applies, giving convergence everywhere.

Addendum: As a rule of thumb, typically a theorem (or lemma) will be stated as an implication: If $F$ is Riemann integral, then this limit of integrals converges to zero. If the conclusion of a theorem gives a nonsense result however, you can use the contrapositive to figure out what went wrong: If that limit of integrals doesn't converge to zero, then it must be that $F$ is not Riemann integrable.

Jade Vanadium
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  • By singularity do you mean here discontinuity? By saying "we should expect F to explode" you mean at this point, as $f$ is unkown, it may be possible that here F has an infinite discontinuity and F is therefore may be unbounded? This problem evidently does not occur at $t = 0$, but at other singularities this problem occurs strictly always, or "ir may or may not occur"? – Patrik Nusszer Mar 09 '22 at 08:12
  • @PatrikNusszer Yes, by singularity I mean discontinuity; we divide by $0$, so $F$ is undefined there. When I said we should expect $F$ to explode, I only meant that usually $F$ will be unbounded there, but it may or may not happen, and it depends on $f$. For example, the singularity at $t=n2\pi$ will be a removable discontinuity so long as $f(0)=f(n2\pi)$. This always happens at $n=0$, and it always happens if $f$ is $2\pi$ periodic. However, if we have $f(0)\neq f(n2\pi)$, then necessarily $F(t)$ is unbounded as $t$ tends to $n2\pi$, so it is an infinite discontinuity. – Jade Vanadium Mar 09 '22 at 22:59
  • One more little subtlety: am I right that for this proof to hold we need the function to be continuous and differentiable as well? Just because as far ad I know, there exist continuous functions that are divergent at some points. – Patrik Nusszer Mar 19 '22 at 13:55
  • @PatrikNusszer indeed, this proof requires the function to be continuous and differentiable (actually I was assuming it was continuously differentiable). There are stronger proofs that weaken these assumptions, but they end up being somewhat harder to work out. – Jade Vanadium Mar 20 '22 at 15:29