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I have asked this question before but I did not get any answers so I hope it is OK if I ask again.

Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ in $C_{\text{st}}$ which satisfies that $$ f(x) = 6x+2 $$ when $-\pi < x < \pi$. Then I have to argue for or against if the Fourier series converges pointwise or uniformly on $\mathbb{R}$. I have asked this question before but as $C_{\text{st}}$ is not common notation I hope I can get some more answers when defining what it means.

I would very much like to know how to tackle these kinds of questions as they most definitely will be a part of my analysis exam in three weeks.

Definition: Let $C_{\text{st}}$ be the set of the functions $f: \mathbb{R} \rightarrow \mathbb{C}$ which satisfies that

  1. $f$ is $2\pi$-periodic
  2. $f$ is piecewise continuous on the interval $[-\pi, \pi]$
  3. $f$ is normalized in its points of discontinunation meaning that $f(x) =\frac{f(x_{-})+f(x_{+})}{2}$

Futhermore we also need the following

Definition: Let $C^1_{\text{st}}$ be the set of the functions $f: \mathbb{R} \rightarrow \mathbb{C}$ which satisfies

  1. $f$ is $2\pi$-periodic
  2. $f$ is piecewise differentiable on the interval $[-\pi, \pi]$
  3. $f$ is normalized in its points of discontinunation

Then my book says that

Definition: The Fourier series for a function $f \in C^1_{\text{st}}$ converges pointwise towards $f$ on $\mathbb{R}$

and

Definition: If $f \in C^1_{\text{st}}$ and continuous on $\mathbb{R}$ then the Fourier series for $f$ converges uniformly on $\mathbb{R}$

Then to prove pointwise convergence, are these definitions sufficient to show that $f$ is piecewise differentiable on $[-\pi,\pi]$ as $f \in C_{\text{st}}$?

Then to prove uniform convergence, are these definitions sufficient to show that $f$ is piecewise differentiable on $[-\pi,\pi]$ as $f \in C_{\text{st}}$ and that $f$ is continuous on $\mathbb{R}$?

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    $\begingroup$ You should always edit your previous question to explain notations/add context instead of asking a new question. $\endgroup$ – Sahiba Arora May 28 at 13:50
  • $\begingroup$ Ok I will keep that in mind. $\endgroup$ – Mathias May 28 at 13:57
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There seems to be a number of questions here, likely covered in your book, but it could be helpful to list non-exhaustively some of the main results regarding the convergence of Fourier series. Apologies if long, but I hope it will be a helpful checklist for you.

We shall assume $ f : \mathbb R \to \mathbb C $ is $2\pi$-periodic. We consider the normed Lebesgue integrable spaces, $L^1(-\pi,\pi)$ and $L^2(-\pi,\pi)$, recalling that on a bounded interval $L^2 \subset L^1$. We can associate any $f$ in either $L^2(-\pi,\pi)$ or $L^1(-\pi,\pi)$ with its Fourier series, writing $$f \sim \sum_{k=-\infty}^{+\infty} a_k e^{ikx} \quad\text{and}\quad S_n(f,x) = \sum_{k=-n}^{n}a_ke^{ikx} $$ where each coefficient is given by $\displaystyle a_k =\frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-ikx} ~dx $. The integrals exists for $ f \in L^2 $ or $L^1$.

The most common covergence results are :

  1. Convergence in $L^2$ norm. For all $f \in L^2(-\pi,\pi)$ $$\left\lVert S_n - f \right\rVert_{L^2} \to 0 \text{ as } n \to \infty $$ where the norm for any $f$ is $ \displaystyle \lVert f \rVert_{L^2} = \int_{-\pi}^{\pi} \lvert f(x) \rvert^2 ~dx $.
  2. Parseval. For all $f \in L^2(-\pi,\pi)$, the sum $\displaystyle \sum_{k=-n}^{n} |a_k|^2 \to \lVert f \rVert^2$ as $n \to\infty$

  3. Pointwise convergence (Jordan). Let $x_0 \in \mathbb R$. If $f \in L^1(-\pi,\pi)$ has bounded variation on an interval $[x_0-r, x_0+r]$ for some $ r > 0$. Then the limits $$f(x_0+) = \lim_{h \searrow 0} f(x+h) \quad\text{and}\quad f(x_0-) = \lim_{h\searrow 0} f(x-h) $$ both exist and $S_n(f,x_0) \to \dfrac{1}{2} ( f(x_0+) + f(x_0-) ) $. This resulte embodies the localisation principle where the convergence of $f$ at $x_0$ depends only on its characteristics in an arbitrarily small interval around $x_0$.

  4. Uniform convergence. If $f$ is $2\pi$-periodic, continuous on $\mathbb R$ (note that implies $f(\pi) = f(-\pi)$) and piecewise continuously differentiable (i.e. the interval $[-\pi,\pi]$ can be divided into a finite number of sub-inetrvals $I_j, j=1, \cdots, m$ and $f$ is continuously differentiable in each $I_j$, with one sided derivatives at the end points) then the Fourier series $S_n(f,x)$ converges absolutely and uniformly to $f(x)$ on $[-\pi,\pi]$.

  5. Gibbs phenomena. For a function $f$ that is piecewise continuous, convergence at points of discontinuity is non-uniform. In fact the the maximum error between $S_n(f,x)$ and $f(x)$ has positive limit.

The function $f(x) = 6x+2$ meets the criteria for 1,2,3 but not 4 because there is no definition at $x = \pm \pi$ that would allow the function to be continuous there. The Fourier series at $\pm \pi$ converges to the mid point $\frac{1}{2}(f(0+)+f(0-)) = 2$.

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  • $\begingroup$ So the Fourier Series converges pointwise but not uniormly In this instance? $\endgroup$ – Mathias May 29 at 20:54
  • $\begingroup$ Furthermore, you say that in 4. uniform convergence on $\mathbb{R}$ implies that $f(\pi) = f(-\pi)$ which in this case is not correct. Is this what you mean when you say there "there is no definition at $x = \pm \pi$ that would allow the function to be continuous there"?. I guess so. Or is it because that $f(x)$ is defined in the open interval from $(-\pi,\pi)$. Thanks for your help! :) $\endgroup$ – Mathias May 29 at 21:14
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    $\begingroup$ Hi. On uniform convergence,: $ f(\pi) = f(-\pi)$ is sufficient when combined with the other properties. I was not saying it is necessary, although in practice it will be. And I added that in your example $f(x) =3x+2$, Which is defined only on the open interval, no values of $f(\pi)$ and $f(-\pi)$ can make it continuous. Hope that helps. $\endgroup$ – WA Don May 29 at 22:30
  • $\begingroup$ Thanks again for your help. If you don't mind I have one last question. This is a question from an Analysis exam which I am studying for. Another example is the given by the function $f \in C_{st}$ satisfying $f(x) = 1 + \cos(x/2)$ for $-\pi < x < \pi$. Here my professor has provided the answers and says that $f$ is continuous on $\mathbb{R}$, thus on $]-\pi,\pi[$ as well, with $f(-\pi) = f(\pi) = 0$ and $C^{1}$ on $\mathbb{R}$ he concludes that the Fourier series converges uniformly on $\mathbb{R}$. Doesn't this contradict what you are saying? Or am I misunderstand something? Thanks. $\endgroup$ – Mathias May 29 at 23:19
  • $\begingroup$ No. In this case the function meets the conditions of No 4. It is periodic and continuous on the real line, which means it has equality at $\pm \pi$ and in this case it is also differentiable everywhere, which is stronger than needed by 4. So it converges uniformly on $[-\pi,\pi]$. Then, as it’s periodic the uniform convergence equally holds true on the whole real line. Is that clearer? $\endgroup$ – WA Don May 30 at 6:16

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