How to show pointwise/uniform converge for Fourier series in general

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 $$\mathbb{R}$$?

• You should always edit your previous question to explain notations/add context instead of asking a new question. – Sahiba Arora May 28 at 13:50
• Ok I will keep that in mind. – Mathias May 28 at 13:57

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 $$\frac{1}{2}(f(0+)+f(0-)) = 2$$.

• So the Fourier Series converges pointwise but not uniormly In this instance? – Mathias May 29 at 20:54
• 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! :) – Mathias May 29 at 21:14
• 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. – WA Don May 29 at 22:30
• 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. – Mathias May 29 at 23:19
• 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? – WA Don May 30 at 6:16