Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ in $C{st}$ 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 $C{st}$ 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 $C{st}$ be the set of the functions $f: \mathbb{R} \rightarrow \mathbb{C}$ which satisfies that
- $f$ is $2\pi$-periodic
- $f$ is piecewise continuous on the interval $[-\pi, \pi]$
- $f$ is normalized in its points of discontinunation
Futhermore we also need the following:
Definition: Let $C^1{st}$ be the set of the functions $f: \mathbb{R} \rightarrow \mathbb{C}$ which satisfies
- $f$ is $2\pi$-periodic
- $f$ is piecewise differentiable on the interval $[-\pi, \pi]$
- $f$ is normalized in its points of discontinunation
Then my book says that:
Definition: The Fourier series for a function $f \in C^1{st}$ converges pointwise towards $f$ on $\mathbb{R}$
and
Definition: If $f \in C^1{st}$ and continuous on $\mathbb{R}$ then the Fourier series for $f$ converges uniformly on $\mathbb{R}$
I have earlier said that $f(x)$ indeed is continuous as it is a composition of well-known continuous functions. Thus the following limits exist $$ \lim{x \rightarrow \pi^-} 6x+2 = 6\pi + 1 $$ and $$ \lim{x \rightarrow -\pi^+} 6x+2 = -6\pi + 1 $$ which means that $f(x)$ is continous in $\pm \pi$. But doesn't this mean that $f(x)$ is not $2\pi$-periodic? As $f(\pi) \neq f(-\pi)$ and how can I use this to argue for or against if the Fourier series for $f$ converges pointwise or uniformly on $\mathbb{R}$.
Do you mind helping me?