Can anyone explain to me, how to show the conditions of the following definitions for any given function:
$\text{Definition 1}:$
$ \text{let}\ \mathcal{C_{st}} \ \text{denote the set of those functions $f:\mathbb{R} \rightarrow \mathbb{C}$},\text{that satisfies the following conditions:} \\(1)\ f \ \text{is periodic with period} \ 2\pi \\(2)\ f \ \text{is piecewise continuous on the interval $[-\pi,\pi]$} \\(3)\ f \ \text{is normalized in its points of discontinuity, i.e.} $ $$ f(x)=\frac{f(x_-)+f(x_+)}{2} $$ at any point of discontinuity $x\in \mathbb{R}$ for $f$
$\text{Definition 2}:$
$\text{let}\ \mathcal{D_{st}} \ \text{denote the set of those functions $f:\mathbb{R} \rightarrow \mathbb{C}$},\text{that satisfies the following conditions:} \\(1)\ f \ \text{is periodic with period} \ 2\pi \\(2)\ f \ \text{is piecewise differentiable on the interval $[-\pi,\pi]$} \\(3)\ f \ \text{is normalized in its points of discontinuity, i.e.} $ $$ f(x)=\frac{f(x_-)+f(x_+)}{2} $$ at any point of discontinuity $x\in \mathbb{R}$ for $f$
Just to let you know $\mathcal{C}_{st}$ is a subspace of the vector space of all complex function on $\mathbb{R}$, so $\mathcal{C}_{st}$ is a complex vector space, that have the following scalar product: $$ \left( f,g\right):=\frac{1}{2\pi} \int_{-\pi}^\pi {f(x) \overline{g(x)}} \ dx \ , f,g\in \mathcal{C}_{st} $$ Where $\mathcal{D}_{st} \subseteq{\mathcal{C}_{st}}$
$f(x_-)=\lim_{y \rightarrow x_-}{f(y)}$ is the limit of the function from the left hand side.
$f(x_+)=\lim_{y \rightarrow x_+}{f(y)}$ is the limit of the function from the right hand side.
Definition 1 and 2 is pretty much the same except for condition number 2, and it could be very nice if someone could explain me how to show that a function is
- Periodic with period $2\pi$
- Piecewise continuous on the interval $[-\pi,\pi]$
- Piecewise differentiable on the interval $[-\pi,\pi]$
- Normalized in its points of discontinuity, i.e. $$f(x)=\frac{f(x_-)+f(x_+)}{2}$$ at any point of discontinuity $x\in \mathbb{R}$ for $f$
I really appreciate it and thanks in advance