A new tag piecewise-continuity has been created by a new user. Currently the tag is used by only 7 questions - all written by the tag creator. All this is typical of a well-meaning user who is not experienced enough about either math in general, or tagging on our site in particular, or both. My...

In several discussions here on meta, one user expressed the opinion, that every new tag (or at least most of them) should be discussed on meta before creating. For example here There is a saying that it's easier to apologize later than get permission first. I disagree with that when it comes...

I'm a little bit confused in piecewise continuity of a function. Say, if we have an odd function like $f(x) = x$ defined on the open interval $(0, \pi)$. We then extend it to a period $2\pi$ function and find its sine Fourier series. Can we say that this function is then piecewise continuous, but...

if I have a piecewise continuously differentiable function. How do I see that on each open interval, where the derivative is continuous, there is a continous extension on the larger closed interval?

Assume $f$ and $g$ are two piecewise continuous functions on an interval $( a , b )$ containing the point $t_0$ . Assume further that $f$ has a jump discontinuity at $t_0$ while $g$ is continuous at $t_0$ .How can i verify that the jump in the product $fg$ at $t_0$ is given by “the jump in $f$ at...

Say we have the piecewise function $f(x) = x^2$ on the interval $0 \le x < 4$; and it equals $x+1$ on the interval $ x \ge 4$. Why is it that, when I take the derivative, the intervals loose their equality and become strictly greater or strictly less than?

Let be $f:\mathbb{R} \rightarrow \mathbb{C}$. Consider the space of piecewise linear curves, with support in the interval [-1,1], sucht that $f(x)= A-|x|$ if $|x|\leq 1$; $f(x)= 0$ otherwise. For this space show that expression $p(f)=\int_{-\infty}^{\infty}|f(x)|^2dx$ is a norm. I am show that $...