2
$\begingroup$

Define $$f(x) = \begin{cases}x^{\alpha}\sin(x^{-\beta}),& 0<x\leq 1 \\ 0,& x=0. \end{cases}$$ Show that $f$ is of bounded variation on $[0,1]$ iff $\alpha > \beta$.

Here is a solution: enter image description here

I didn't understand the following implication:

"... $f$ is of bounded variation on $[0,x_{0}]$, hence of bounded variation on $[0,1]$."

Could someone explain to me?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.