Let be a function defined on some interval and define for . Show that exists if and only if exists in which case they are equal.
My attempt of proof:
Let and suppose is finite. Then for each there exists such that implies . If , then , so implies . Let . So it follows that there exists , namely , such that implies , and so .
The converse is pretty much the reverse of this proof and if , I'd just simply replace the epsilons with and , and do the similar if
I think this proof is correct, but I just want to make sure, thanks.