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.