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.