Ok, I'd like to try:

Prove that if

on a subset of the real axis

, then

throughout

.

Since

is analytic, then we can write for some

:

where

is analytic and bounded in some neighborhood of

, that is

for

, and

is the first non-zero term. Now:

so that:

Now recall that we have a dense set of points over which

. Therefore, we can find a

such that

. This means:

This would imply

since

and

but by assumption

. This must mean we've reached a contradiction and all

so that

.

I'll leave it to you to show the function

on the same domain implies

since

can still be written as

for some (new) set of

.

This may need a little more work . . .