How to extend a logarithmic function from the rationals to the reals? Question.  Say I have a continuous function $H$ with the established property that $H(p^y) = y H(p)$ for all rational numbers $y$.  How can I extend it to all real numbers?
My attempt.  Being continuous we have (by definition) that $\lim H(p^x) = H(p^y)$ where $x \to y$.  Let $(X_n)$ be a convergent sequence of rational numbers whose limit is $y$, a fixed real number.  If $x$ is an element of $(X_n)$ then $H(p^x) = x H(p)$ by the already-established property for all rational numbers.  Then we can deduce
$$H(p^y) = \lim_{x \to y} H(p^x) = \lim x H(p) = (\lim x) H(p) = y H(p).$$
I'm not concerned with proving such construction is clear and possible given current standards of rigor.  But I wonder if this strategy is at all possible.  My main concern is with the possibly wild statement "Let $(X_n)$ be a convergent sequence of rational numbers whose limit is $y$, a fixed real number." Does this work?  If not, how is this problem solved more or less properly?  (Feel free to just point me to a reference when the extension is done.  That'd be great actually.  Thank you.)
 A: Let's see if I have the actual question and assumptions laid out?
You have a continuous $H:\mathbb R \to \mathbb R$.
We have the rather strange condition that for any given $w = p^q$ where $q$ is a rational number that $H(w) = H(p^q) = q H(p)$.
You want to know if we can conclude that for any $\omega = p^x$ where $x$ is real that $H$ has the property that $H(\omega) = H(p^x) = xH(p)$?
ANd this all assumes that we have defined what $p^x$ where $x$ is irrational is?[1]
You are correct.  By the very definition of the reals if $x$ is real then then there are always sequences of rational, $q_i$ so that $q_i \to x$.  That is what the reals are.
So.....  As $H$ is continuous $\lim_{q\to x;q\in \mathbb Q} H(p^q) = H(p^x)$.
And as $H(p^q) = qH(p)$ for all rational $q$ then  $\lim_{q\to x;q\in \mathbb Q} H(p^q) =\lim_{q\to x;q\in \mathbb Q}qH(p) = [\lim_{q\to x;q\in \mathbb Q}q]H(p) = xH(p)$.
Our claim is proven exactly as you claimed.
.....
[1]  But is that the question and have you defined what $p^x$ if $x$ is irrational?
Is the question actually if $S = \{p^q| p,q \in \mathbb Q\}$ and $H:S\to \mathbb R$ with the property $H(p^q) = qH(p)$ can we extent the definition to $H'(x) = \lim_{q\to x} H(q)$ and if we define $p^x$ so $\lim_{q\to x} p^q$ then will $H'(p^x) = xH'(p)$?
Was that the actual question?
