I have been dusting the cobwebs off of my math knowledge and in working through Rudin's PMA, I have come to a rather small detail that I have been unable to supply an explanation for. In constructing the exponential and logarithmic functions in chapter 8, Rudin eventually writes that if $x>0$, $n$ is an integer, and $m$ is a positive integer, we have
$$ x^{\frac{1}{m}}=E\bigg(\frac{1}{m}L(x)\bigg) $$
I find myself unable to confidently supply the missing details using anything in the section on exponents and logarithms up to that point. Is it simply a consequence of the fact that Rudin has already constructed $E(p)=e^p$ for rational $p$ and defined $E(x)=e^x=\mathrm{sup}\ e^p$, where $x$ is real, and $p<x$, in which case I would suppose that $$E\bigg(\frac{1}{m}L(x)\bigg)=E\bigg(L(x)\frac{1}{m}\bigg)=(e^{ln(x)})^\frac{1}{m}=x^\frac{1}{m}$$ I feel as though I am overlooking something.
Best Answer
Note that $E\left(\frac{1}{m}L(x)\right) > 0$ and $$\left(E\left(\frac{1}{m}L(x)\right)\right)^m = E\left(m\cdot \frac{1}{m}L(x)\right) = E(L(x)) = x$$ where the first equality follows from (29) with $z_1=\ldots=z_n = \frac{1}{m}L(x)$ and $n=m$.
It follows from the theorem-definition 1.21 that $$E\left(\frac{1}{m}L(x)\right) = x^{1/m}$$