Power Function to Rational Power permits Unique Continuous Extension
Theorem
Let $a \in \R_{> 0}$.
Let $f: \Q \to \R$ be the real-valued function defined as:
- $\map f q = a^q$
where $a^q$ denotes $a$ to the power of $q$.
Then there exists a unique continuous extension of $f$ to $\R$.
Proof
Consider $I_k := \openint {-k} k$ for arbitrary $k \in \N$.
Let $I_k' = I_k \cap \Q$.
Note that, for all $x, y \in I_k'$:
| \(\ds \size {a^x - a^y}\) | \(=\) | \(\ds \size {a^{x - y + y} - a^y}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \size {a^{x - y} a^y - a^y}\) | Powers of Group Elements | |||||||||||
| \(\ds \) | \(=\) | \(\ds \size {a^y} \size {a^{x - y} - 1}\) | Absolute Value Function is Completely Multiplicative | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^y \size {a^{x - y} - 1}\) | Power of Positive Real Number is Positive over Rationals | |||||||||||
| \(\ds \) | \(\le\) | \(\ds M \size {a^{x - y} - 1}\) | Power Function is Monotone over Rationals |
where $M = \map \max {a^{-k}, a^k}$.
Fix $\epsilon \in \R_{> 0}$.
From Power Function tends to One as Power tends to Zero: Rational Number:
| \(\ds \exists \delta \in \R_{>0}: \, \) | \(\ds 0\) | \(<\) | \(\, \ds \size {x - y} \, \) | \(\, \ds < \, \) | \(\ds \delta\) | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds \size {a^{x - y} - a^0} \, \) | \(\, \ds < \, \) | \(\ds \dfrac \epsilon M\) | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds \size {a^{x - y} - 1} \, \) | \(\, \ds < \, \) | \(\ds \dfrac \epsilon M\) | Definition of Power of Zero |
where $M$ is defined as above.
Therefore, $\exists \delta \in \R_{> 0}$ such that:
| \(\ds \size {x - y}\) | \(<\) | \(\ds \delta\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \size {a^x - a^y}\) | \(\le\) | \(\ds M \size {a^{x - y} - 1}\) | |||||||||||
| \(\ds \) | \(<\) | \(\ds M \dfrac \epsilon M\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \epsilon\) |
That is, $f$ is uniformly continuous on $I_k'$.
From Rationals are Everywhere Dense in Topological Space of Reals, $I_k'$ is everywhere dense in $I_k$.
From Continuous Extension from Dense Subset, there exists a unique continuous extension of $f$ to $I_k$.
Call this function $f_k$.
Define:
- $\ds F = \bigcup \set {f_k : k \in \N}$
Note that $\sequence {I_k}$ is an exhausting sequence of sets in $\R$.
By the Union of Functions Theorem, $F$ defines a function $\R \to \R$
From the Pasting Lemma, $F$ is continuous on $\R$.
Hence the result.
$\blacksquare$