ProofWiki:Sandbox

Theorem
Let $a \in \R$ be a real number such that $a > 0$.

Let $f : \Q \to \R$ be the mapping defined by $f \left({ x }\right) = a^x$.

That is, let $f \left({ x }\right)$ denote $e$ to the power of $x$, for  rational $x$.

Then there exists a unique continuous extension of $f$ to $\R$.

Proof
Consider $I_k := \left({ -k \,.\,.\, k }\right)$ for arbitrary $k \in \N$.

Note that, for all $x, y \in I_k$:

where $M = \max \left({ a^{-k}, a^{k} }\right)$.

Before we proceed, a lemma will be needed.

Lemma

 * $\displaystyle \lim_{x \to 0} a^x = 1$

where x runs through the rationals.

Proof of Lemma
Suppose for now that $a > 1$ and $x \to 0^{+}$.

We first show that, for $0 < x < 1$:
 * $1 \leq a^x \leq 1 + ax$

Fix $x$ and define:
 * $g \left({ a }\right) = 1 + ax - a^x$

We have that: $D_a \left[{ 1 + ax - a^x }\right] = x \left({ 1 - a^{x-1} }\right)$

We show now that the derivative above is positive for all $a > 1$:

So $D_a g$ is increasing for all $a > 1$.

Further, $D_a g \left({ 1 }\right) = 0$.

Thus $D_a g \left({ a }\right)$ is positive for all $a > 1$.

Whence, by Derivative of Monotone Function, $g$ is increasing for all $a > 1$.

Now, $g \left({ 1 }\right) = x > 0$.

So $g \left({ 1 }\right)$ is positive for all $a > 1$.

That is:
 * $ a^x \leq 1 + ax$

Finally,

So the desired inequality holds.

By the Squeeze Theorem:
 * $\displaystyle \lim_{ x \to 0^{+} } a^x = 1$

We now quickly treat the other cases.

If $0 < a < 1$ and $x \to 0^{+}$, we look at $0 < x < 1$.

From above, we have that $\dfrac{1}{a} > 1$ and the desired inequality becomes:
 * $ \dfrac{1}{1 + ax} < a^x < 1$

If $a > 1$ and $x \to 0^{-}$, we look at $-1 < x < 0$.

From above, we have that $0 < -x < 1$ and the desired inequality becomes:
 * $ \dfrac{1}{1 - ax} < a^x < 1$

If $0 < a < 1$ and $x \to 0^{-}$, we look at $-1 < x < 0$.

From above, we have that $\dfrac{1}{a} > 1$ and $0 < -x < 1$. Then the desired inequality becomes:
 * $ 1 < a^x < 1 - \dfrac{x}{a}$

Applying squeeze theorem in each case brings us to the result.

We now complete the proof. Fix $\epsilon \in \R_{> 0}$.

By the lemma,
 * $\exists \delta \in \R_{> 0} : \vert{ x - y }\vert < \delta \implies \vert{ a^{x - y} - 1 }\vert < \dfrac{\epsilon}{M}$

where M is defined as above.

Therefore, $\exists \delta \in \R_{> 0}$ such that:
 * $\vert{ x - y }\vert < \delta \implies \vert{ a^x - a^y }\vert \leq M \vert{ a^{x - y} - 1 }\vert < M \dfrac{\epsilon}{M} = \epsilon$

That is, $f$ is uniformly continuous on $I_k \cap \Q$.

From Rationals are Everywhere Dense in Reals, $I_k \cap Q$ 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 $\displaystyle F = \bigcup \left \{ f_k : k \in \N \right \}$

Note that $ \left \langle{ I_k }\right \rangle$ 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.