No need to prove the continuity for $e^x$?

Question

I've read the wikipedia article of "Limits of rational exponents". And it says the real exponents are defined as the limit of rational exponents. So, for a positive real number $b$ and any abitrary real number $x$, $b^x$ is well-defined continuously, is what they say. In short, is the exponential function defined as a continuous function by design? You don't need to prove its continuity because it's already continuous?

Answer 1

The article says that the rational exponential, the exponential defined on the rationals, is extended by continuity on the real line by $$ b^x=\lim_{r(\in \mathbb Q)\to x}b^r. $$ There is a list of things that should be proved:

• The function is well-defined: you need to show that the above limit exists, since it might not be the case. The proof might in principle split into the cases $x\in \mathbb Q$ and $x\in\mathbb R\setminus\mathbb Q$.
• Consistency: you might want to show that the above definition is actually extending the rational exponential, so when $x\in\mathbb Q$ you recover $b^x$. Luckily, this is a consequence of the previous point.
• Continuity:
  • You need to show that the limit exists without the restriction $r\in \mathbb Q$
  • You need to show that the limit coincides with the value of the function at the point $x$ (this will be actually a consequence of the definition of the function and all the previous points).

I will not give details on the proof, since this is not what you are asking, I found this post which gives what looks like a partial answer. In any case, the above points are definitely not an immediate consequence of the definition. There is definitely a few things to check to ensure that the definition is well-posed and the resulting function is continuous.

Answer 2

There's a general approach going on here:

• you define a function, let's call it $p_1: \mathbb Q \to \mathbb R$, where $p_1(x) = 2^x$ (I'm using "2" as an example here). The right-hand side is defined for integer $x$, and then using various bits of cleverness, for all rational $x$.

• You show that $p_1$, as a function from the rationals to the reals, is continuous. That's a bit of a pain, but it can be done.

• You apply a general theorem: if $A$ is a dense subset of $B$, and $f : A \to \mathbb R$ is uniformly$^{*}$ continuous, then there's a unique function $g: B \to \mathbb R$ such that (a) for $x \in A$, we have $g(x) = f(x)$, and (b) $g$ is continuous. In simpler words: any continuous function $f$ on a dense subset admits a unique continuous extension on the whole set. Applying this theorem to $p_1$ gives you a continuous function $p_2: \mathbb R \to \mathbb R$ with the property that $p_2(x) = p_1(x)$ for $x \in \mathbb Q$, as desired.

• If $q_1, q_2, \ldots$ is a sequence of rationals that converges to a rational $q$, then by continuity of $p_1$, you know that $$p_1(q) = \lim_{n \to \infty} p_1(q_n).$$ But because $p_2$ and $p_1$ agree on $\mathbb Q$, you also know that $$p_2(q) = \lim_{n \to \infty} p_2(q_n).$$

• On the other hand, if the sequence of $q_i$s converges to a non-rational, $s$, then continuity of $p_2$ (on all of $\mathbb R$!) once again ensures that $$p_2(s) = \lim_{n \to \infty} p_2(q_n).$$

• Indeed, if $r_1, r_2, \ldots \in \mathbb R$ converges to $r$, we have $$p_2(r) = \lim_{n \to \infty} p_2(r_n),$$ so the various concerns expressed in the comments about different kinds of limits are all taken care of by continuity of $p_2$.

The good news is that the general theorem is not too hard; it appears early in most topology texts.

Post-comment addition

The bad news is that I stated the theorem wrong --- you need uniform continuity. (Thanks, Lukas Heger, for pointing that out).

To make the result work with this more restrictive (but correct!) theorem, I'll consider an interval of the form $I_n = [n-1, n+1]$ where $n$ is an integer. Such an interval is compact (by Heine-Borel) so continuity and uniform continuity are the same thing. So now we know that there's a unique continuous extension of $p_1$ on such intervals. And the same is true for intervals of the form $[n, n+1]$. By considering $I_1$ and $I_2$, we can see that the extensions of $p_1$ for these two intervals can each be restricted to the (unique!) extension on $[0, 1]$, hence they must agree on that interval. A similar argument shows that the extensions on all possible $I_n$ agree on their overlaps, giving a function on all of $\mathbb R$ that extends $p_1$.

An alternative approach is to find a continuous extension over $[0, 1]$, and then note that for $x \in [1, 2]$, we have $$ p_1(x) = 2 * p_1(x-1) $$ and therefore define $p_2$ on $[1, 2]$ by $$ p_2(x) = 2 * p_2(x-1) $$ A similar trick lets us define $p_2$ on the interval $[2, 3]$, etc.

Answer 3

We define

$$a^q:=\sqrt[n]{a^m}$$ for a rational $q=\frac mn$. It is an easy matter to show that with this definition, $a^{p+q}=a^pa^q$.

It is enough to reason with $a>1$. In $\mathbb Q$, we have

$$|a^{q+\delta}-a^q|=|a^{q+\delta}-a^{q}|=a^{q}|a^\delta-1|<\epsilon$$

if

$$\log_a\left(1-\frac{\epsilon}{a^{q}}\right)<\delta<\log_a\left(1+\frac{\epsilon}{a^{q}}\right),$$ or for small $\epsilon$,

$$\delta<\frac{\epsilon}{a^{q}log(a)}.$$

Hence for all $\epsilon$ there is always a $\delta$ that fits, which proves continuity in $\mathbb Q$. If you don't like the logarithm, $\delta<\dfrac\epsilon{a^{q+1}}$ can do.

---

In $\mathbb R$, we define

$$a^r:=\lim_{q\to r,q\in\mathbb Q}a^q.$$ This limit indeed exists, by the above continuity argument. And we can show that $a^{r+s}=a^ra^s$, and that the real exponential so defined is a monotonic function.

Remains to show that there is a $\delta$ such that

$$\left|\lim_{d\to\delta,d\in\mathbb Q}a^d-1\right|$$ can be made smaller than $\epsilon$. This is certainly true because we can find a rational $d$ such that $|a^d-1|<\epsilon$, and take $\delta\le d$.