# Local uniform continuity of a^q

1. Aug 19, 2014

### jostpuur

Let $a\in\mathbb{R}$, $a>0$ be fixed. We define a mapping

$$\mathbb{Q}\to\mathbb{R},\quad q\mapsto a^q$$

by setting $a^q=\sqrt[m]{a^n}$, where $q=\frac{n}{m}$. How do you prove that the mapping is locally uniformly continuous? Considering that we already know what $q\mapsto a^q$ looks like, we can define the local uniform continuity by stating that the restriction to $[-R,R]\cap\mathbb{Q}$ is uniformly continuous for all $R>0$. The continuity is considered with respect to the Euclidian metric, which $\mathbb{Q}$ inherits from $\mathbb{R}$.

The use of a mapping

$$\mathbb{R}\to\mathbb{R},\quad x\mapsto a^x$$

and its derivative is not allowed, because the claim is elementary, and may be needed in the proofs of the most basic results concerning the $a^x$.

edit: Actually I don't know how to prove that $q\mapsto a^q$ is merely continuous in the ordinary way either, so I wouldn't mind some information on that too.

Last edited: Aug 19, 2014
2. Aug 19, 2014

### pasmith

Let $f : \mathbb{Q} \to \mathbb{R} : x \mapsto a^x$. Then we have that $f$ is continuous at $x$ if and only if $$\lim_{h \to 0} |f(x + h) - f(x)| = 0.$$ But $$|f(x + h) - f(x)| = |f(x)f(h) - f(x)| = |f(x)||f(h) - 1|.$$
Thus continuity at $x \in \mathbb{Q}$ follows from continuity at $0 \in \mathbb{Q}$, ie. $a^q \to 1$ as $q \to 0$. It suffices to show that $$\lim_{m \to \infty} a^{1/m} = 1$$ since for $q \neq 0$ we have $$\lim_{q \to 0} a^q = \lim_{m \to \infty} (a^{1/m})^n = \left( \lim_{m \to \infty} a^{1/m} \right)^n$$ using the facts that $x \mapsto x^n$ is continuous for positive integer $n$, and that if $g$ is continuous at $L$ and $x_n \to L$ then $$\lim_{n \to \infty} g(x_n) = g(L).$$

3. Aug 19, 2014

### jostpuur

I managed to solve my problem within few hours after posting it, but I didn't rush back with "never mind" comments, because I thought it would be better to see what responses appear.

In my solution I proved the following claim: Assume that $n_1,n_2,n_3,\ldots\in\mathbb{Z}$ and $m_1,m_2,m_3,\ldots\in\mathbb{Z}$ are such sequences that $m_i>0$ for all $i$ and

$$\frac{n_i}{m_i}\underset{i\to\infty}{\to} 0$$

Then

$$\sqrt[m_i]{a^{n_i}} \underset{i\to\infty}{\to} 1$$

Keeping the $n$ as constant in the limit looks like a mistake.