Local uniform continuity of a^q

[jostpuur]

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

<br /> \mathbb{Q}\to\mathbb{R},\quad q\mapsto a^q<br />

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&gt;0. The continuity is considered with respect to the Euclidian metric, which \mathbb{Q} inherits from \mathbb{R}.

The use of a mapping

<br /> \mathbb{R}\to\mathbb{R},\quad x\mapsto a^x<br />

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.

 

[pasmith]

Let a\in\mathbb{R}, a&gt;0 be fixed. We define a mapping

<br /> \mathbb{Q}\to\mathbb{R},\quad q\mapsto a^q<br />

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&gt;0. The continuity is considered with respect to the Euclidian metric, which \mathbb{Q} inherits from \mathbb{R}.

The use of a mapping

<br /> \mathbb{R}\to\mathbb{R},\quad x\mapsto a^x<br />

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.

Let f : \mathbb{Q} \to \mathbb{R} : x \mapsto a^x. Then we have that f is continuous at x if and only if <br /> \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 <br /> \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 <br /> \lim_{n \to \infty} g(x_n) = g(L).<br />

 

[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&gt;0 for all i and

<br /> \frac{n_i}{m_i}\underset{i\to\infty}{\to} 0<br />

Then

<br /> \sqrt[m_i]{a^{n_i}} \underset{i\to\infty}{\to} 1<br />

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