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Local uniform continuity of a^q

  1. Aug 19, 2014 #1
    Let [itex]a\in\mathbb{R}[/itex], [itex]a>0[/itex] be fixed. We define a mapping

    [tex]
    \mathbb{Q}\to\mathbb{R},\quad q\mapsto a^q
    [/tex]

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

    The use of a mapping

    [tex]
    \mathbb{R}\to\mathbb{R},\quad x\mapsto a^x
    [/tex]

    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 [itex]a^x[/itex].

    edit: Actually I don't know how to prove that [itex]q\mapsto a^q[/itex] is merely continuous in the ordinary way either, so I wouldn't mind some information on that too.
     
    Last edited: Aug 19, 2014
  2. jcsd
  3. Aug 19, 2014 #2

    pasmith

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    Let [itex]f : \mathbb{Q} \to \mathbb{R} : x \mapsto a^x[/itex]. Then we have that [itex]f[/itex] is continuous at [itex]x[/itex] if and only if [tex]
    \lim_{h \to 0} |f(x + h) - f(x)| = 0.[/tex] But [tex]|f(x + h) - f(x)| = |f(x)f(h) - f(x)| = |f(x)||f(h) - 1|.[/tex]
    Thus continuity at [itex]x \in \mathbb{Q}[/itex] follows from continuity at [itex]0 \in \mathbb{Q}[/itex], ie. [itex]a^q \to 1[/itex] as [itex]q \to 0[/itex]. It suffices to show that [tex]\lim_{m \to \infty} a^{1/m} = 1[/tex] since for [itex]q \neq 0[/itex] we have [tex]
    \lim_{q \to 0} a^q = \lim_{m \to \infty} (a^{1/m})^n = \left( \lim_{m \to \infty} a^{1/m} \right)^n[/tex] using the facts that [itex]x \mapsto x^n[/itex] is continuous for positive integer [itex]n[/itex], and that if [itex]g[/itex] is continuous at [itex]L[/itex] and [itex]x_n \to L[/itex] then [tex]
    \lim_{n \to \infty} g(x_n) = g(L).
    [/tex]
     
  4. Aug 19, 2014 #3
    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 [itex]n_1,n_2,n_3,\ldots\in\mathbb{Z}[/itex] and [itex]m_1,m_2,m_3,\ldots\in\mathbb{Z}[/itex] are such sequences that [itex]m_i>0[/itex] for all [itex]i[/itex] and

    [tex]
    \frac{n_i}{m_i}\underset{i\to\infty}{\to} 0
    [/tex]

    Then

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

    Keeping the [itex]n[/itex] as constant in the limit looks like a mistake.
     
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