Local uniform continuity of a^q

In summary: If the limit is taken as n_1,n_2,\ldots,n_k\to\infty then the sequence m_i converges to a constant and this contradicts the assumption that m_i>0 for all i. When n_1,n_2,\ldots,n_k\to\inftyI proved the following claim: Assume that n_1,n_2,\ldots,n_k\in\mathbb{Z} and m_1,m_2,\ldots,m_k are such sequences that m_i>0 for all i and\frac{n_i}{m_i}\underset
  • #1
jostpuur
2,116
19
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:
Physics news on Phys.org
  • #2
jostpuur said:
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.

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]
 
  • #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.
 

1. What is "Local Uniform Continuity"?

Local uniform continuity refers to the property of a function to be uniformly continuous within a specific interval or neighborhood of a point. It means that for any small change in the input within that interval, the output will also change by a small amount.

2. What is "a^q"?

"a^q" is a mathematical notation for expressing the exponential function where the base is "a" and the exponent is "q". It is also known as the power function.

3. How is "Local uniform continuity of a^q" determined?

The local uniform continuity of a^q can be determined by checking if the function satisfies the definition of uniformly continuous function within a specific interval or neighborhood of a point. This can be done by using the epsilon-delta definition of uniform continuity or by using the intermediate value theorem.

4. What is the importance of "Local uniform continuity of a^q"?

The local uniform continuity of a^q is important in analysis and calculus as it allows us to make precise statements about the behavior of the exponential function within a specific interval or neighborhood of a point. It also helps us to prove the continuity of a^q and its derivatives in that interval.

5. How does "Local uniform continuity of a^q" differ from "Uniform continuity of a^q"?

The main difference between local uniform continuity and uniform continuity is that local uniform continuity is defined within a specific interval or neighborhood of a point, whereas uniform continuity is defined for the entire domain of the function. This means that a function can be locally uniformly continuous but not uniformly continuous, and vice versa.

Similar threads

  • Calculus
Replies
12
Views
369
Replies
1
Views
1K
Replies
2
Views
1K
  • Math POTW for University Students
Replies
15
Views
885
Replies
18
Views
2K
  • Topology and Analysis
Replies
8
Views
394
  • Calculus and Beyond Homework Help
Replies
26
Views
821
Replies
5
Views
270
Back
Top