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fmam3
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I spent at least 2-3 hours thinking about this "deceivingly" (at least to me) simple problem but I just don't know how to proceed. Any hints would be greatly appreciated!
Directly from the [tex]\varepsilon - \delta[/tex] definition of uniform continuity, prove that [tex]2^x[/tex] is uniformly continuous on [tex] [0, n] [/tex] for all [tex] n \in \mathbb{N}[/tex].
(i.e. Do not use the fact that continuous functions on a closed interval are uniformly continuous).
Definition of a function [tex]f[/tex] uniformly continuous on a set [tex]S[/tex]: For all [tex]\varepsilon > 0[/tex] there exists some [tex]\delta > 0[/tex] such that for all [tex]x,y \in S[/tex], [tex] |x - y| < \delta[/tex] implies [tex]|f(x) - f(y)| < \varepsilon[/tex].
There must be some sort of algebraic trick that I'm missing here, but here's an attempt.
Fix [tex]n \in \mathbb{N}[/tex] and let any [tex]x,y \in [0,n][/tex]. Clearly, we have [tex]0 \leq x,y \leq n[/tex]. Now, consider [tex]|2^x - 2^y| \leq |2^x| + |2^y| = 2^x + 2^y \leq 2^n + 2^n = 2 \cdot 2^n[/tex].
It is at this precise point where I'm stuck --- how am I supposed to "introduce" the [tex]|x-y|[/tex] term in this expression? That is, I've already hit the point where [tex]2 \cdot 2^n[/tex] is already a constant; at least I've proved the obvious fact that this function (in this case) is bounded above. Or perhaps my above thinking is totally incorrect? Or is it that there's something so obvious that I'm oblivious to? I've also thought about using the binomial theorem but it doesn't seem to work very well either.
Thanks!
Homework Statement
Directly from the [tex]\varepsilon - \delta[/tex] definition of uniform continuity, prove that [tex]2^x[/tex] is uniformly continuous on [tex] [0, n] [/tex] for all [tex] n \in \mathbb{N}[/tex].
(i.e. Do not use the fact that continuous functions on a closed interval are uniformly continuous).
Homework Equations
Definition of a function [tex]f[/tex] uniformly continuous on a set [tex]S[/tex]: For all [tex]\varepsilon > 0[/tex] there exists some [tex]\delta > 0[/tex] such that for all [tex]x,y \in S[/tex], [tex] |x - y| < \delta[/tex] implies [tex]|f(x) - f(y)| < \varepsilon[/tex].
The Attempt at a Solution
There must be some sort of algebraic trick that I'm missing here, but here's an attempt.
Fix [tex]n \in \mathbb{N}[/tex] and let any [tex]x,y \in [0,n][/tex]. Clearly, we have [tex]0 \leq x,y \leq n[/tex]. Now, consider [tex]|2^x - 2^y| \leq |2^x| + |2^y| = 2^x + 2^y \leq 2^n + 2^n = 2 \cdot 2^n[/tex].
It is at this precise point where I'm stuck --- how am I supposed to "introduce" the [tex]|x-y|[/tex] term in this expression? That is, I've already hit the point where [tex]2 \cdot 2^n[/tex] is already a constant; at least I've proved the obvious fact that this function (in this case) is bounded above. Or perhaps my above thinking is totally incorrect? Or is it that there's something so obvious that I'm oblivious to? I've also thought about using the binomial theorem but it doesn't seem to work very well either.
Thanks!