# Uniform continuity of 2^x over [0,n]

1. Aug 13, 2009

### fmam3

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!

1. The problem statement, all variables and given/known data
Directly from the $$\varepsilon - \delta$$ definition of uniform continuity, prove that $$2^x$$ is uniformly continuous on $$[0, n]$$ for all $$n \in \mathbb{N}$$.

(i.e. Do not use the fact that continuous functions on a closed interval are uniformly continuous).

2. Relevant equations
Definition of a function $$f$$ uniformly continuous on a set $$S$$: For all $$\varepsilon > 0$$ there exists some $$\delta > 0$$ such that for all $$x,y \in S$$, $$|x - y| < \delta$$ implies $$|f(x) - f(y)| < \varepsilon$$.

3. The attempt at a solution
There must be some sort of algebraic trick that I'm missing here, but here's an attempt.

Fix $$n \in \mathbb{N}$$ and let any $$x,y \in [0,n]$$. Clearly, we have $$0 \leq x,y \leq n$$. Now, consider $$|2^x - 2^y| \leq |2^x| + |2^y| = 2^x + 2^y \leq 2^n + 2^n = 2 \cdot 2^n$$.

It is at this precise point where I'm stuck --- how am I supposed to "introduce" the $$|x-y|$$ term in this expression? That is, I've already hit the point where $$2 \cdot 2^n$$ 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!

2. Aug 13, 2009

### fmam3

I just had a "random idea" and I just want to post it here and see if I could answer my own question. Please let me know if it works (or not)! Thanks all again!

Since $$0 \leq x,y \leq n$$, let $$M = \sup\{|x - y|: x,y \in [0,n]\}$$. Then clearly, we have that $$|x - y| \leq M$$. Let any $$\varepsilon > 0$$.

Then we have $$2^{|x-y|} \leq 2^M$$, rearrange then get $$2^x \leq 2^M \cdot 2^y$$. Now, see that $$|2^x - 2^y| \leq |2^x| + |2^y| = 2^x + 2^y \leq 2^M \cdot 2^y - 2^y = 2^y(2^M - 1) \leq 2^n (2^M - 1)$$. Since $$M \geq 0$$, we must have $$2^M - 1 \geq 0$$. We want the expression above to be less than $$\varepsilon$$. So, $$2^n(2^M - 1) < \varepsilon$$, rearrange, so we have $$2^M < \varepsilon / 2^n + 1$$. Hence, we can choose $$\delta(\varepsilon) = \delta = \min \{M, \varepsilon / 2^n + 1\}$$.

Hence, for all $$\varepsilon > 0$$, $$|x - y| < \delta = \min \{M, \varepsilon / 2^n + 1\}$$ implies $$|2^x - 2^y| < \varepsilon$$ and hence $$f(x) = 2^x$$ is uniformly continuous over $$[0,n]$$ for all $$n \in \mathbb{N}$$.

Last edited: Aug 13, 2009
3. Aug 13, 2009

### JG89

I don`t think you can just say that the expression $$2^n(2^M - 1) < \epsilon$$ for every positive epsilon. You need to actually show that the expression gets arbitrarily small. This may be too informal for you, but you can note that $$2^M - 1$$ goes to 0 for M getting smaller and smaller, and since $$2^n$$ is a constant, then $$2^n(2^M - 1)$$ must go to 0 for M = sup|x-y| going to 0.

4. Aug 13, 2009

### fmam3

@JG89

Thanks for catching that! I was actually staring at that exact place where you pointed out immediately I made the posting. It seems that I did make an error in the choice of $$\delta$$.

I think it's possible to "fix" this argument by simply setting $$\delta (\varepsilon) = \delta = \min\{M, \log_2 (\varepsilon / 2^n + 1)\}$$. I'm actually not sure if this is the best argument --- as I'm using the logarithm base 2 function, which is technically a more "advanced" function than the technical background requirement implied by this question.

Question -- can you clarify what you mean by "for M = sup|x - y| going to 0"? It's not necessarily true that M tends to zero; since by the definition of uniform continuity, I can pick any x,y in the domain specified. That is, I could technically set x = 1, y = 10000 and in that sense, the supremum of |1 - 10000| will never go to zero. But I could be completely wrong so please do point out where I've made a mistake in any of the above as I'm still learning this material! Thanks for your reply!

5. Aug 13, 2009

### JG89

$$\delta$$ HAS to go to 0. Remember, the definition of uniform continuity is that a function f is uniformly continuous in an interval if for every positive epsilon, there exists a delta such that we have $$|f(x) - f(y)| < \epsilon$$ for ANY x and y in the interval satisfying $$|x-y| < \delta$$. Obviously if $$|f(x) - f(y)|$$ is to go to 0, so must |x-y|. M = sup|x-y| < delta, so sup|x-y| goes to 0 as well.

I think your argument is getting too complex. If I were you, I’d fix a delta so that if x and y = x +h be any two points in the interval [0,n] then $$|h| < \delta$$. It’s obvious that $$2^h - 1 < 2^{\delta} - 1 \rightarrow 0$$ as $$\delta \rightarrow 0$$. Finish it off.

6. Aug 13, 2009

### JG89

The last sentence of the previous post should be:

It’s obvious that $$0 < 2^h - 1 < 2^{\delta} - 1 \rightarrow 0$$ as $$\delta \rightarrow 0$$

7. Aug 13, 2009

### fmam3

Hi JG89,

I think I see what you mean --- but let me know if I'm not exactly getting your point and I would appreciate any corrections! The argument you had about various things going is essentially using a sequential definition of continuity, not necessarily the epsilon-delta version; i.e. using Cauchy's condition for uniform convergence. And unfortunately (for me), while I completely understand what you mean by writing things "going to zero" with the arrow signs, if I were to write that on an assignment / exam I would get destroyed by my instructor; for epsilon-delta proofs, we are required to construct that exact expression for delta, epsilon, etc. But I agree that, intuition wise, yours is much clearer than writing epsilon-delta mumbo jumbo. Thanks again!

8. Aug 13, 2009

### JG89

If you are required to construct an exact expression, then using $$\delta = log_2(\frac{\epsilon}{2^n} + 1)$$ is okay. That’s the same delta I found. You don’t need to use the minimum of sup|x-y| and $$log_2(\frac{\epsilon}{2^n} + 1)$$. The sup|x-y| part is unnecessary.

9. Aug 13, 2009

### fmam3

Sorry guys for making so many (stupid) mistakes on this proof. But I realized that I had a "minus" instead of a "plus" in my proof and thus voiding the entire argument above (mine and JG89's). I hope this proof, instead, works! And I would appreciate it if anybody can point out a way to avoid using the log base 2 function! Thanks and I apologize for the inconvenience!

Let any $$x,y \in [0,n]$$ and WLOG, let $$x \geq y$$. Then consider, $$|2^x - 2^y| = |2^y(2^{x-y} - 1)| = |2^y| \cdot |2^{x-y} - 1| \leq 2^n \cdot (2^{x-y} - 1) = 2^n \cdot (2^{|x-y|} - 1) < \varepsilon$$. Then, rearrange, get $$\dfrac{\varepsilon}{2^n} + 1} > 2^{|x-y|}$$ and taking the log base 2, get $$\log_2 (\varepsilon / 2^n + 1) > |x-y|$$. Thus, we can take $$\delta (\varepsilon) = \delta = \log_2 (\varepsilon / 2^n + 1)$$.

Therefore, for all $$\varepsilon > 0$$, for any $$x,y \in [0,n]$$, $$| x - y | < \delta = \log_2 (\varepsilon / 2^n + 1)$$ implies $$|2^x - 2^y| < \varepsilon$$.