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It seems intuitively very clear that $e^{x}$ is not uniformly continuous on $\mathbb{R}$. I'm looking to 'prove' it using $\epsilon$-$\delta$ analysis though. I reason as follows:

Suppose $\epsilon > 0$; in fact, fix it to be $\epsilon=1$.

For contradiction, suppose that $\exists \delta >0$ s.t. $$ (\star) \ |x-y|<\delta \Rightarrow |e^{x}-e^{y}|<\epsilon=1 \text{ for } x,y \in \mathbb{R}.$$ Note that $e^{x+\delta}-e^{x}=e^{x}(e^{\delta}-1)$. So, for $x$ large enough (so that RHS $>1$), the relation $(\star)$ does not hold.

This is our contradiction, and so the exponential function is not uniformly continuous on $\mathbb{R}$.

Is this reasoning correct and sufficient?

Thanks.

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It looks mostly good to me, but I would suggest instead of fixing $\varepsilon>0$ to be $\varepsilon=1$, just let it be arbitrary, and then calculate exactly how large $x$ has to be... something like $\ln\left(\frac{2\varepsilon}{e^\delta-1}\right)$. –  Clayton Jan 23 '13 at 1:36
    
Thanks @Clayton for the help. –  Mathmo Jan 23 '13 at 1:39
    
You're welcome. –  Clayton Jan 23 '13 at 1:42

2 Answers 2

up vote 3 down vote accepted

If you are truly looking for a rigorous answer then you need to justify the "So, for $x$ large enough...".

For instance, here is a very rigorous solution along the lines you suggest:

Assume that $e^x$ is uniformly continuous on $\mathbb {R}$. Let $\epsilon = 1$. Thus there is $\delta >0$ such that for all $x,y\in \mathbb R$ if $|x-y|<\delta $ then $|e^x-e^y| < 1$. Let $a=\delta/2$. Since $\lim_{x\to\infty }e^x=\infty$ and since $e^a-1>0$ it follows that $\lim_{x\to \infty }e^x(e^a-1)=\infty$. Consequently, there is some $x\in \mathbb {R}$ such that $e^x(e^a-1)>1$. However, taking $y=x+a$ we have $|x-y|<\delta$ while $|e^x-e^y|=e^x(e^a-1)>1$, a contradiction.

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Thanks, I see it. This is the rigour I was looking for. –  Mathmo Jan 23 '13 at 1:41
    
you are welcome! –  Ittay Weiss Jan 23 '13 at 1:42

Your proof looks good to me. There are only a few technicalities:

  1. You should consider an arbitrary $\delta>0$ because the contrapositive of $$\forall \epsilon>0,\ \exists\delta>0,\ \forall|x-y|<\delta,\ |f(x)-f(y)|<\epsilon$$ is $$\exists \epsilon>0,\ \forall\delta>0,\ \exists|x-y|<\delta,\ |f(x)-f(y)|\ge\epsilon.$$
  2. You should consider something like $e^{x+\delta/2}-e^x$ rather than $e^{x+\delta}-e^x$ because $|(x+\delta)-x|$ is not smaller than $\delta$.
  3. As Clayton pointed out in the comment section, if you claim that $(\star)$ does not hold when $x$ is large enough, you'd better explicitly write down an instance of $x$.
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