real analysis - Proving that the exponential function is not uniformly continuous

I'm getting stuck on this proof to show that $\exp(x)$ is not uniformly continuous. My approach: find a way to make $\epsilon$ dependent on $\delta$ such that if you fix that $\delta$, the property would not hold for all $\epsilon\gt0$.

To do this I wrote out some things, but kept on getting stuck on equations;

We want to show that this does NOT hold for all $\epsilon\gt0$ : $|x-a|\lt\delta\implies |\exp(x)-\exp(a)|\lt\epsilon.$
We can write $$|\exp(x)-\exp(a)|=|\exp(x)(1-\exp(a-x))|$$ If we then take the $\log$ of this:
$$|\log(\exp(x)(1-\exp(a-x))|=|x+\log(1-\exp(a-x))|$$ With the $x$ returning, it starts to look like $|x-a|$ if we set $\log(1-\exp(a-x))=-a$. However, this equation only gets me more stuck.

In the end, I thought to be aiming for something like $\epsilon\geq\exp\delta$, so that if $0<\epsilon<\exp\delta$ the property doesn't hold and thus fails to hold for all $\epsilon>0$ which means the exponential function cannot be uniformly continuous.

How can I proceed to make this point? Is it right to find $\epsilon$ dependent on $\delta$ such that you can find $\epsilon>0$ for which the property doesn't hold, or is that a wrong approach?

Answer

For the function $e^x$, to disprove uniform continuity, any choice of $\epsilon > 0$ will work.

Thus, let $\epsilon=1$, and suppose for some $\delta > 0$, we have
$$|a-b| < \delta \implies |e^a-e^b| < 1$$
Leaving $b$ free, choose $\theta\in (0,\delta)$, and let $a=b+\theta$.
$$\begin{align*} \text{Then}\;\;&a=b+\theta\\[4pt] \implies\;&|a-b| < \delta\\[4pt] \implies\;&|e^a-e^b| < 1\\[4pt] \implies\;&e^a-e^b < 1\\[4pt] \implies\;&(e^b)(e^{a-b}-1) < 1\\[4pt] \implies\;&(e^b)(e^\theta-1) < 1\\[4pt] \implies\;&e^b < \frac{1}{e^\theta-1}\\[4pt] \end{align*}$$
Thus, we have
$$e^b < \frac{1}{e^\theta-1}$$

for all $b\in\mathbb{R}$, contradiction, since the function $e^x$ is not bounded above.

It follows that $e^x$ is not uniformly continuous.

As regards your question

"can you make $\epsilon$ dependent on $\delta$?"

the answer is "no, definitely not".

To disprove uniform continuity, it's you who must prove the existence of a fixed (constant) $\epsilon > 0$ for which no $\delta > 0$ works.

Think of it as a game with two players, you, and an opponent.

You claim $e^x$ is not uniformly continuous; your opponent disagrees.

Thus, your opponent is claiming that no matter what $\epsilon > 0$ you choose, he or she can find a $\delta > 0$ such that
$$|a-b|<\delta \implies |e^a - e^b| < \epsilon$$

But note the order of selections.

Your opponent doesn't need to choose $\delta$ until after you've chosen $\epsilon$. Thus, the value of $\delta$ can depend on the declared value of $\epsilon$, but the value of $\epsilon$ can't depend on $\delta$, since it hasn't necessarily yet been declared.