Proving uniform continuity of the exponential function on a semi-finite interval

Question

Suppose $A\in\mathbb{R}$.

Prove that the exponential function is uniformly continuous on the interval $(-\infty, A]$.

Proof:

Suppose $a,b\in(-\infty,A]$, and choose $a=A\geq b$ without loss of generality. We know that the exponential function is continuous everywhere, so that: $$|A-b|<\delta\implies |e^A-e^b|<\epsilon$$

The choice of $\delta$ is then only dependent on $A$, and not on $b$ as the only restriction on $b$ is that it is less than or equal to $A$, which simply means $b\in(-\infty,A]$. So for all $\epsilon>0$ there is a $\delta>0$ such that for all $b\in(-\infty,A]$, the property holds. So, the exponential function is uniformly continuous on $(-\infty,A]$.

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At first I split the cases $(-\infty,0]$ and $[0,A]$ but halfway through typing this post I realised it might not have been necessary, though I must say I'm greatly unsure about my proof. What I initially wrote was along the lines of the following;

Suppose $a,b\in(-\infty,A]$, and choose $a\geq b$ without loss of generality, so that $|a-b|=a-b$. Suppose $|a-b|=a-b<\delta$ for any $\delta>0$. We can write $|e^a-e^b| =|e^b(e^{a-b}-1)|$ so that whenever $b\to-\infty$, $|e^b(e^{a-b}-1)|\to0<\epsilon$ for all $\epsilon>0$. We have that for all $ \epsilon>0$ that there is a $\delta>0$ such that for all $a\in(-\infty,A]$, whenever $b\to-\infty$ that $|a-b|<\delta\implies |e^a-e^b|<\epsilon$ so that the exponential function is uniformly continuous as $b\to-\infty$...

Answer 1

It doesn't looks right to me. Why do you say that we can choose $a=A$ without loss of generality? I can see how to justify that assertion, but how do you justify it?

And it's strange that you write that “the exponential function is always continuous”. Did you mean that it is continuous everywhere?

Answer 2

Hint: By the mean value theorem, $f(y) - f(x) = f'(c)(y-x).$ How big can $|f'(c)|$ be here, and what does this imply?

Answer 3

Suppose $a,b\in(-\infty,A]$, and choose $a=A\geq b$ without loss of generality.

Just to expand on José's point: You cannot “choose” $a$ to be equal to $A$; you are given $A$, $a$, and $b$. You're right that you can assume $a \geq b$, because one of the two must be greater than or equal to the other; rename $a$ to be the greater and $b$ the lesser. But how do you know that it is safe to assume $a=A$?

We know that the exponential function is continuous everywhere, so that: $$|A-b|<\delta\implies |e^A-e^b|<\epsilon$$

You just dropped in some symbols; they need to be quantified. After “so that” you should write: “given $\epsilon > 0$, there exists a $\delta > 0$ such that:” But are you using the fact that the exponential function is continuous at $A$? or at $b$?

The choice of $\delta$ is then only dependent on $A$, and not on $b$ as the only restriction on $b$ is that it is less than or equal to $A$, which simply means $b\in(-\infty,A]$.

You have also imposed the restriction that $|A-b| < \delta$. You can't simultaneously send $b$ to $-\infty$.

Here's a flag: At what point did you use anything specific to the exponential function? It seems like all you used was the fact that it's continuous on $(\infty,A]$. If that is all you need, then any function satisfying that property is also uniformly continuous. For example, $f(x) = x^2$ on $(-\infty,0]$. But that function is not uniformly continuous.