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Is the complex function $f(z)=\exp(-|z|)$ uniformly continuous on $\mathbb{C}$?

The complex exponential function $\exp(z)$ is defined as $$ \exp(z) := \lim_{n\to \infty} \Bigl(1+ \frac{z}{n}\Bigr)^n $$

I tried solving this using the definition of the uniform continuity and got this far
\begin{align} |f(w)-f(z)| &= \Biggl|\lim_{n\to \infty} \biggl(1-\frac{|w|}{n}\biggr)^n - \lim_{n\to \infty} \biggl(1-\frac{|z|}{n}\biggr)^n\Biggr| \\[0.5ex] &=\Biggl|\lim_{n\mapsto \infty} \biggl(\frac{|z|-|w|}{n}\biggr)^n\Biggr| \end{align}

What should I do next? Is this function even uniformly continuous?

Zntzozt
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2 Answers2

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First off, I believe that the right thinking and intuitive base for these concepts helps significantly in finding the core approach - see my post on the subject here:

What is the intuition behind uniform continuity?

An intuitive understanding of uniform continuity - and continuity concepts more generally - is, as I've pointed out, related to the notion of "approximability": the ability to approximate what the output value of a function will be for an input you cannot obtain exactly, but can only obtain approximately - something which is necessary in order for approximate empirical measurements, and devices like electronic calculators which have limited precision, to be useful. Continuity basically asserts that, if you measure the input precisely enough, you can be assured of a desired accuracy in the output.

The desired accuracy in the output is $\epsilon$: think of it as a "tolerance". If I choose, say, $\epsilon = 0.001$, then what I'm asking is that I want to know the output value of the function to within $\pm 0.001$ (think about the confidence or uncertainty intervals in scientific measurements) of the true output value. This is what

$$|f(x_\mathrm{meas}) - f(x_\mathrm{true})| < \epsilon$$

means (where I've used somewhat more suggestive alternate notations for the values in question): the value of $f$ evaluated at the measured approximating value $x_\mathrm{meas}$ of the true, "desired" input $x_\mathrm{true}$, is not different from that true by any more than $\epsilon$. Indeed, even more suggestively, we may write

$$f(x_\mathrm{meas}) \in \left(f(x_\mathrm{true}) - \epsilon, f(x_\mathrm{true}) + \epsilon\right)$$

i.e. that $f(x_\mathrm{meas})$ is literally within the $\pm \epsilon$ uncertainty interval about $f(x_\mathrm{true})$. The $\delta$, then, is how accurately I need to measure the input: again note that we can write

$$|x_\mathrm{meas} - x_\mathrm{true}| < \delta$$

which has the same interpretation.

Now the key here is that in ordinary continuity, the value of $\delta$ needed to get $\epsilon$ may vary from one point $x_\mathrm{true}$ to another: if we know that, say, $\delta = 0.005$ suffices to get our $\pm 0.001$ level of tolerance from before when trying the input $x_\mathrm{true} = \pi$, for example, i.e. take $x_\mathrm{meas} = 3.141$, and the $f$ is merely continuous, we cannot be assured that that same $\delta$ will work at, say, $x = \pi^5$. That is, taking $x_\mathrm{meas} = 306.020$, which is within 0.005, may not give us $\pm 0.001$ tolerance on the output!

Clearly, this is a less than ideal state of affairs. What uniform continuity, then, tells us is that we are in a better state of affairs than this: if $f$ is uniformly continuous, then for each tolerance $\epsilon$, the same $\delta$ will work at every input! In other words, in the above example, it doesn't matter if we're trying to figure the value at $\pi$, $\pi^5$, or anything else - $\delta = 0.005$ will always suffice to get that desired $\pm 0.001$ accuracy.

So, now, how does this work in the case you are after? Well, it's simple: if you want to figure out how to prove the statement in question, you can begin by empirically analyzing how the function behaves near some "easy" point - say, for example, $z = 0$, plus or minus some $\epsilon$. And when you do this, you may see that it seems that if you take $\exp(-\epsilon)$ near $0$, that

$$\delta(\epsilon) := \epsilon$$

works: consider $\epsilon = 0.001$. We have $\exp(-0.001) \approx 0.99900050$, which you can check is just a hair under $0.001$ from $\exp(0) = 1$. Likewise, you will see it holds if you use your calculator with $\epsilon = 0.0001$: the needed $\delta$ seems to drop by a tenth. (In fact, this is an even better property, called Lipschitz continuity, but we won't talk that here.)

Because we believe the function is uniformly continuous, simply take that as your ansatz: that the above choice of $\delta$ works for any $\epsilon$. That is, that to get

$$|f(z_\mathrm{meas}) - f(z_\mathrm{true})| = |e^{-|z_\mathrm{meas}|} - e^{-|z_\mathrm{true}|}| < \epsilon$$

it suffices to take

$$|z_\mathrm{meas} - z_\mathrm{true}| < \epsilon$$

too. Now consider your limit: the above becomes

$$\left| \left[\lim_{n \rightarrow \infty} \left(1 - \frac{|z_\mathrm{meas}|}{n}\right)^n\right] - \left[\lim_{n \rightarrow \infty} \left(1 - \frac{|z_\mathrm{true}|}{n}\right)^n\right] \right| < \epsilon$$

or

$$\lim_{n \rightarrow \infty} \left|\left(1 - \frac{|z_\mathrm{meas}|}{n}\right)^n - \left(1 - \frac{|z_\mathrm{true}|}{n}\right)^n\right| < \epsilon$$

. And you want to show that that inequality holds good true whenever $|z_\mathrm{meas} - z_\mathrm{true}| < \epsilon$.

Now, I will stop here for now, and give two more pieces. We have from algebra:

$$a^n - b^n = (a - b)(a^{n-1} + a^{n-2} b + a^{n-3} b^2 + \cdots + a^2 b^{n-3} + ab^{n-2} + b^{n-1})$$

and we also have:

$$\left| |a| - |b| \right| \le |a - b|$$

Can you now see something about the limit expression

$$\lim_{n \rightarrow \infty} \left|\left(1 - \frac{|z_\mathrm{meas}|}{n}\right)^n - \left(1 - \frac{|z_\mathrm{true}|}{n}\right)^n\right|$$

and

$$|z_\mathrm{meas} - z_\mathrm{true}| < \epsilon$$

? Note that both have $\epsilon$; there is no $\delta$ - which is the key point regarding specifically uniform continuity in all this.

The_Sympathizer
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  • Thank you for the answer, I'll look into it as soon as I get back from work. – Zntzozt Dec 19 '19 at 08:16
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    So I got to say. This is the best answer I have ever got on this site so far. It's well explained, well structured and it made me finally understand :). So I think what you're trying to show me in the last section, is that if we set $\delta := \epsilon$ then continue with the definition of uniform continuity: $|z_{meas} - z_{true}| < \delta = \epsilon$ then write $|f(z_{meas}) - f(z_{true})| = |e^{-|z_{meas}|} - e^{-|z_{true}|}|$ and according to the inequality : $| |a| - |b| = | \le |a - b|$ that tells us that $|f(z_{meas}) - f(z_{true})| < \epsilon$. Is this right or am I missing something? – Zntzozt Dec 19 '19 at 16:45
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    Would you also please explain more about Lipschitz continuity in my example. – Zntzozt Dec 19 '19 at 19:24
  • @Zntzozt : Note that in the limit, it has the form $a^n - b^n$ where $a = 1 - \frac{|z_\mathrm{meas}|}{n}$ and $b$ is likewise for $z_\mathrm{true}$. Hence you can pull out a factor of the form $(a - b) = \frac{|z_\mathrm{true}| - |z_\mathrm{meas}|}{n}$, and then through the absolute value to get $\frac{\left| |z_\mathrm{true}| - |z_\mathrm{meas}| \right|}{n}$ which is now seen to be no larger than $\frac{\epsilon}{n}$. – The_Sympathizer Dec 20 '19 at 04:09
  • @Zntzozt : Also tweaked the wording a little, I think it was a bit less clear than it could have been, following a second look. – The_Sympathizer Dec 20 '19 at 04:11
  • I see now. So what about the rest of the term that you pulled the factor out of ? – Zntzozt Dec 20 '19 at 08:01
  • @Zntzozt : Once $n$ has grown sufficiently large - namely when $n > \mathrm{max}(|z_\mathrm{meas}|, |z_\mathrm{true}|)$, then both $a := 1 - \frac{|z_\mathrm{meas}|}{n}$ and $b := 1 - \frac{|z_\mathrm{true}|}{n}$ will be strictly less than $1$. Hence, so will all terms in the expression $a^{n-1} + a^{n-2} b + \cdots + ab^{n-2} + b^{n-1}$. Thus, since there are $n$ terms, such expression is strictly less than $n$, hence the limit value as a whole is strictly less than $\frac{\epsilon}{n} \cdot n = \epsilon$, which is what we sought to prove. – The_Sympathizer Dec 20 '19 at 08:07
  • And thus $f$ is uniformly continuous on $\mathbb{C}$. – The_Sympathizer Dec 20 '19 at 08:07
  • This makes a lot more sense now. – Zntzozt Dec 20 '19 at 08:16
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    Thank you for your help. You're a life saver – Zntzozt Dec 20 '19 at 08:17
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The definition of uniform continuity requires for all $\epsilon > 0$ that there exists $\delta > 0$ such that if $|z-w| < \delta$ then $|f(z)-f(w)|<\epsilon$ for all $z, w \in \mathbb{C}$. Given your starting point, I'm guessing you want to prove uniform continuity from this definition. You've unfortunately complicated the problem by substituting the limit definition of $\exp$, and your simplification is incorrect; $(a + b)^n - (c + d)^n \neq (a + b - (c + d))^n$. (Incidentally, your approach from there would go on to conclude that if $|z - w| < \delta < 1$ then $|f(z)-f(w)| = 0$, which is clearly false.)

Fix $\epsilon > 0$ and note $$ |f(z)-f(w)| = |e^{-|z|} - e^{-|w|}| = |e^{-|z|}(1 - e^{-(|w|-|z|)})|. $$

Somehow you want $|z-w|$ to be introduced, since this is what you are allowed to control. You can note $e^{-|z|} \leq 1$ and $$ |w|-|z| \leq ||w|-|z|| \leq |w-z| $$ by the reverse triangle inequality to conclude $$ |f(z)-f(w)| =|e^{-|z|}(1 - e^{-(|w|-|z|)})| \leq 1 - e^{-|w-z|} = 1 - e^{-|z-w|}. $$

You want this to be at most $\epsilon$, and so either appeal to continuity of $1 - e^{-x}$ or directly solve for a $\delta$ such that if $|z - w| < \delta$ then $1 - e^{-|z-w|} < \epsilon$. Note $\delta$ is independent of $z$ (and dependent on $\epsilon$), as it should be in the case of uniform continuity.

Riley
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  • Thank you sir :). I'll take a look at this directly after work. – Zntzozt Dec 19 '19 at 08:17
  • How would you suggest solving for a $\delta$ ? I tried setting $1-e^{-\delta} = \epsilon$ would it work if set $\delta = \ln(\epsilon + 1)$ ? – Zntzozt Dec 19 '19 at 17:16
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    Edit: I meant $\delta = -ln(1-\epsilon)$ – Zntzozt Dec 19 '19 at 19:05
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    That's fine yes. I'm more inclined to just write $1 - e^{-|z-w|} < \epsilon$ implies $1-\epsilon < e^{-|z-w|}$ so $\log(1-\epsilon) < -|z-w|$, and of course this gives $|z-w| < -\log(1-\epsilon)$ as you deduced. Thus if $\delta<-\log(1-\epsilon)$ then $|z-w| < \delta < -\log(1-\epsilon)$.This gives the same answer as solving $1 - e^{-\delta} = \epsilon$, but working with inequalities is (usually) nicer since you can take shortcuts; if your function is not "nice" to work with, solving $f(\delta) = \epsilon$ might not have an explicit solution. – Riley Dec 20 '19 at 10:01
  • Thank you so much. It's really weird how clear stuff seems after it has been solved by someone else. – Zntzozt Dec 20 '19 at 15:14