My real analysis professor mentioned in passing that there exist functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy all of the following conditions for all $a,b \in \mathbb{R}$:

  1. $$f(1)=e$$

  2. $$f(a+b)=f(a)f(b)$$

  3. $$\exists\, x \in \mathbb{R} \,\,\,\,\text{ such that } \,\,\,f(x)\neq e^{x}$$

I was trying to find such a function and I encountered some difficulty. I tried finding out which values of $x$ necessitated $f(x) = e^x$ ...

I started with $0$ and got to all of the integers, then to the rationals.

From property $2$ we can let $a=0$ to find $f(0)$ as follows:

$$f(b) = f(0+b) = f(0)f(b)$$


For any positive integer $n \in \mathbb{N}$ we can find $f(n)$ as

$$f(n) = f(1+(n-1)) = f(1)f(n-1) =\cdots=f(1)^n = e^n$$

We can find $f(-1)$ as follows:

$$1 = f(0) = f(-1+1) = f(-1)f(1) = f(-1) \cdot e$$

$$f(-1) = \frac{1}{e} = e^{-1}$$

This lets us find $f(-n)$ for any $n \in \mathbb{N}$:

$$f(-n) = f(-1 + -(n-1)) = f(-1)f(-(n-1)) = \cdots = f(-1)^n = e^{-n}$$

Thus, for all $m \in \mathbb{Z}$, $f(m)=e^m$.

We can find $f(1/2)$ as follows:

$$e = f(1) = f\left(\frac{1}{2}+\frac{1}{2}\right) = f\left(\frac{1}{2}\right)^2$$

$$f\left(\frac{1}{2}\right) = \pm \sqrt{e} = \pm \,e^{1/2}$$


$$f\left(\frac{1}{3}\right) = \sqrt[3]{e} = e^{1/3}$$

We can also get

$$f\left(\frac{2}{3}\right) = e^{2/3}$$

It follows quite quickly that for all $p/q \in \mathbb{Q}$ with $p \in \mathbb{Z}$, $q \in \mathbb{N}$ (so that $q > 0$), and $\gcd(p,q)=1$, we have $$f\left(\frac{p}{q}\right) = \pm \,e^{p/q}$$

when $q$ is even, and

$$f\left(\frac{p}{q}\right) = e^{p/q}$$

when $q$ is odd. Since these work for all rationals, we can say that if $f$ is continuous, then $f(x) = e^x$ for all $x \in \mathbb{R}$.

But we don't know that $f$ is continuous, so there can definitely still be values of $x$ for which $f(x) \neq e^x$. Namely, $x \in \mathbb{R}$ with $x \not\in\mathbb{Q}$. Let's take the first example of an irrational which comes to mind: $\sqrt{2}$

Consider $f\left(\sqrt{2}\right) \in \mathbb{R}$. Just as we used $f(1) = e$ above, for any rational $p/q$ we can derive $$f\left(\frac{p}{q}\sqrt{2}\right) = e^{p/q}f\left(\sqrt{2}\right)$$

This could have a $\pm$ in front if $q$ is even.

In any case, if we ignore the possible negative values, we can look at the graph of this function, plotting the points we know so far. At any rational $x$, it is $e^x$. At $\sqrt{2}$ it is some real number, call it $k$. At any rational multiple of $\sqrt{2}$, $x\sqrt{2}$, it is $ke^x$.

So if we only look at rational multiples of $\sqrt{2}$, the function looks like a stretched copy of $e^x$, where the extent of stretching depends on the value of $f\left(\sqrt{2}\right)$.

At this point, I thought I'd found a function that satisfied all three properties, and that I could explicitly write its values. Suppose $f\left(\sqrt{2}\right) = 5$ and $f\left(\frac{p}{q}\sqrt{2}\right) = 5e^{p/q}$ for any rational $p/q$. For any other $x \in \mathbb{R}$, we let $f(x) = e^x$.

Unfortunately, this falls apart quickly. Consider $$a=\frac{\sqrt{2}+\sqrt{3}}{2}$$

$$b = \frac{\sqrt{2}-\sqrt{3}}{2}$$

Neither are multiples of $\sqrt{2}$ but their sum is. $f(a+b) = f\left(\sqrt{2}\right)$ has to simultaneously be $e^\sqrt{2}$ and $5$, which is a problem.

My subsequent efforts all resulted in failure. I may have found some functions that satisfied the three conditions, exploiting the $\pm$ that appears with even-denominator fractions, but if I added a fourth condition for $f(x)$ to be positive, I couldn't write out any functions that worked.

I asked my professor about this, and he said that it wasn't possible to write out a function, and that was why I wasn't able to. He said that mathematicians only know they exist because of something called a Hamel basis.

In a fortunate turn of events, the textbook for my set theory class had a page or two on the existence of a Hamel basis, and it turns out that the proof of their existence requires the Axiom of Choice. We haven't gotten close to talking about the Axiom of Choice in my set theory class, so I'm not too clear on it, and I didn't really understand it.

This brings me, at long last, to my question(s). If such functions are impossible to write out explicitly, how do we know they exist? It's not just that we've proven that they exist and simply haven't managed to find one yet, it's that we've proven (somehow!?) that they can't be explicitly written out.

Is the proof of the existence of a Hamel basis uncontroversial? Is the assumption of the Axiom of Choice something that every mathematician thinks is a good one? How do we know that it's impossible to write out a function that has positive values and satisfies properites $1$, $2$ and $3$?

  • $\begingroup$ Someone get @AsafKaragila to shine light on this Hamel basis stuff :) $\endgroup$ – Zubin Mukerjee Nov 21 '14 at 16:06
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    $\begingroup$ Get who to do where? $\endgroup$ – Asaf Karagila Nov 21 '14 at 16:08
  • $\begingroup$ @AsafKaragila That question is just for you mate ;) $\endgroup$ – AlexR Nov 21 '14 at 16:09
  • $\begingroup$ @Alex: Surely there are a handful of other people that can write an excellent answer. I just happen to be omnipresent. $\endgroup$ – Asaf Karagila Nov 21 '14 at 16:19
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    $\begingroup$ You can find links to several related posts in: Overview of basic facts about Cauchy functional equation. $\endgroup$ – Martin Sleziak May 13 '17 at 6:48

First let's make things slightly simpler. Note that if $f(x+y)=f(x)f(y)$, then $\log f(x+y)=\log f(x)+\log f(y)$. So it sufficient to talk about $g(x+y)=g(x)+g(y)$. Moreover, $f$ is continuous if and only if $g$ is continuous.

Why is this simpler? Because now it's not difficult to show that $g$ is a linear function from $\Bbb R$ to itself, as a vector space over $\Bbb Q$.

We can show, in fact, that if $g$ is continuous, or measurable, or Baire measurable, then $g(x)=ax$, which means that $f(1)=e^a$, and for $a=1$ we simply get $e^x$.

But what about a discontinuous solution? That's harder to come up by. The reason is that it is consistent with the failure of the axiom of choice that all solutions are continuous. In which case, there are no other functions satisfying the first two conditions except $e^x$.

But with the axiom of choice we can show that $\Bbb R$ has a basis, as a vector space, over $\Bbb Q$. This basis is called a Hamel basis. Moreover, this basis, as any basis would, has the property that any function from it to $\Bbb R$ has a unique extension to a linear function on $\Bbb R$. And there's even more that we can do with the axiom of choice. We can extend any set of $\Bbb Q$-linearly independent real numbers into a basis.

Combining all these facts together, we extend $\{1,e\}$ to a Hamel basis, and consider any function such that $g(1)=1, g(e)=0$. Extend this $g$ to a linear function, and finally take $f(x)=e^{g(x)}$. And this indeed satisfying the first two conditions, but $f(e)=1\neq e^e$.

Let me also add somewhat of a historical perspective on this.

How can we prove something exists without writing it down explicitly? The axiom of choice allows us to conclude that from certain assumptions there exists a function with certain properties. That is what the axiom is for, letting us make this inference. And so we can make a lot of similar inferences as a consequence.

But that's fine. You know that there is no largest natural number. But can you write one which has more than $10^{100000000}$ digits explicitly? Probably not. How do you know it exists, then? Because you have a list of axioms, and you used them to conclude that there is no largest natural number, and that if $x,y$ are natural numbers then $x^y$ is a natural number, and therefore there is a natural number with so many digits, and more.

Historically, however, the axiom of choice caused some controversy with its consequences, but nowadays it is generally considered part of mathematics, and as someone whose research is models where the axiom of choice fails, I get more strange looks than supportive looks, which should indicate how far we've come from the early 20th century.

  • $\begingroup$ Thanks for the quick response! I still have a few questions. What do "measurable" and "Baire measurable" mean? The wikipedia article defined measurable functions in terms of measurable spaces, and defined those in terms of "$\sigma$-algebras" and I don't know what those are either ... maybe I should just not worry about that. Also, I thought I proved that $f(n)$ for any integer $n$ had to equal $e^n$, using just the first two properties. How can $f(2)=1$? Is there something wrong with my proof? $\endgroup$ – Zubin Mukerjee Nov 21 '14 at 16:31
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    $\begingroup$ First let me reassure you that if you don't understand what measurable and Baire measurable mean, then you can just ignore that. You are also correct that $f(2)=e^2$, I meant to write $g(e)$ and therefore $f(e)$ instead. I'll fix that. Finally, the fact that $\Bbb R$ has a Hamel basis is probably the most well known application of Zorn's lemma. Every vector space has a basis. And of course it cannot be proved without appealing to the axiom of choice. In this answer I show that from a Hamel basis we get a discontinuous solution, and I mentioned that such solution might not exist without choice. $\endgroup$ – Asaf Karagila Nov 21 '14 at 16:39
  • $\begingroup$ Thank you. One more question, sorry. What about the whole "it's impossible to write an explicit example"? I know it's quite common to prove that an example of something exists without explicitly finding it, but in this case can we prove it's not possible to write out a Hamel basis? As in, literally list or precisely describe the set of vectors (whose entries are in $\mathbb{R}$ and whose scalars are rational numbers) that form a basis for $\mathbb{R}$? EDIT: Thanks for the addition. $\endgroup$ – Zubin Mukerjee Nov 21 '14 at 16:53
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    $\begingroup$ I've added some discussion as to how it is possible to prove something exists without writing it down. $\endgroup$ – Asaf Karagila Nov 21 '14 at 16:57
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    $\begingroup$ Thanks you again. I think I understand better now. Also, I just did a search on mathoverflow for this topic and found this question. It looks very related. $\endgroup$ – Zubin Mukerjee Nov 21 '14 at 16:59

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