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Suppose that the function $f:\mathbf R\to\mathbf R$ satisfies $f(n)=e^n$, where $n\in \mathbf Z$, and $f(x)f(y)=f(x+y)$ for all $x,y\in\mathbf R$. Must it be the case that $f=\exp$?

What I have tried so far:

If we can prove that $\lim_{h\to0}f(h)=1$, then it follows that $$\lim_{h\to0}f(a+h)=\lim_{h\to0}f(a)f(h)=f(a)\lim_{h\to0}f(h)=f(a) \, ,$$ and so $f$ is continuous. With the extra hypothesis of $f$ being continuous, it can then be shown that $f=\exp$. So if the statement is true, then the problem reduces to proving that $\lim_{h\to0}f(h)=1$. If the statement is false, then there is a function $f$ satisfying the hypotheses of the question, but not satisfying $\lim_{h\to0}f(h)=1$.

Arctic Char
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Joe
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    We can prove with induction on $n$ that for all $x_1,\ldots,x_n$, we have $\prod_{i=1}^nf(x_i)=f\left(\sum_{i=1}^nx_i\right)$. Now put $x_1=\dots=x_n=\frac1n$, then we find that $f(1/n)^n=f(1)=e$. For odd $n$ this gives $f(1/n)=e^{1/n}$. So if the limit $\lim_{h\to 0}f(h)$ exists, it must be equal to $f(1)=e$. One problem I see is that, in theory, this still allows for $f(1/2n)=-e^{1/2n}$ – Mastrem Sep 18 '21 at 09:15
  • So if we restrict the codomain to $\mathbb{R}_{\ge 0}$, then $f(x)=e^x$ for all $x\in\mathbb{Q}$. But you probably already knew this. – Mastrem Sep 18 '21 at 09:18
  • What you can prove is that $f(q)=e^q$ for all $q\in \mathbb Q$. If $f$ is continuous at $0$, then indeed you can prove that $f(x)=e^x$, but if such assumption is not fulfilled, I don't see any reason to get $f(x)=e^x$ for all $x\in \mathbb R$. – Surb Sep 18 '21 at 09:18
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    Could the downvoter please explain why they have downvoted? – Joe Sep 18 '21 at 09:46

3 Answers3

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There exist pathological (specifically non-measurable) functions $g$ such that $g(x+y)=g(x)g(y)$ for all $x,y$. For such a function $g$ we have $g(n)=a^{n}, n \in \mathbb Z$ for some $a$. Let $f(x)=(\frac e a)^{x}g(x)$. The $f$ satisfies the hypothesis of your question. So some continuity assumption is essential.

Joe
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Kavi Rama Murthy
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  • That is indeed the simplest answer. OP already knows that there are non-continuous solutions of the Cauchy exponential functional equation, and any solution satisfies $f(n) = e^n$ (for integer arguments) after some scaling. – Martin R Sep 18 '21 at 09:36
  • Hi Kavi. Thanks for this answer. What is a "non-measurable function"? – Joe Sep 18 '21 at 09:51
  • If you know that the given functional equation has a dis-continuous solution then my answer will make complete sense: Just replace 'non-measurable' by 'dis-continuous'. @Joe – Kavi Rama Murthy Sep 18 '21 at 09:54
  • @KaviRamaMurthy: Ok, thanks. Is there a discontinuous solution to the equation $g(x)g(y)=g(x+y)$ which is actually possible to write down? – Joe Sep 18 '21 at 10:06
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    @Joe: See https://math.stackexchange.com/q/423492/42969 for an overview about the Cauchy functional equation, with links to the construction of discontinuous solutions. – See also https://mathoverflow.net/a/57532/116247: “It is not possible to provide an explicit expression for a non-linear solution.” – Martin R Sep 18 '21 at 10:08
  • @MartinR: Thanks for the links. Does this mean that is consistent within $\mathsf{ZF}$ that there are no solutions to the equation other than $f(x)=e^x$? – Joe Sep 18 '21 at 19:26
  • @Joe: I am the wrong person to ask, set theory is definitely not my area of expertise, I have only quoted what I found in https://math.stackexchange.com/q/423492/42969. – Martin R Sep 18 '21 at 19:29
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Let $\{b_i\}_{i\in I}$ be a basis for $\mathbb{R}$ over $\mathbb{Q}$ which contains $1$, and let $\{c_i\}_{i\in I}$ be any collection of real numbers. Then $$f\left(\sum_{i\in I}\lambda_i b_i\right):=\prod_{i\in I}e^{c_i\lambda_i}$$ satisfies the functional equation. Set the $c_i$ corresponding to $b_i=1$ to $1$, and take some other $c_j$ and set it to anything but $1$, say $2$. Then $f$ agrees with $\exp$ on $\mathbb{Q}$, but not on $\mathbb{R}$.

Mastrem
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Must it be the case that $f=\exp$?

No.

From the functional equation you can conclude that $f=\exp$ on $\Bbb Q$ and that $f(0)=1=e^0$. But it's completely fine to have $$f(y)=2^y \quad\text{where}\ y\in \sqrt{2}\Bbb Q$$

The functional equation implies the definition on the additive group $G=(\Bbb Q+\sqrt{2}\Bbb Q,+)$ $$ f(x+\sqrt{2}y) = e^x2^y\quad\text{ where }\quad x,y\in\Bbb Q$$

The condition you are missing is that $f$ must be continuous to conclude for all of $\Bbb R$. Without continuity, your limits are not defined. Or at least it must be continuous on some open interval.

emacs drives me nuts
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  • It's not completely obvious to me that this function $\mathbb{Q}(\sqrt2)\to\mathbb{R}$ can be extended to a function $\mathbb{R}\to\mathbb{R}$ which still satisfies the functional equation. Can you prove that? – Mastrem Sep 18 '21 at 09:24
  • @Mastrem: $\sqrt{2}\Bbb Q$ is not $\Bbb Q(\sqrt{2})$. The latter is a number field that contains $\Bbb Z$. The former is not a field, it's just a group under addition. And that closedness is much less restrictive than the operations permitted in $\Bbb Q(\sqrt{2})$. – emacs drives me nuts Sep 18 '21 at 09:26
  • Why is $f(n)=e^{n}$ in your example? – Kavi Rama Murthy Sep 18 '21 at 09:27
  • @emacsdrivesmenuts I'm aware of that, but by defining $f$ on $\mathbb{Q}$ and $\sqrt2\mathbb{Q}$, you define it on $\mathbb{Q}(\sqrt{2})$ – Mastrem Sep 18 '21 at 09:27
  • @Mastrem: I edited my answer and defined it on all of $(\Bbb Q+\sqrt{2}\Bbb Q,+)$. Thanks for spotting my thinko. – emacs drives me nuts Sep 18 '21 at 09:58