(Edited) Let $f:R\to R^+$ a continuous function where $f(x)f(-x)=1$ for every $x \in R$. Is $f$ necessarily the exponential?

Question

Additionally, we suppose $f'(x)\neq 0$ in any point, so it is not constant at any interval.

Well, your first implication isn't strictly true. the function $f(x)=0$ for all $x$ satisfies the functional equation. Beyond that...I'm not sure what you are asking. There are exotic functions that satisfy the functional equation, see this question Did you at least want to assume continuity? — lulu, Jun 02 '23 at 18:10
You're wright, but if $f(x)\neq0$, then, if $f(x)f(y)=f(x+y)$, so $f(x)f(0)=f(x+0)=f(x)$, so $f(0)=1$, then $f(x)f(-x)=f(0)=1$, so the first implication is true. I just want to know if the inverse is true, so if $f(x)f(-x)=f(0)=1$ then $f(x)f(y)=f(x+y)$. The question you pointed out is about $f(x)+f(y)=f(x+y)$, which is not the case. And yes, we assume continuity. — QuantumBrachistochrone, Jun 02 '23 at 18:39

score 2 · Accepted Answer · answered Jun 02 '23 at 18:10

2

Take any function $f$ on $[0,\infty)$ with $f(0)=1$ that doesn't take the value $0$, and define $f$ on $(-\infty, 0)$ by $f(x) = 1/f(-x)$.

answered Jun 02 '23 at 18:10

Robert Israel

You could also do the same thing if $f(0) = -1$ of course... – Daniel Schepler Jun 02 '23 at 18:14
But how this proves the question? – QuantumBrachistochrone Jun 02 '23 at 18:34
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@QuantumBrachistochrone The question, as I understand it, is whether the equation $f(x) f(-x) = 1$ implies $f(x) f(y) = f(x+y)$ (or maybe that $f(x) = \exp(c x)$ for some $c$). This shows that the answer is no. – Robert Israel Jun 02 '23 at 21:40

score 0 · Answer 2 · answered Jun 02 '23 at 18:11

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We have this result if $f$ is continuous: Prove if $f(x+y)=f(x)f(y)$ then $f(x)=a^x$. For $f$ not necessarily continuous, consider for example:

$$f(x) = \begin{cases}1 & x \in \mathbb Q \\ -1 & x \not \in \mathbb Q\end{cases}$$

since $x$ is rational iff $-x$ is.

answered Jun 02 '23 at 18:11

George C

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This does not answer the question because the assumption is $f(x)f(-x)=1$, not $f(x)f(y)=f(x+y)$, the second implies the first, obviously, but the question is in fact if the first implies the second. – QuantumBrachistochrone Jun 02 '23 at 18:33
Well for this function $f(\sqrt 2 + \sqrt 2) = -1$ while $f(\sqrt 2)f(\sqrt 2) = 1$, yet $f(x)f(-x) = 1$ for all $x$... The question title and the body ask two different things – George C Jun 02 '23 at 18:38
If $f$ is continuous and satisfies $f(x+y) = f(x) f(y)$, then $f$ is an exponential. But $f$ can be continuous and satisfy $f(x) f(-x) = 1$ and not be an exponential: see my answer. – Robert Israel Jun 02 '23 at 21:43
@RobertIsrael sure, was just not very clear what the OP was asking and continuity was thrown in after I posted the answer – George C Jun 02 '23 at 21:55

2 Answers2