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I'm having some trouble solving the following equation for $f: A \rightarrow B$ where $A \subseteq \mathbb{R}$ and $B \subseteq \mathbb{C}$ such as:

$$f(x)f(y) = f(x+y) \quad \forall x,y \in A$$

The solution is supposed to be the form $f(x) = e^{kx}$ where k can be complex but I don't see how to show it.

Note: I'm trying to solve this equation to find irreductible representation of U(1) but it doesn't really matter I think

Samuel
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  • Which set is $A$? – ajotatxe Jan 29 '15 at 21:58
  • For my case, it's $U(1) = \lbrace e^{i\varphi} \mid \varphi \in \left[ 0, 2\pi \right] \rbrace$ which can be represented by $ \varphi \in \left[ 0, 2\pi \right]$ so I guess that A is $\left[ 0, 2\pi \right]$ – Samuel Jan 29 '15 at 22:02
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    @Samuel I would disagree with you on that, $U(1)$ is the circle, which is certainly different from $[0,2\pi]$. In particular, $0$ and $2\pi$ have the same image under $\varphi\mapsto e^{i\varphi}$. – Hayden Jan 29 '15 at 22:04
  • So $\left[ 0, 2\pi \right[$ ? – Samuel Jan 29 '15 at 22:05
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    Put $f(x) = e^{g(x)}$ then the equation becomes $g(x) + g(y) = g(x+y)$. The solution to this last equation can be found here (which shows that the only continuous solution is $g(x)=kx$) – Winther Jan 29 '15 at 22:09
  • Is A a subset of real numbers? –  Jan 29 '15 at 22:46
  • yes, it is a subset of real numbers – Samuel Jan 29 '15 at 22:48
  • We can prove that the function (log) o f is linear. It follows that there's a constant k for which: ((log) o f)(x) = kx. Hence f(x) = e^kx –  Jan 29 '15 at 22:54
  • This is true, if you assume that $f$ is continuous, see here. in fact, it is enough to require continuity in one point. But there are also non-continuous solutions. You can find at least a few pointers for other related questions here. (Although the question I linked to deal with real-valued functions, not complex-valued. But I still think that they might be useful.) – Martin Sleziak Jan 30 '15 at 14:13

3 Answers3

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If we assume differentiability on the outset, there is one approach that I like that is quite general to turn it into a differential equation: consider the map $\Bbb R^2\to\Bbb R^2$ defined by

$$(x,y)\mapsto (f(x)f(y), f(x+y)).$$

The functional equation implies that the image is one dimensional, so the Jacobian is everywhere degenerate. Working it out we see:

$$f'(x)f(y)f'(x+y) = f'(x+y)f(x)f'(y)$$

from which you can easily argue that

$$\frac d{dx}\log f(x) = \frac{f'(x)}{f(x)} = \frac{f'(0)}{f(0)} = constant$$

so that the result follows (in the differentiable case).

doetoe
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Let $y=1$ and $f(1)=e^k$ for some $k\in\mathbb{C}$, then $$ f(x+1)=e^kf(x)\\ f(x)=e^kf(x-1)\\ f(x)=e^{2k}f(x-2)\\ f(x)=e^{3k}f(x-3)\\ \vdots\\ f(x)=e^{(x-1)k}f(1)\\ f(x)=e^{kx}\\ $$

  • This derivation holds only for integer $x$. – Winther Jan 30 '15 at 08:09
  • @Winther. No, because you do not have to assume that $f(1)=e^k$. You can assume that for some $a\in\mathbb{R},f(a)=e^k$. –  Jan 30 '15 at 09:18
  • I'm referring to the $f(x-3)\cdots f(1)$ step which only works for $x$ integer. – Winther Jan 30 '15 at 09:33
  • @Winther. It is just an example. Because $f(x+y)=f(x)f(y)$ is true for all $x,y$, we can let $y$ have some positive value very close to $0$, so we have $y\to0^+$ and $f(y)=e^k$. –  Jan 30 '15 at 09:50
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    That is not correct. There are other (strange) solutions to this equation. It's only when continuity is assumed that this becomes the only solution. You can read more about that for a similar equation here – Winther Jan 30 '15 at 09:53
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The general solution of the given functional equation depends on conditions imposed to $f$. We have:

The general continuous non vanishing complex solution of $f(x+y)=f(x)f(y)$ is $\exp(ax+b\bar x)$, where $\bar x$ is the complex conjugate of $x$. Note that this function is not holomorphic.

If we request that $f$ be differentiable then the general solution is $f(x)=\exp (ax)$, as proved in the @doetoe answer.

Added:

here I give a similar but a bit simpler proof:

Note that $f(x+y)=f(x)f(y) \Rightarrow f(x)=f(x+0)=f(x)f(0) \Rightarrow f(0)=1$ ( if $f$ is not the null function). We have also: $f'(x+y)=f'(x)f(y)$ (where $f'$ is the derivative with respect to $x$). Now, setting $x=0$ and dividing by $f(y)$ yields: $$ f'(y)=f'(0)f(y) \Rightarrow f(y)=k\exp\left(f'(0)y\right) $$ and, given $f(0)=1$ we must have $k=1$.

The continuity request can be weakened but, if $f$ is not measurable then, with the the aid of an Hamel basis, we can find more general ''wild'' solutions.

For a proof see: J.Aczél : lectures on functional equations and their applications, pag. 216.

Emilio Novati
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