Continuous, 1-periodic $f$ with $f(x+y) = f(x) f(y)$ for $x, y \in \mathbb{R}$

Question

The problem is to find all $f : \mathbb{R} \to \mathbb{C}$ that is continuous, has $f(x) = f(x+1) \forall x$, and

$$f(x+y) = f(x) f(y) \quad x, y \in \mathbb{R}$$

Plug in $y=0$, we find $f(x) = f(0)f(x)$. We write down $f = 0$ as a solution, and move on assuming $f(0) = 1$.

We have $f(n) = 1$ for integer $n$ by periodicity. Furthermore, we can get that $f(p/q) = \sqrt[q]{1}$, which leads me to believe that the solutios are of the form $f_m = e^{2 \pi i m t}$ for each integer $m$. How to finish this proof?

Umberto P. · Accepted Answer · 2015-03-11T01:57:34.870

3

Suppose that $f(0) = 1$. Then $|f(x) - 1| < 1$ for all $x$ in some neighborhood $I$ of $0$ so that $\log f$ is defined and continuous on $I$ and for all $x,y \in I$ you have $$\log f(x+y) = \log f(x) + \log f(y).$$ Thus $\log f$ is additive and continuous in $I$. It is well known that the only functions satisfying this are $\log f(x) = Cx$, i.e. $$ f(x) = e^{Cx},\quad x \in I.$$

The functional relationship $f(x+y) = f(x)f(y)$ implies that $f$ extends uniquely to the entire line with the same definition: $f(x) = e^{Cx}$, $x \in \mathbb R$.

Finally to take periodicity into account, we insist $e^{Cx} = e^{C(x + 1)}$ so that $e^C = 1$. Thus $C = 2\pi i m$ for some integer $m$ giving you $f(x) = e^{2\pi i m x}$.

edited Mar 11 '15 at 01:57

answered Mar 10 '15 at 23:47

Umberto P.

52,165

The values of $f$ are complex, so what does $f(x)>0$ mean? – hmakholm left over Monica Mar 10 '15 at 23:53
Could you provide a proof (or a link to one) about the $f(x+y) = f(x) f(y) \Leftrightarrow f(x) = e^{Cx}$ statement given only continuity? According to http://math.stackexchange.com/a/1126444/92774 the general solution is $e^{Ax + B\overline{x}}$. – MT_ Mar 10 '15 at 23:55
@Soke: The domain of $f$ is $\mathbb R$, where $x=\bar x$ anyway. – hmakholm left over Monica Mar 11 '15 at 00:06
@HenningMakholm True. – MT_ Mar 11 '15 at 00:11
@HenningMakholm good point. Editing... – Umberto P. Mar 11 '15 at 01:54

score 1 · Answer 2 · answered Mar 11 '15 at 00:01

You have shown that $|f(x)|=1$ when $x$ is rational. By continuity this must also be true for irrational $x$.

Because $f$ is continuous and nowhere zero, it has a continuous argument $g:\mathbb R\to\mathbb R$, such that $$ f(x) = e^{ig(x)}$$ Without loss of generality we can take $g(0)=0$.

By the functional equation for $f$ we have that $$ g(x+y)-g(x)-g(y)=\text{some multiple of }2\pi $$ everywhere, and since the right-hand-side is $0$ when $x=y=0$ and is continuous, it must be $0$ everywhere. So $g(x+y)=g(x)+g(y)$, and the only continuous functions that satisfy that are $g(x)=kx$ (set $k=g(1)$ and prove this first for rational $x$, and then for irrationals by continuity).

Because $f(1)=f(0)=1$, we have that $g(1)=k$ must be a multiple of $2\pi$, which completes the proof.

Continuous, 1-periodic $f$ with $f(x+y) = f(x) f(y)$ for $x, y \in \mathbb{R}$

2 Answers2