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Is there a nice description of all nonzero measurable functions $f:\mathbb{C}\to\mathbb{C}$ such that $f(ab)=f(a)f(b)\ $ for all $a$ and $b$ in $\mathbb{C}$?

This is inspired by the question Multiplicative Analytic Functions, Theo Buehler's comments there, and idle curiosity. The only examples I know have the form $f(z)=|z|^cz^k$ or $f(z)=|z|^c\overline z^k$ for some $c\in[0,\infty)$ and nonnegative integer $k$. A subquestion is: Are these all of them? (Answer: No, see update.)

I tried searching, and found that nonzero multiplicative maps from a semigroup to the multiplicative semigroup of complex numbers are sometimes called semicharacters, and other times called characters, but so far this hasn't helped me find anything that answers this question.

Update: I found from reading a remark in Grillet's Commutative semigroups that if $f$ is an example, then so is the function $g$ defined by $g(0)=0$, $g(z)=f(z)/|f(z)|$ if $z\neq 0$, and $g$ maps into the unit circle unioned with $\{0\}$. This then made me realize that if $f$ and $g$ are examples, then so is the function $h$ defined by $h(0)=0$, $h(z)=f(z)/g(z)$ if $z\neq 0$. So there are examples I did not include above. I would have to include $f(0)=0$, $f(z)=|z|^cz^k$ where $c$ is any real number and $k$ any integer.

Also, I'm not even sure about what all of the continuous examples are.

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So I am pretty sure that all pathological solutions to the Cauchy functional equation (en.wikipedia.org/wiki/Cauchy's_functional_equation) are non-measurable. If that's true then $|f(z)| = |z|^n$ for some $n$ and it shouldn't be hard to conclude from here. –  Qiaochu Yuan Jul 19 '11 at 3:34
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Yes, it is true that the only measurable solutions of the Cauchy equation are of the form $cx$. For the complex version, can one always split off the action on the norm from the action on the argument? –  André Nicolas Jul 19 '11 at 7:21
    
^ The relationship with Cauchy's functional equation can be seen by looking at the function $\ln|f(e^z)|$. Also, it looks more symmetrical to describe the general form as $f(z) = z^a\bar{z}^b$ in my opinion. (I'm just saiyan.) –  anon Jul 19 '11 at 7:40
    
@anon Yes $g(z)=\ln{|f(e^z)|}$ gives rise to $g(z+w)=g(z)+g(w)$, but this is a function on $\mathbb{C}$; usually Cauchy's functional equation is spoken about in the context of (real) functions on $\mathbb{R}$. That's how I interpret Andre's inquiry at least. –  Nick Strehlke Jul 19 '11 at 14:55
    
@anon: What are $a$ and $b$? –  Jonas Meyer Jul 19 '11 at 17:27

2 Answers 2

up vote 16 down vote accepted

(Haar) measurable endomorphisms of a locally compact Polish group are always continuous. In this case, unless $f$ is the constant $0$ map, that means $f$ is already continuous on $\mathbb{C}\backslash \{0\}$. Then, by continuity, there is some real $c$ such that for all nonzero real $r$ we have $|f(r)| = |r|^c$.

We first examine how $f$ behaves on positive real numbers. In general, for real $r$ we have $f(r) = r^c e^{2 \pi i g(r)}$ for some continuous function $g: \mathbb{R}^+ \to \mathbb{R}/\mathbb{Z}$. Multiplicativity of $f$ ensures that $g(1) = 0$ and $g(rs) = g(r) + g(s)$; in other words $g$ is a continuous homomorphism from $(\mathbb{R}^+,\times)$ to $(\mathbb{R}/\mathbb{Z}, +)$. There are nontrivial examples of such homomorphism: for example $g(r) = \log(r) + \mathbb{Z}$. This seems to introduce new solutions. Does anyone know of a reasonable classification of such homomorphisms?

Edit: it turns out that all such homomorphisms have the form $g(r) = a \log(r) + \mathbb{Z}$ for some real $k$, as indicated by Artin's Algebra text (p 332).

So writing $z$ as $|z| \cdot z/|z|$, it suffices to understand what $f$ looks like on the unit circle.

Writing $f(e^{2\pi i x}) = e^{2 \pi i h(x)}$, we see that what's left is understanding continuous endomorphisms of the group $(\mathbb{R}/\mathbb{Z},+)$, which all have the form $f(r) = kr$ for some integer $k$.

Unfolding all of this, we see that all continuous (and thus all measurable) endomorphisms of $(\mathbb{C}\backslash \{0\},+)$ are indeed of the form $f(z) = e^{2 \pi i g(|z|)} |z|^c z^k$ for some real $c$ and integer $k$. Adding $0$ back in, we see that the grand list of solutions is:

  • $f(z) = e^{2 a \pi i \log|z|} |z|^c z^k$, $f(0) = 0$
  • $f(z) = f(0) = 0$
  • $f(z) = f(0) = 1$,

where $z$ above represents a nonzero complex number, $a$ and $c$ are real parameters, and $k$ is an integer parameter. In particular, this adds a new function $f(z) = e^{2 \pi i \log|z|}$, $f(0) = 0$ to the initial list presented in the problem.

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If you're willing to assume all such measurable $f(z)$ are continuous on ${\mathbb C} - \{0\}$ then it's doable. (I prove this for locally integrable $f(z)$ below). First observe $f(1) = f(1)f(1)$ so $f(1) = 0$ or $1$. If $f(1) = 0$, then $f(z) = f(z)f(1) = 0$ for all $z$ and there's not much to say. So assume $f(z) = 1$ in the following.

Next, Since $f(z^n) = f(z)^n$ for all $z$ and all integers $n$, it suffices to find the form of $f(z)$ in a neighborhood of $z = 1$. Equivalently, it suffices to find the form of $g(r,s) = f(e^re^{is})$ for all $(r,s)$ in a neighborhood of the origin. Since $g(0,0) = 1$ and we are assuming $f(z)$ to be continuous, there is a well-defined $h(r,s) = \log(g(r,s))$ on a neighborhood of the origin. The functional equation translates into $h(r_1 + r_2,s_1 + s_2) = h(r_1,s_1) + h(r_2,s_2)$, where as before the variables are near the origin. So the real and imaginary parts of $h(r,s)$ must also satisfy this functional equation.

Since $Re(h)$ and $Im(h)$ are continuous real-valued functions satisfying the functional equation, by the well-known additive case (whose proof works even for functions only defined in a neighborhood of the origin) we have that $Re(h)(r,s) = ar + bs$ and $Im(h)(r,s) = cr + ds$ for some real $a,b,c,$ and $d$. In terms of $g(r,s)$, this gives $$g(r,s) = e^{ar + bs}e^{icr + ids}$$ Not every $g(r,s)$ of this form will work globally; since $g(r,s) = f(e^re^{is})$ one must have that $g(r,s+2\pi) = g(r,s)$. This forces $b = 0$ and $d$ to be a multiple of $2\pi$. So we have $$f(e^re^{is}) = g(r,s) = e^{ar}e^{icr + ins}$$ Here $n$ is an integer. Equivalently where $m$ denotes $a - n$, we have (for $z \neq 0$): $$f(z) = |z|^me^{ic\ln|z|}z^n$$ Conversely, since each of the factors $|z|^m$, $e^{ic\ln|z|}$ and $z^n$ satisfy the functional equation so do their product. Thus these are exactly the functions continuous on ${\mathbb C} - \{0\}$ satisfying the functional equation, other than the zero function.

Now back to why measurable examples are always continuous. Based on ccc's answer it seems there is a general theory that gives this. For locally integrable $f(z)$ it can be proven directly. Let $\phi(z)$ be a bump function with integral 1 supported on a neighborhood of the origin. Then for any $z$ and $w$ you have $$f(z)f(w)\phi(w) = f(zw)\phi(w)$$ Integrate both sides with respect to $w$. You get $$f(z) = \int f(zw)\phi(w)\,dw$$ Doing a change of variables this becomes $$= {1 \over |z|^2}\int f(w)\phi(z^{-1}w)\,dw$$ (We assume $z \neq 0$). By the dominated convergence theorem the right-hand side is continuous (even smooth) as a function of $z$, so that $f(z)$ is a continuous function.

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For the general theory see this MO-thread. (It is more than a bit shameful that I participated there myself and didn't see it, but I got myself horribly confused last night...) –  t.b. Jul 19 '11 at 19:37
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Ah, this appeared while I was editing. It looks like we came to the same conclusion in any case. I don't know how I missed the last step in my argument for so long, but it's nice to see a different approach. –  user83827 Jul 19 '11 at 19:46

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