Re: Solutions to q(x y) = q(x) q(y)

In article <Pine.BSI.4.58.0506212040070.18648@xxxxxxxxxxxxxxxxx>,
William Elliot <marsh@xxxxxxxxxxxxxxxxxx> wrote:

> On Tue, 21 Jun 2005, Ronald Bruck wrote:
> > William Elliot <marsh@xxxxxxxxxxxxxxxxxx> wrote:
> > > From: bazley@xxxxxxxxx
> > >
> > > > Suppose the function q: \R \to \R$ satisfies $q(x y) = q(x) q(y)$
> > > > for all $x, y \in \R$. Note that $q(x) = x^n$ ($n$ a positive
> > > > integer) and $q(x) = |x|^r$ ($r$ any real number) satisfy this
> > > > ``functional equation''. So does $q(x) = 0$ ($r = - \infty$?). Show
> > > > that if $q$ satisfies the functional equation and $q(x) > 1$ for $x >
> > > > 1$, then there is a real number $r > 1$ such that $q(x) = x^r$ for
> > > > all positive $x$.
> > >
> > > q(x) = sqr x also fulfills the premise. Do you mean r > 0 ?
> > >
> > > > I figured out so far that q(x^{m/n}) = q(x)^{m/n}, and that
> > > > it suffices to prove that q(x)^r = q(x^r), but now I'm stuck
> > >
> > > It does? I doubt it unless q is continuous.
> > >
> > > > Any help would be greatly appreciated
> > >
> > > q(x) = q(e^log x) = q(e)^log x
> > > = e^(log q(e) * log x) = x^log q(e)
> >
> > sqr(x)? You mean the square ROOT of x? That's not defined for x < 0.
>
> Yes. So what?
> The question is asking about q with positive domain.
>
> > (q maps R to R.) You mean the square? That's a subcase of q(x) = x^n.
>
> No. So is q(x) = sqr(x) = x^(1/2)
>
> > Not sure what you're getting at. (Of course r must be >0. 0^r is
> > undefined when r < 0.)
> >
> That the requirement r > 1 is in error and that r > 0 is how the problem
> should read. (Last sentence of problem description)
>
> > The proposition is proved by noting that q(1) = q(1*1) = q(1)^2 hence
> > q(1) = 0 or 1. If it's 0, then q(x) = q(x*1) = q(x)*q(1) = 0 for all
> > x. If it's 1, note that for 0 < x < y,
> >
> > q(y) = q(x*(y/x)) = q(x)*q(y/x), so q(y/x) = q(y)/q(x). Therefore
> >
> > 1 < y/x ==> 1 < q(y/x) by hypothesis, hence 1 < q(y)/q(x),
> > i.e. 0 < x < y ==> q(x) < q(y). This proves q is increasing on
> > (0,\infty). Combined with q(x^(m/n)) = q(x)^(m/n) you can use
> > continuity to prove q(x^r) = q(x)^r.
>
> Ok, nice. That's how continuity is established.
>
> > Negatives can be handled similarly. Hmmm. Could be a problem at 0.
> > No, it's OK there too, you can let a sequence decrease (or increase)
> > exponentially to 0 to get continuity at 0.

Well, you got me. I don't see anywhere it says the domain is the
POSITIVE reals. I took \R to be a home-grown macro for \mathbb{R}, or
maybe \mathbf{R}.

And if he DOES mean R^+, why does he explicitly use |x|?

As for q(x) = sqr(x) also being a subcase of x^n, he explicitly said,
"n a positive integer", and 1/2 ain't an integer in Nuevo California.
That, presumably, was to avoid the sign problem for x < 0 (otherwise
you have to write |x|^{r-1}x or |x|^r sgn(x)--the latter is better for
r > 0 but < 1). Hmm. That's another solution, so my rumblings about
zero being a problem were right on as it turns out. You have to be
more careful about x < 0. (Or solve the problem on R^+, and reflect
backward.)

Come to think of it, isn't sgn(x) a solution? Ah. No. x > 1 doesn't
imply sgn(x) > 1. But it is a DISCONTINUOUS example of the original.
(But there are surely lots of discontinuous examples.)

--Ron Bruck
.

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