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

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.

.

Relevant Pages

  • Re: Solutions to q(x y) = q(x) q(y)
    ... > Newsgroups: sci.math ... > Please skip the excessive LaTex, ... You mean the square ROOT of x? ... continuity at 0. ...
    (sci.math)
  • Re: OT: Illegal immigration into Canada
    ... is in Canada for the same stuff made here,,, hmmm,,, ... bring back that old continuity. ... bring back Clerk Maxwell to me. ...
    (alt.support.mult-sclerosis)
  • Re: Are the ready welders any good?
    ... then it's probably in top operating ... Hmmm. ... bring back that old continuity. ... bring back Clerk Maxwell to me. ...
    (sci.engr.joining.welding)