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        riddles >> putnam exam (pure math) >> f(xf(y)) = yf(x)
        
(Message started by: Aryabhatta on Dec 17^th, 2004, 10:56am)

Title: f(xf(y)) = yf(x)
Post by Aryabhatta on Dec 17^th, 2004, 10:56am

This is from some math olympiad (probably an old IMO).

Find all functions f: [bbr]⁺ [mapsto] [bbr]⁺ such that
1) f(xf(y)) = yf(x) [forall] x [in] [bbr]⁺
2) Lt_x[to]0 f(x) = 0

[bbr]⁺ is the set of postive reals.

Title: Re: f(xf(y)) = yf(x)
Post by Icarus on Dec 18^th, 2004, 11:36am

::[hide]Let a=f(1). Then it quickly follows that f(aⁿx) = f(x) for all x, and all integers n. If a [ne] 1, then either aⁿx [to] 0 as n [to] [infty] or as n [to] -[infty]. In either case, the limit condition gives us f(x) = 0 for all x.

If a = 1, then letting x=1 gives f(f(y)) = y. I.e. f is it's own inverse. Let u = f(y) (so y = f(u)) in the original equation, and we have f(xu) = f(x)f(u). Thus, f(x^r) = f(x)^r for all rational numbers r. if x<1, letting r [to] [infty], we must have f(x)^r [to] 0, which requires f(x) < 1. So if x < y, f(x) = f(x/y)f(y) < f(y), so f is strictly increasing.

Let f(e) = b. Since e > 1, b > 1. The values of e^r with r rational are dense in [bbr]⁺. The values of b^r = f(e^r) are also dense. The only way that f can be strictly increasing is for it to be continuous. And so f(e^t) = b^t for all real t. So f(x) = f(e^{ln x}) = b^{ln x} = x ^{ln b} = x^c. But since f(f(x)) = x, we must have c² = 1. And since f(x) is increasing, c > 0. I.e. c = 1.

So the only functions satisfying the conditions are f(x) = 0, and f(x) = x.[/hide]::

Title: Re: f(xf(y)) = yf(x)
Post by Aryabhatta on Dec 19^th, 2004, 11:01am

on 12/18/04 at 11:36:17, Icarus wrote:

Let f(e) = b. Since e > 1, b > 1. The values of e^r with r rational are dense in [bbr]⁺. The values of b^r = f(e^r) are also dense. The only way that f can be strictly increasing is for it to be continuous .

Can you give a brief proof of this assertion? It does not look very obvious to me. The name of the theorem will do, if it is a known fact.

Also, f(x) = 0 is not a solution as the range is the set of positive reals.

Title: Re: f(xf(y)) = yf(x)
Post by Eigenray on Dec 20^th, 2004, 12:48pm

Suppose f,g are increasing, g is continuous, and f(x)=g(x) for x in D, a dense set. Then for any a,
f(a) [le] inf { f(x) : x>a } = inf { f(x) : x>a, x in D } = inf { g(x) : x>a, x in D } = g(a);
similarly f(a) [ge] g(a), and it follows f=g is continuous.

Title: Re: f(xf(y)) = yf(x)
Post by Aryabhatta on Dec 20^th, 2004, 1:56pm

Nicely done Icarus and Eigenray.

Here is the solution I had in mind.

[hide]
Let us first show that f(1) = 1.
Now f(xf(y)) = yf(x) means f(f(y)) = yf(1)
We also have that
f(f(xf(y)) = f(yf(x)) = xf(y)
This implies that f(1) = 1.

So we have that f(f(x)) = x.

We can easily see that:
i) f(xⁿ) = f(x)ⁿ for all natural numbers n.
ii) f(x)f(y) = f(xy)

if 0 < x < 1, the xⁿ -> 0 as n-> oo. Thus we must have that f(x) < 1.

Similarly if x > 1 then f(x) > 1.

Now we have that f(1/x) = 1/f(x)

Now assume x < f(x).

Consider f(x/f(x)) = f(xf(1/x)) = f(x)/x.
This is not possible as x/f(x) < 1 while f(x/f(x)) = f(x)/x > 1.

Similarly we see that f(x) < x is not possible.
Hence f(x) = x.

[/hide]

Title: Re: f(xf(y)) = yf(x)
Post by Icarus on Dec 20^th, 2004, 3:35pm

on 12/19/04 at 11:01:35, Aryabhatta wrote:

Yes, after coming up with the f(x) = 0 solution in the a [ne]1 case, I overlooked that it was artificially excluded.

EigenRay's argument works fine for what I needed, but it is possilbe to show that f is continuous without presupposing the existance of a continuous function that matches f on a dense subset:

If U and V are intervals in [bbr] and f: U [to] V is monotone, and if there exists D [subset] U such that f(D) is dense in V, then f is continuous.

Proof: Assume f is increasing (the argument for f decreasing is similar). For all x [in] U, f(x-) = sup {f(y) : y <x}, and f(x+) = inf {f(y): y>x} (both infimum are supremum must exist, as the sets are bounded above and below respectively by f(x)). So we have f(x-) [le] f(x) [le]f(x+). If f(x-) < f(x), then there exists d [in] D such that f(d) [in] (f(x-), f(x)), but f(d) < f(x) [bigto] d < x [bigto] f(d) [le] f(x-), a contradiction. So f(x-) = f(x). Similarly, f(x+) = f(x), and therefore f is continuous at x. Since x was arbitrary, f is continuous on all of U.

Title: Re: f(xf(y)) = yf(x)
Post by Icarus on Dec 20^th, 2004, 3:48pm

So, how many solutions are possible if you weaken the restrictions?

In particular, find all functions f: U [to] [bbr] for some maximal interval U, such that
1) f(xf(y)) = yf(x) for all x, y in U
2) f is continuous.

(The maximal condition on U means that U is the largest possible interval to which f can be extended without violating the other conditions.)

Obviously f(x) = x and f(x) = 0 work. Are there any other solutions?