       Re: Re: What kind of math problem is this?

• To: mathgroup at smc.vnet.net
• Subject: [mg21045] Re: [mg20950] Re: [mg20896] What kind of math problem is this?
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Sun, 12 Dec 1999 23:51:12 -0500 (EST)
• References: <199911200607.BAA04739@smc.vnet.net> <199912010650.BAA07636@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Daniel Lichtblau wrote:
>
> Seth Chandler wrote:
> >
> > This may be a little off topic, but I was hoping someone here could help.
> >
> > Let r and f be real numbers between 0 and 1 inclusive. How does one
> > determine if there is a function u that satisfies this relation and, if so,
> > what the function is.
> >
> > InverseFunction[u][f u[x] + (1 - f)u[y]] == (1 - r)(f x + (1 - f)y) + r x
> >
> > [ ... ]
>
> The answer depends to some extent on what exactly you are willing to
> use. Before going into more detail, I'll reformulate your problem a bit.
> You require a function u[x], invertible on some domain of interest, such
> that
>
> f*u[x] + (1 - f)*u[y] == u[(1 - r)*(f* x + (1 - f)*y) + r*x]
>
> By letting f+r-f*r == k, we may rewrite the rhs as
>
> u[k*x + (1-k)*y]
>
> The case where k==f is trivial (just take u[x]==x) and it arises when
> either r==0 or f==1. We'll assume neither of these occurs.
>
> One can show there is no differentiable function u[x] that satisfies
> this functional equation. To prove this, assume u is differentiable at x
> and chose y = x + dx/(1-k). We then have
>
> u[x+dx] == f*u[x] + (1-f)*u[x+dx/(1-k)]
>
> Invoking differentiability at x we obtain
>
> u[x] + u'[x]*dx + o(dx) ==
>         f*u[x] + (1-f)*(u[x]+u'[x]*dx/(1-k)+o(dx/(1-k)))
>
> where in both places o(...) vanishes faster than dx (think of dx as
> small, i.e. we take a limit as it goes to zero). Rewriting the rhs we
> have
>
> u[x] + (1-f)/(1-k)*u'[x]*dx + (1-f)/(1-k)*o(dx/(1-k))
>
> As the small-o terms vanish faster than the linear terms, sufficiently
> close to x these cannot be equal because the linear terms in dx
> disagree.
>
> I have not successfully convinced myself whether or not there can be a
> nowhere differentiable function u[x] that does what you want. But
> already we realize there will be no useful computable u[x] (how many
> nowhere-differentiable functions can you compute?).
> [ ... ]

Okay, I can do a bit better on this score. Call a function u[x] valid if
it is invertible and satisfies the functional equation

u[k*x + (1-k)*y] == m*u[x] + (1 - m)*u[y]

where we assume k!=m.

I'll indicate why u[x] cannot be continuous. Say it is. First note that
invertibility and continuity together imply u[x] must be (strictly)
monotonic. Clearly we can subtract a constant and still satisfy the
functional equation (and retain continuity and invertibility), hence we
may assume u = 0. Then we know u != 0 (or it would not be
invertible). As we can clearly multiply by a nonzero scalar without
invalidating the function, we may furthermore assume u = 1.

Letting x = 1/k and y = 0 we get 1 == u == m*u[1/k] so u[1/k] == 1/m.
We let x = 1/k^2, y = 0 and similarly obtain u[1/k^2] == 1/m^2. Likewise
we find u[1/k^s] == 1/m^s for positive integer exponents s, and the same
idea works for negative integer exponents as well.

Now take x = 0 and y = 1/(1-k) to get u[1/(1-k)] == 1/(1-m). Taking y =
1/(1-k)^t similarly gives u[1/(1-k)^t] == 1/(1-m)^t for integer
exponents t.

Right away we see that NO function u[x] will work if k is 1/2. This is
because u[1/2] == u[1/(1-1/2)] == 1/(1-m) != 1/m == u[1/2]. (This really
did not require an assumption that u[x] be continuous, you just need to
modify the argument a bit.)

In the general case just note that we have two sequences in the x-u[x]
plane

{k^s,m^s} and {(1-k)^t,(1-m)^t}

that change at different rates. Perhaps it is easier to see this by
taking logarithms. The sequences now lie on lines, the slope of the
first is

s*Log[m] / (s*Log[k]) == Log[m]/Log[k]

and likewise the slope of the second is Log[1-m]/Log[1-k]. These slopes
cannot be equal by assumption that m!=k. From this one can conclude that
eventually there will be a point {x1,y1} = {s1*Log[k],s1*Log[m]} on the
first line, and a corresponding point {x2,y2} =
{t1*Log[1-k],t1*Log[1-m]} on the second, such that x1<x2 but y1>y2 or
vice versa (depending on which slope is larger). This implies that u[x]
must violate monotonicity.

I believe there cannot be a nonconstant function u[x] satisfying the
basic functional equation, but so far have been unable to prove that
(even assuming invertibility).

Daniel Lichtblau
Wolfram Research

```

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