ca.analysis and odes - Are two probability distributions uniquely constrained by the sum of their p-norms?

Here is a proof that Steve's rescaling gives you all solutions, together with the trivial operation of permuting the components of $A$, $B$, and $C$ if you view them as vectors with positive coeifficients. (If you view them this way, then Steve's notation $||A||_p$ is just the usual $p$-norm.)

I first tried what Alekk tried: You can take the limit as $p to infty$ and eventually obtain certain power series expansions in $1/p$. Or you can take the limit $p to 0$ and obtain certain power series expansions in $p$. The problem with both approaches is that the information in the terms of these expansions is complicated. To help understand the second limit, I observed that the two sides of Steve's equation are analytic in $p$, but it only helped so much.

Then I realized that when you have a complex analytic function of one variable, you can get a lot of information from looking at singularities. So let's look at that. Let
$alpha_k = ln a_k$, so that
$$||A||_p = expleft( frac{ln bigl[exp(alpha_1 p) + exp(alpha_2 p) + cdots + exp(alpha_d p) bigr]}{p} right).$$
The expression inside the logarithm has been called an exponential polynomial in the literature, which I'll call $a(p)$. As indicated, $||A||_p$ has a logarithmic singularity when $a(p) = 0$. $||A||_p$ has another kind of singularity when $p = 0$, but won't matter for anything. Also $a(p)$ is an entire function, which means in particular that it is univalent and has isolated zeroes. Also, none of the zeroes of $a(p)$ are on the real axis. Let $b(p)$ and $c(p)$ be the corresponding exponential polynomials for $B$ and $C$.

Suppose that you follow a loop that starts on the positive real axis, encircles an $m$-fold zero of $a(p)$ at $p_0$, and then retraces to its starting point. Then the value of $||A||_p$, which is non-zero for $p > 0$, gains a factor of $exp(2mpi i/p_0)$. Thus Steve's equation is not consistent unless all three of $a(p)$, $b(p)$ and, $c(p)$ have the same zeroes with the same multiplicity. (Since $exp(2mpi i/p_0)$ cannot have norm 1, geometric sequences with this ratio but with different values of $m$ are linearly independent.)

At this point, the problem is solved by a very interesting paper of Ritt, On the zeros of exponential polynomials. Ritt reviews certain results of Tamarkin, Polya, and Schwengler, which imply in particular that if an exponential polynomial $f(z)$ does not have any zeroes, then it is a monomial $f_alpha exp(alpha z)$. Ritt's own theorem is that if $f(z)$ and $g(z)$ are exponential polynomials, and if the roots of $f(z)$ are all roots of $g(z)$ (with multiplicity), then their ratio is another exponential polynomial. Thus in our situation $a(p)$, $b(p)$, and $c(p)$ are all proportional up to a constant and an exponential factor. Thus, $A$, $B$, and $C$ must be the same vectors up to permutation, repetition, and rescaling of the coordinates. Repetition is an operation that hasn't yet been analyzed. If $A^{oplus n}$ denotes the $n$-fold repetition of $A$, then $||A^{oplus n}||_p = n^{1/p}||A||_p$. Again, since geometric sequences with distinct ratios are linearly independent, Steve's equation is not consistent if $A$, $B$, and $C$ are repetitions of the same vector by different amounts.

The same argument works for the generalized equation
$$x_1||A_1||_p + x_2||A_2||_p + cdots + x_n||A_n||_p = 0.$$
The result is that any such linear dependence trivializes, after rescaling the vectors and permuting their coordinates.

Update (by J.O'Rourke): Greg's paper based on this solution was just published:

"Norms as a function of $p$ are linearly
independent in finite dimensions," Amer. Math. Monthly, Vol. 119, No. 7, Aug-Sep 2012, pp. 601-3
(JSTOR link).

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