Let A, B and C be finitely supported probability distributions with at most d nonzero probabilities each. Now consider the following simultaneous equations using p-norms, for each value of p≥1, given by


||A||p + ||B||p = ||C||p


where A, B and C are still non-negative, but we relax normalization on A and B. Imagine that C is fixed and, without loss of generality, normalized. We want to solve for A and B.


First, note that one obvious family of solutions is 


A = (1-x) C , B = x C , 0≤x≤1 .


Question: Ignoring the obvious permutation symmetries, are these the *only* solutions?


Edit: By p-norm, I mean the vector p-norm: ||A||p = (∑j |aj|p )1/p. Although we don't really need the absolute values, since the aj are all non-negative.



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
$$||A||\_p = \exp\left( \frac{\ln \bigl[\exp(\alpha\_1 p) + \exp(\alpha\_2 p) + \cdots + \exp(\alpha\_d p) \bigr]}{p} \right).$$
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 $\exp(2m\pi i/p\_0)$. 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 $\exp(2m\pi i/p\_0)$ 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](http://www.jstor.org/pss/1989556). 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](http://www.jstor.org/stable/10.4169/amer.math.monthly.119.07.601)).
> 
> 
> 


