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In my research (I'm a physicist) I encountered an interesting expression that I'll describe to you. A lot of numerical evidence makes me believe that this expression always gives a positive number, but I'm out of ideas for how to prove that.

Let $r \in \mathbb R^N$ be a sorted probability vector ($r_k \in [0,1]$ and $\sum_k r_k = 1$ and $r_1 \geq \cdots \geq r_N$) and $x \geq 0$ is a real number. Let also $U \in \mathrm U(N)$ be a unitary matrix and $$ S = \begin{pmatrix} 0 & \cdots \\ 1 & 0 & \cdots \\ 0 & 1 & 0 & \cdots \\ & \ddots & \ddots & \ddots \end{pmatrix} \in \mathbb R^{N,N} $$ the right shift.

I have found the expression $$ \sum_{m,n} \left( r^n \mathbb e^x - r^m \mathbb e^{-x} \right) (1 + m - n)\, \left| (USU^\ast)_{nm} \right|^2 $$ to be positive for all choices of $x$, $r$ and $U$.

Note that for $U=1$, the expression is zero (because $S_{nm} = \delta_{n,1+m}$), and for $x=0$, it is trivially positive.

Any ideas? Counterexamples?

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    and it doesn't hold for an unsorted matrix? will $|(USU^*)_{nm}|^2$ ever be $<0$ since you have an absolute sign there and a square for good measure. Depending on how you answer above, I am guessing its the first two products that you need to understand? – 2017-02-08
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    You mean unsorted *probability vector*? No, then it doesn't hold. I agree that the first two factors are important, but it also doesn't work if the third factor is something arbitrary. – 2017-02-08

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Okay, this apparently didn't find too much interest, but I finally found a proof and I thought I'd post it here to give myself closure.

  • First of all, the whole expression has the form $$ A e^x - B e^{-x} \;, $$ this being positive for all $x \geq 0$ is equivalent to $A>0$ and $B

  • Proving $B

  • We define the partial sums $$ R^n = \sum_{k=1}^n r^k $$ and note that because of the ordering, $$ R^m - R^n \leq (m-n) r^n \tag{1} . $$ Using (1), we get immediately $$ A \geq \sum_{m,n} (r^n + R^m - R^n) |(USU^\ast)_{mn}|^2 . $$

  • We can now in each term evaluate one of the sums. \begin{align} \sum_n |(USU^\ast)_{mn}|^2 &= |(U\, S^\ast S\, U^\ast)_{mm}|^2 , \\ \sum_m |(USU^\ast)_{mn}|^2 &= |(U\, S S^\ast\, U^\ast)_{nn}|^2 . \end{align} (This is easy to see if you use Dirac bra-ket notation.)
    After renaming one summation index, we have $$ A \geq \sum_n r^n |(U\, S S^\ast\, U^\ast)_{nn}|^2 + \sum_n R^n |(U\, [S^\ast, S]\, U^\ast)_{nn}|^2 . $$

  • Finally, since $r$ and $R$ are ordered, those sums are minimized by the unitary $U$ that diagonalizes $SS^\ast$ (or, respectively, $[S^\ast,S]$) in such a way that their eigenvalues are ordered ascending (descending).
    The spectrum of $SS^\ast$ is $(0, 1, 1, \cdots, 1)$ and that of $[S^\ast,S]$ is $(1, 0, \cdots, 0, -1)$. Therefore $$ A \geq \sum_{n=2}^N r^n + (R^1 - R^N) = 0 , $$ qed.