Inequality involving matrices that are unitarily equivalent to a shift matrix 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?
 A: 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 $B<A$.

*Proving $B<A$ is easy because the ordering of the $r^n$ implies
$$ (r^n - r^m) (1+m-n) \geq 0 . $$
It remains to prove that
$$ A = \sum_{m,n=1}^N r^n (1+m-n) |(USU^\ast)_{mn}|^2 $$
is positive.

*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.
