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I am having some trouble proving an inequality involving the trace norm or the operator $f{\cal{F}}_Ng$ where $f, g$ are diagonal matrices, f is positive semidefinite and $\cal{F}_N$ is the Discrete Fourier transform matrix given by $$ ({\cal{F}_N})_{n,m}= \frac{1}{\sqrt{N}}e^{-\frac{2\pi i}{N}nm} $$ a unitary Vandermonde matrix. The inequality I am trying to show is this: $$ \|f{\cal{F}_N} g{\cal{F}_{N}}^{\ast}\|_{tr}\leq \frac{1}{\sqrt{N}} \|f\|_{tr}\|g\|_{tr} $$ where $(\cdot)^{\ast}$ denotes the Hermitian adjoint.

Has anybody got an idea? Thanks very much for your help.

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  • $\begingroup$ Why do you think the inequality is true? $\endgroup$ Aug 26, 2013 at 15:41
  • $\begingroup$ I tried to find a counterexample to this, but none has emerged so far. Obviously the counterexamples I was trying to construct are not trivial, i.e. I am not plugging in numbers. :) $\endgroup$
    – John
    Aug 26, 2013 at 15:43
  • $\begingroup$ Maybe rewrite $fFg$ using the Schur product and then using inequalities for matrix norms involving Schur products helps. $\endgroup$
    – Suvrit
    Aug 26, 2013 at 16:17
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    $\begingroup$ That unfortunately would not give me the factor of $\frac{1}{\sqrt{N}}$ at the front! :) Also, by doing that I wouldn't have a sharp bound on $\|\cdot\|_{tr}$, which is the main aim of this. The only thing I could do for sure is reducing this to $\|f{\cal{F}}_{N}g\|$. Thanks for your help anyway! :) $\endgroup$
    – John
    Aug 26, 2013 at 20:35
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    $\begingroup$ Erm... Once $f$ and $g$ are diagonal (so the trace norm for them is just the sum of absolute values of the diagonal entries), isn't it true that it suffices to check the inequality just for two single coordinate projections, where it is obvious? Am I missing something subtle? $\endgroup$
    – fedja
    Aug 26, 2013 at 22:02

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