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For the absolute value $|C|=(C^*C)^\frac{1}{2}$ and the Hilbert-Schmidt norm $\parallel C\parallel_{HS}=(trC^*C)^\frac{1}{2}$ of the operator $C$. The following inequality is shown by Araki et al in 'An Inequality for Hilbert-Schmidt Norm, Commun. Math. Phys. 81, 89-96 (1981)'

For any two bounded linear operators $A$ and $B$ on a Hilbert space $\mathbb{H}$, $$\parallel |A|-|B| \parallel_{HS}\le \sqrt{2}\parallel A-B \parallel_{HS},$$ and the factor $\sqrt{2}$ is best possible.

What is the best constant factor $c$ such that $$\parallel |A|+|B| \parallel_{HS}\ge c\parallel A+B \parallel_{HS}?$$

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  • $\begingroup$ infinity. Take A to be a non-zero positive operator and B = -A. Incidentally, either you meant to say "finite dimensional Hilbert space" or you meant "two Hilbert-Schmidt linear operators" since not all bounded linear operators on a Hilbert space have a Hilbert-Schmidt norm. $\endgroup$
    – Andrew Stacey
    Commented Jun 21, 2010 at 7:40
  • $\begingroup$ Andrew, that doesn't work - the inequality is "the other way round" from what your comment seems to say. $\endgroup$
    – Yemon Choi
    Commented Jun 21, 2010 at 7:47
  • $\begingroup$ @Stacey: $A, B$ are arbitrary. In this case, I actually want to determine the maximum of $c$ such that the inequality holds for arbitrary bounded linear operators $A, B$. $\endgroup$
    – Russel
    Commented Jun 21, 2010 at 7:53
  • $\begingroup$ @Yemon: Whoops! Just goes to show that "best is in the eye of the beholder". @Russel: The identity operator on an infinite dimensional Hilbert space does not have a Hilbert-Schmidt norm. So for your question to make sense, you need to include some conditions to exclude this case. $\endgroup$
    – Andrew Stacey
    Commented Jun 21, 2010 at 8:02
  • $\begingroup$ @Stacey: The condition here is the same as in the paper by H.Araki. We allow an operator with norm infinity. $\endgroup$
    – Russel
    Commented Jun 21, 2010 at 8:13

1 Answer 1

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edit: there is a problem with this, see the comments; I'm leaving this up for the moment in case it helps someone else come along to write a better answer.

I'm pretty sure that the inequality $$ \vert {\rm tr}(A^*B) \vert \leq {\rm tr}( |A| |B| ) $$ holds whenever $A$ and $B$ are Hilbert-Schmidt, just by using the polar decompositions of $A$ and $B$.

If this is the case, then we'd have

$$ \eqalign{ {\rm tr} ((A+B)^*(A+B)) & = {\rm tr}(A^*A) + {\rm tr}(A^*B) + {\rm tr}(B^*A) + {\rm tr}(B^*B) \\ & \leq {\rm tr}(\vert A\vert^2) + {\rm tr}(\vert A\vert \vert B\vert) + {\rm tr}(\vert B\vert \vert A\vert) + {\rm tr}(\vert B\vert^2) = {\rm tr}((\vert A\vert + \vert B\vert)^2) } $$ which implies that you can get away with $c=1$. Taking $A=B$ to be positive shows that this is sharp.

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  • $\begingroup$ I'm not seeing, right now, why $|tr(A^*B)|\leq tr(|A| |B|)$. Can you give a hint... $\endgroup$
    – Matthew Daws
    Commented Jun 21, 2010 at 8:19
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    $\begingroup$ It is true $|tr(AB)|\leq tr(|AB|)$, but it is not true in general $tr(|AB|)\leq tr(|A| |B|)$. $\endgroup$
    – Russel
    Commented Jun 21, 2010 at 8:29
  • $\begingroup$ Hmm, I may have been careless here: I was just thinking of putting $A= U \vert A\vert$ and $B=V \vert B\vert$ for partial isometries $U$ and $V$, then ${\rm tr}(A^*B) = {\rm tr} (\vert A\vert U^*V \vert B\vert)$ -- but then I've overlooked the fact that $\vert A\vert \vert B\vert$ might not be positive... Will have to think about this more! $\endgroup$
    – Yemon Choi
    Commented Jun 21, 2010 at 8:33

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