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It is well known that the absolute value on operators does not satisfy the triangle inequality.

My question is whether for all positive operators $P,Q \in B(\mathcal H)$ is there a universal constant $C$ such that $$ |P + iQ| \leq C(P + Q)? $$

I am doubtful that this is true but cannot find a counterexample.

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  • $\begingroup$ By "universal", you mean C should not depend on P and Q, right? $\endgroup$ – Yemon Choi Nov 3 '17 at 22:31
  • $\begingroup$ @YemonChoi That's right. $\endgroup$ – Chris Ramsey Nov 3 '17 at 23:04
  • $\begingroup$ There are counterexamples but I don't know any trivial one. They all require letting the dimension of the space go to infinity; in any fixed finite dimension the statement is true. $\endgroup$ – fedja Nov 8 '17 at 1:29
  • $\begingroup$ @fedja Any papers that you can point me to? I really only need to know that it isn't possible for a paper I'm writing. Do you have an argument or source for the fixed finite-dimensional argument? I would be very happy to know this. $\endgroup$ – Chris Ramsey Nov 8 '17 at 1:38
  • $\begingroup$ I'll post the argument a bit later (maybe even today, but not now). References are my weak point: I never remember any :-) However this should be well-known (to people who know it well) $\endgroup$ – fedja Nov 8 '17 at 1:55
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OK, here goes.

We start with a reformulation of the problem. Assume that everything is invertible and $P+iQ=UBB^*$ where $U$ is unitary and $BB^*=|P+iQ|$. We can also re-parameterize $P$ and $Q$ as $BPB^*$ and $BQB^*$. Then the question reduces to whether for two positive operators $P,Q$, the relation $P+iQ=B^{-1}UB$ implies $P+Q\ge\delta I$ for some universal $\delta>0$. In finite dimension, this condition merely means that all eigenvalues of $P+iQ$ equal $1$ in absolute value (and, say, are distinct for the counterexample purposes to avoid boring discussions about Jordan blocks and approximation arguments).

If the dimension $n$ is fixed, then it implies that $|\operatorname{Tr}(P+iQ)|=|\operatorname{Tr}P+i\operatorname{Tr}Q|\le n$, so, since the traces are positive, we can bound them both by $n$, which, in turn, implies (due to positive definiteness) that $\|P\|,\|Q\|\le n$. Thus $\|P+iQ\|\le 2n$ but $|\det(P+iQ)|=1$, so $\|(P+iQ)^{-1}\|\le C_n$. In particular, for every $x$, we have $\|Px\|+\|Qx\|\ge\|(P+iQ)x\|\ge C_n^{-1}\|x\|$. On the other hand, if we have a vector $x$ with $\langle Px,x\rangle+\langle Qx,x\rangle\le\delta\|x\|^2$, then, by Cauchy-Schwarz, for any unit vector $y$, we have $$ |\langle Px,y\rangle|\le\langle Px,x\rangle^{1/2}\langle Py,y\rangle^{1/2}\le \sqrt{\delta n}\|x\|\,, $$ i.e., $\|Px\|\le \sqrt{\delta n}\|x\|$ and similarly for $Qx$, so if $\delta>0$ is too small, we get a contradiction.

This is a terribly crude bound but it suffices to explain why Christian's approach with $2\times 2$ matrices had no chance to work.

Now the example in high dimension. We need to know that the $n\times n$ real antisymmetric matrix $H$ with diagonal entries $0$ and $H_{i,j}=\frac 1{i-j}$ for $i\ne j$ (the truncated discrete Hilbert transform) has norm bounded by some universal constant $\Psi>0$ regardless of $n$ while the $n\times n$ real symmetric matrix $L$ with $0$ diagonal entries and $L_{i,j}=\frac 1{|i-j|}$ for $i\ne j$ has norm $\Phi_n\approx\log n$. Now put $$ P=a[D_1-(\Phi_n+\Psi)^{-1}(L-iH)], Q=a[D_2-(\Phi_n+\Psi)^{-1}(L+iH)] $$ where $D_1, D_2$ are diagonal with distinct positive entries slightly above $1$ so that $D_1^2+D^2_2=a^{-2}I$ with $a$ just a tiny bit less than $1/\sqrt 2$.
Since $(1-i)+i(1+i)=0$, we conclude that $P+iQ$ is triangular (zero entries for $i>j$) with the diagonal $a(D_1+iD_2)$, so the spectral condition is satisfied. On the other hand, $P+Q$ is norm close to the degenerate matrix $2a(I-\Phi_n^{-1}L)$.

This construction shows that in dimension $n$ the constant in the original problem should be at least of order $\log n$. I believe it is the right order of growth, but I'll leave it to the "people who know all that stuff well" to comment on that and to provide relevant references.

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