# Comparison of the absolute value of an operator with its positive parts

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.

• By "universal", you mean C should not depend on P and Q, right? Nov 3, 2017 at 22:31
• @YemonChoi That's right. Nov 3, 2017 at 23:04
• 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. Nov 8, 2017 at 1:29
• @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. Nov 8, 2017 at 1:38
• 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) Nov 8, 2017 at 1:55

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.