When does analytic in the operator norm imply analytic in the trace class norm?

This is a crosspost from MSE. It's been up there for a few weeks now. A 200 rep bounty yielded no results (or even comments). I'm hoping someone here has some helpful ideas. See this post for the original.

Consider $U$ a nice compact region in $\mathbb{C}$ with boundary $\Gamma$. Let $S_1$ b the ideal of trace class operators on a separable complex Hilbert space $H$. We will let $\|\cdot \|$ be the operator norm and $\|\cdot \|_1$ be the trace norm. Suppose $W:U\to S_1$ is complex analytic in the operator norm.

Under what conditions is $W(\lambda)$ analytic in the $\| \cdot \|_1$ norm?

I have proved that the following are equivalent when $W(\lambda)$ is operator analytic:

1. $W(\lambda)$ is continuous in the operator norm
2. for a fixed $M$, we have $\|W(\lambda)\|_1 <M$ for each $\lambda \in \Gamma$
3. tr $W(\lambda)B$ is analytic for each bounded operator $B$.
4. $W(\lambda)$ is analytic in the $\|\cdot\|_1$ norm.

I can provide some ideas for these proofs if that's be helpful.

This leads us to

Question 1: What if we know that tr $W(\lambda)$ is analytic?

Is there a nice way to compare tr $W(\lambda)$ with tr $W(\lambda)B$? I would love an inequality like $$|\text{tr }AB| \leq\|B\||\text{tr }A|$$ for $A \in S_1$ and $B$ bounded. Although it would probably be greedy to expect this in general.

Also, I'm willing to impose even stronger assumptions on $W$ if necessary. One very strong constraint is to assume that $W$ has a rank bound along $\Gamma$, I.E. for a fixed $N$ we have rank $W(\lambda)<N$ for each $\lambda \in \Gamma$. This actually guarantees analyticity as $$\|W(\lambda)\|_1 \leq N\sup_{\lambda \in \Gamma} \|W(\lambda)\|$$

Attempting to weaken this condition, we arrive at

Question 2: What happens if $W(\lambda)$ is finite rank for each $\lambda$, but has no rank bound?

I suspect that finite rank and analytic actually implies rank bounded, but I do not know.

Edit 1 Here's a fun idea to that might help prove finite rank implies rank bounded. The set $$S_n = \{\lambda : W(\lambda) \text{ has rank at most}n\}$$

is closed (continuity of $W$ tells us singular values are continuous). So Baire Category theorem tells us that some $S_n$ is dense somewhere. So in some open set, $W$ is rank bounded. So can I use an analytic extension in the trace norm to do something? This looks like the proofs of the open mapping theorem and whatnot...

Edit 2 Here's another fact that may be helpful. Consider a sequence of complex analytic functions $f_n:U\to \mathbb{C}$. Suppose they converge pointwise to an analytic function $f$. Then $f$ is analytic on an open dense neighborhood of $U$. This is potentially helpful because for any orthonormal basis $\phi_i$, $$\text{tr} W(\lambda) B = \sum_{i=1}^\infty \langle W(\lambda)B\phi_i,\phi_i\rangle$$ And because $W(\lambda)B$ is analytic in the operator norm, so will the each inner product be analytic.

I am fairly familiar with Gohberg's work on trace class operators. Unfortunately, despite all of the great theorems on bounds for singular values, knowing that tr $W(\lambda)$ is analytic gives no information about the singular values.

• I'm confused by your opening remarks: if you've proven that continuity of $W(\lambda)$ in operator norm implies holomorphy in trace norm, then since that continuity is implied by analyticity in the operator norm, you'd be done... ? Is there a typo? Am I misunderstanding something? – paul garrett Jun 22 '15 at 22:24
• No, it's not true that $\|\cdot \|_1$ continuity implies operator analyticity. What we know is that if $W$ is operator analytic and $\|\cdot \|_1$ continuous, then it's $\|\cdot \|_1$ analytic. What I tried to say in the opening remarks is that if $W(\lambda)$ is operator analytic and any other condition on that list, then it's all of them. – Zach Stone Jun 22 '15 at 22:31
• I'm sorry, I didn't mean to suggest that trace-norm continuity implies operator analyticity. Maybe I misunderstood what hypothesis you're taking and hoping to prove trace-norm analyticity... – paul garrett Jun 22 '15 at 23:08
• The inequality $\textrm{tr}\,AB\le \|B\|\textrm{tr}\,A$ is hopeless, as you already suspected ($\textrm{tr}\, A=0$). I'm not sure I completely understand the rest of your question, but there is a general principle that all reasonable definitions of analyticity for Banach space valued functions are equivalent, I wonder if this is relevant. – Christian Remling Jun 23 '15 at 0:28
• Extending a bit Christian Remling's comment: Grothendieck proved that for a complete locally convex space $X$ a function $f:U\to X$ is holomorphic if $\varphi\circ f:U\to \mathbb C$ is holomorphic for all continuous linear functionals $\varphi$ on $X$. As far as I remember, one does not even need all $\varphi$. – Jochen Wengenroth Jun 23 '15 at 15:55