# Bochner Integral of trace class operator-valued function

Suppose I have a separable Hilbert space $\mathcal{H}$, a measure space $(M,\mu)$ and a strongly measurable operator-valued function $f:M \rightarrow \mathcal{S_{1}}(\mathcal{H})$ whose values are trace class operators on this Hilbert space. Suppose further that its Bochner integral $\int_{M}f(s)d\mu(s)$ exists.

1) Is it also a trace class operator, maybe with additional data?

2) Does something like $\|\int_{M}f(s)d\mu(s)\|_{\mathcal{S_{1}}}\leq \int_{M}\|f(s)\|_{\mathcal{S_{1}}}d\mu(s)$ (where $\| \cdot \|_{\mathcal{S_{1}}}$ is the trace class norm) hold, maybe with additional data?

3) If these questions both have a positive answer, do their appropriate modifications also have a positive answer for other $\mathcal{S_{p}}$ (Schatten p-class) operator-valued functions?

I know the appropriate modification of the first question has a positive answer for compact operator-valued functions and various papers point out to both the first two questions having a positive answer (e.g. page 13 of https://arxiv.org/pdf/1402.0763.pdf). I also suspect the third should be an easy generalisation.

I think this should be easy but can't seem to approach it well. I would be very grateful for every hint.

The properties listed on the wiki page for Bochner integration are useful here.

The answer to 1 is yes. In particular: $Tr$ is a continuous linear operator with respect to the trace-norm, which means that for any trace class operator $f$, we have $$Tr\left[\int f \,d\mu\right] = \int Tr(f)\,d\mu$$ The tricky bit here is to say that $$\left| \int f \,d\mu\right| \overset{!}{\leq} \int |f| \, d\mu$$ Which is to say that $\int |f| \, d\mu - \left| \int f \,d\mu\right|$ is a positive operator. In particular, we need to show that the map $f \mapsto f^*f$ is convex, then apply Jensen's inequality. With that, we would have $$\left\| \int f \,d\mu\right\| = Tr\left| \int f \,d\mu\right| \overset{!}{\leq} Tr\int |f| \, d\mu = \int Tr |f|\,d\mu = \int \|f\|\,d\mu$$

The answer to 2 is yes; no additional data required.

The answer to 3 is also yes.

With regards to the inequality: note that $f \leq g$ if and only if for every $x \in \mathcal H$, we have $(x,fx) \leq (x,gx)$. Now, note by linearity that $$\left( x,\left[\int f \,du\right]x \right) = \int (x,f(x))\,d\mu$$ Thus: for convexity, we note that for any fixed $x$, operators $f_1,f_2$, and $t \in [0,1]$ we have $$(x,[(1-t)f_1 + tf_2]^*[(1-t)f_1 + tf_2] x) = \\ \|[(1-t)f_1 + tf_2]x\|^2 = \\ \|(1-t)f_1x + tf_2x\|^2 \leq\\ ((1-t)\|f_1x\| + t\|f_2x\|)^2 \leq\\ (1-t)\|f_1x\|^2 + t\|f_2x\|^2 = \\ (1-t)(x,f_1^*f_1 x) + t(x,f_2^*f_2 x)$$ since this holds for all $x$, we have $$[(1-t)f_1 + tf_2]^*[(1-t)f_1 + tf_2] \leq (1-t)f_1^*f_1 + t f_2^*f_2$$ as desired.

• Thank you for your answer! Unfortunately, I do not understand some parts of the argument. We need continuity (so, boundedness with respect to the sup norm) in order to exchange the trace and the integral, right? So boundedness w.r.t. the trace norm is not enough. Also, how is that map convex, and can we apply Jensen's for the Bochner integral? – folouer of kaklas Dec 28 '16 at 20:05
• But I do understand that this argument would then prove 2), which would prove 1) if the integral on the right hand side of the inequality is finite. – folouer of kaklas Dec 28 '16 at 20:11
• Continuity in this case means bounded with respect to the trace norm, since we're looking at the trace class operators. – Omnomnomnom Dec 28 '16 at 21:18
• But isn't the integral defined using the sup norm? – folouer of kaklas Dec 28 '16 at 21:25
• Not necessarily; the Bochner integral can be (and usually is) defined on any Banach space with respect to the associated norm. The trace class operators form a Banach space under the trace norm. – Omnomnomnom Dec 28 '16 at 21:34

I realised that this has a very simple answer and I will share it here for future reference, hoping it is correct. By the Bochner Integrability theorem, a strongly measurable function $f:M→\mathcal{S}_{p}(\mathcal{H})$ that satisfies $\int_{M}\|f(s)\|_{\mathcal{S_{p}}(\mathcal{H})}d\mu(s) < \infty$ converges by definition in the appropriate norm, and furthermore satisfies 2) above. So if this absolute integrability condition is satisfied, the integral is in $\mathcal{S_{p}}(\mathcal{H})$. Fortunately, for the concrete case I want to use this, this holds, so it solves my question.