# 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 $$\left\| \int f \,d\mu\right\| = Tr\left| \int f \,d\mu\right| \overset{!}{\leq} Tr\int |f| \, d\mu = \int Tr |f|\,d\mu = \int \|f\|\,d\mu$$

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.
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.