Quadratic velocity tensors are Bochner integrable

By [A1] and [F4], scalar $L^2(D,\lambda )$ and $L^1(D,\lambda )$ are complete. The $H$ norm is the finite ${\mathrm{\ell }}^2$ product of the component $L^2$ norms. For a matrix field $A$, $\mathrm{\parallel }A{\mathrm{\parallel }}_K={\int }_D\mathrm{\mid }A{\mathrm{\mid }}_{F}d\lambda$, and $\mathrm{\parallel }A{\mathrm{\parallel }}_K\le {\sum }_{i,j}\mathrm{\parallel }{A}_{ij}{\mathrm{\parallel }}_1\le d\mathrm{\parallel }A{\mathrm{\parallel }}_K\mathrm{.}$ Thus a Cauchy sequence in either $H$ or $K$ is Cauchy in each scalar component; the component limits exist by [F4], and the displayed norm formulas show convergence to their assembled vector or matrix field. Hence $H$ and $K$ are Banach spaces under [A1].

For representatives $v,w$, the Frobenius norm satisfies $\mathrm{\mid }v(x)\otimes w(x){\mathrm{\mid }}_F=\mathrm{\mid }v(x)\mathrm{\mid }\mathrm{\mid }w(x)\mathrm{\mid }$. Holder's inequality [F3] therefore gives $\mathrm{\parallel }v\otimes w{\mathrm{\parallel }}_K={\int }_D\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }d\lambda \le \mathrm{\parallel }v{\mathrm{\parallel }}_H\mathrm{\parallel }w{\mathrm{\parallel }}_H<\mathrm{\infty }$. Changing either representative on a null set changes the product only on a null set [F1], so the tensor is well-defined as a $K$-class. The same estimate applied to differences proves continuity of the bilinear tensor map.

By [F2], choose simple $H$-valued maps $U_n$ with $U_n(\omega )\to U(\omega )$ in $H$ outside a measurable null set. Each $U_n\otimes U_n$ is a finite-valued measurable $K$-valued simple map. Step 1.2 gives $\mathrm{\parallel }{U}_n\otimes U_n-U\otimes U{\mathrm{\parallel }}_K\le \mathrm{\parallel }{U}_n-U{\mathrm{\parallel }}_H(\mathrm{\parallel }{U}_n{\mathrm{\parallel }}_H+\mathrm{\parallel }U{\mathrm{\parallel }}_H)⟶0$ outside that null set, since norm convergence makes $(\mathrm{\parallel }{U}_n{\mathrm{\parallel }}_H)$ bounded there. Hence $U\otimes U$ is strongly measurable.

Holder's inequality on the probability space gives $\mathbb{E}\mathrm{\parallel }U{\mathrm{\parallel }}_H\le (\mathbb{E}\mathrm{\parallel }U{\mathrm{\parallel }}_H^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$. Together with the given strong measurability, [F5] makes $U$ Bochner integrable. The identity $\mathrm{\mid }v\otimes v{\mathrm{\mid }}_F=\mathrm{\mid }v{\mathrm{\mid }}^2$ gives $\mathrm{\parallel }U(\omega )\otimes U(\omega ){\mathrm{\parallel }}_K=\mathrm{\parallel }U(\omega ){\mathrm{\parallel }}_H^2$ for almost every $\omega$, so its norm is integrable as well; [F5] and step 2.1 make $U\otimes U$ Bochner integrable and yield the displayed norm identity.