If $x_n \to x$ in $L^1(X,H)$ then $\langle x_n, h \rangle \to \langle x, h \rangle$ in $L^1(X,\mathbb{R})$

Question

Suppose $H$ is a Hilbert space. Is it true that if $x_n \to x$ in $L^1(X,H)$ then $\langle x_n, h \rangle \to \langle x, h \rangle$ in $L^1(X,\mathbb{R})$ for any fixed $h\in H$?

Certainly if $x_n\to x$ in $L^2(X,H)$ then I can appeal to Cauchy-Schwarz but this textbook I am reading claims it for $L^p$ for any $p\geq 1$.

Answer 1

It's ok. Letting $\xi\in X$ denote the integration dummy variable we have $$\left\lvert \int_X \langle x_n(\xi)-x(\xi), h\rangle\, d\xi\right\rvert\le \lVert h\rVert_H\int_X\lVert x_n(\xi)-x(\xi)\rVert_H \, d\xi = \lVert x_n-x\rVert_{L^1(X;H)}\lVert h\rVert_H\to 0.$$