Let $\mu_n\to\mu$ weakly as prob. measures on compact $X$. If $B$ is Banach and $f:X\to B$ is continuous, does $\int fd\mu_n\to\int fd\mu$ in norm?

Question

Fix a compact metric space $X$, and let $B$ be a separable Banach space (a separable Hilbert space will also do, if that makes things simpler). Suppose $\mu_n$ and $\mu$ are measures on $X$ such that $\mu_n\to\mu$ weakly, i.e. for all continuous $f:X\to\mathbb R$, we have that $$\int_X\!f\,\mathrm d\mu_n\to\int_X\!f\,\mathrm d\mu.$$ Then suppose $f:X\to B$ is a continuous function. Clearly we have weak convergence $$\int_X\!f\,\mathrm d\mu_n\to\int_X\!f\,\mathrm d\mu$$ but can we improve this to get convergence in norm?

Answer 1

Yes and this follows immediately from Shorohod's Theorem https://en.wikipedia.org/wiki/Skorokhod%27s_representation_theorem and Bounded Convergence Theorem.

Answer 2

Because $X$ is compact and $f:X\rightarrow B$ is continuous, $f(B)$ is compact.

Fix $\delta>0$.

Then there exist $v_1,...,v_K\in B$ such that $f(X)\subset B_\delta(v_1)\cup...\cup B_\delta(v_K)$. Define: $$E_1:=f^{-1}(B_\delta(v_1)),....,E_K:=f^{-1}(B_\delta(v_K)).$$ Then $E_1,...,E_K$ is an open cover of $X$. Being $X$ a metric space, there exists a partition of unity $\varphi_1,...,\varphi_K$ of $X$ subordinated to $E_1,...,E_K$. Then: $$\|\int_Xf\operatorname{d}\mu_n-\int_Xf\operatorname{d}\mu\|_B\le\sum_{k=1}^K\|\int_X\varphi_kf\operatorname{d}\mu_n-\int_X\varphi_kf\operatorname{d}\mu\|_B\\\le\sum_{k=1}^K\|\int_X\varphi_k(f-v_k)\operatorname{d}\mu_n\|_B+\sum_{k=1}^K\|\int_X\varphi_kv_k\operatorname{d}(\mu_n-\mu)\|_B+\sum_{k=1}^K\|\int_X\varphi_k(v_k-f)\operatorname{d}\mu\|_B\\=\sum_{k=1}^K\|\int_{E_k}\varphi_k(f-v_k)\operatorname{d}\mu_n\|_B+\sum_{k=1}^K\|\int_X\varphi_kv_k\operatorname{d}(\mu_n-\mu)\|_B+\sum_{k=1}^K\|\int_{E_k}\varphi_k(v_k-f)\operatorname{d}\mu\|_B\\=\sum_{k=1}^K\|\int_{E_k}\varphi_k(f-v_k)\operatorname{d}\mu_n\|_B+\sum_{k=1}^K\|v_k\|_B\left|\int_X\varphi_k\operatorname{d}(\mu_n-\mu)\right|+\sum_{k=1}^K\|\int_{E_k}\varphi_k(v_k-f)\operatorname{d}\mu\|_B\le\delta\sum_{k=1}^K\int_{E_k}|\varphi_k|\operatorname{d}|\mu_n|+\sum_{k=1}^K\|v_k\|_B\left|\int_X\varphi_k\operatorname{d}(\mu_n-\mu)\right|+\delta\sum_{k=1}^K\int_{E_k}|\varphi_k|\operatorname{d}|\mu|\\=\delta\sum_{k=1}^K\int_{E_k}\varphi_k\operatorname{d}\mu_n+\sum_{k=1}^K\|v_k\|_B\left|\int_X\varphi_k\operatorname{d}(\mu_n-\mu)\right|+\delta\sum_{k=1}^K\int_{E_k}\varphi_k\operatorname{d}\mu\\=\delta\sum_{k=1}^K\int_{X}\varphi_k\operatorname{d}\mu_n+\sum_{k=1}^K\|v_k\|_B\left|\int_X\varphi_k\operatorname{d}(\mu_n-\mu)\right|+\delta\sum_{k=1}^K\int_{X}\varphi_k\operatorname{d}\mu\\=\delta\mu_n(X)+\sum_{k=1}^K\|v_k\|_B\left|\int_X\varphi_k\operatorname{d}(\mu_n-\mu)\right|+\delta\mu(X)\rightarrow 2\delta\mu(X), n\rightarrow\infty.$$ Then: $$\limsup_{n\rightarrow\infty}\|\int_Xf\operatorname{d}\mu_n-\int_Xf\operatorname{d}\mu\|_B\le2\delta\mu(X).$$ Taking the limit for $\delta\rightarrow 0^+$, you get: $$\limsup_{n\rightarrow\infty}\|\int_Xf\operatorname{d}\mu_n-\int_Xf\operatorname{d}\mu\|_B\le0,$$ and so the conclusion.