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Banach spaces where all weakly convergent sequences are norm convergent are said to have the Schur property. A classical Theorem of Schur says that $\ell^1(I)$ has the Schur property for every set $I$ …

I think that you can construct a counterexample if $X$ is the space of measurable functions on a nice probability space (like the unit interval with the Lebesgue measure) endowed with stochastic conve …

It seems that this can be deduced from the Portmanteau theorem: Assume the convergence of the intergals $\int fd\mu_n$ for all $f\in C_b$ vanishing in a neighbourhood of $0$ and fix $E\in C_\mu$ with …

I am not completely satisfied by the accepted answer because the functor which characterizes the convergence of a filter depends on the limit. I therefore add another quite simple answer (written for …

Due to periodicity in $x$ you can write $f$ as a function on $(0,\infty)\times S^1$ so that your assumption in 1. becomes just uniform convergence on $S^1$ and this implies $f(t/\varepsilon,\cdot)\to …

A trivial counterexample is $\Omega=\mathbb R$, $\mu_n=\delta_{1/n}, \mu=\delta_0$, and $F=\{0\}$. A standard result is $\mu_n(F)\to \mu(F)$ if $\mu(\partial F)=0$. This looks somehow opposite to what …

The result (even for LF-spaces) is due to J. Dieudonné and L. Schwartz La dualité dans les espaces (F) et (LF), Annales de l’institut Fourier, tome 1 (1949), p. 61-101, propositions 2 and 4. (Proposi …

Disclaimer: The question has been edited several times so that this may not apply to actual version. This seems obvious (after the edit, you consider only one Dirichlet form, right?): Take a subsequen …