I am trying to nail down the relation between probability and functional analysis. In particular, how the notion of weak convergence used in probability theory is related to the weak-* convergence of functional analysis.

In probability, we have the notion of weak convergence: Let $X$ be a metric space and $\{\mu_{n}\}$ be a sequence of probability measures in $P(X)$. Then $\{\mu_{n}\}$ is said to converge weakly to a measure $\mu \in P(X)$ if $\int_{X}d\mu_{n}(x)f(x) \rightarrow \int_{X}d\mu(x)f(x)$ for every bounded continuous function $f:X \rightarrow \mathbb{R}$ (one gets an equivalent definition by restricting to bounded Lipschitz-continuous functions (from Wikipedia convergence of measures)

On the other hand, we have the functional analysis weak-* convergence: Let $Y$ be a normed vector space. Let $\{\ell_{n}\}$ be a sequence in the continuous dual $Y^{'}$. Then $\{\ell_n\}$ converges to $\ell \in Y^{'}$ if $\ell_n(y) \rightarrow \ell(y)$ for every $y\in Y$.

I think it will be a great help if I can find a correct statement of this "theorem", which starts with a complete metric space $X$ that is not necessarily compact. Is it the right way to state the result?


Let $X$ be a metric space. Let $C_{b}(X)$ be space of bounded continuous functions from $X$ to $\mathbb{R}$ with the sup-norm. Let $C_{b}(X)^{'}$ be the continuous dual of $C(X)$, endowed with the weak-* topology.

Define the map $\phi : P(X) \rightarrow C_{b}(X)^{'}$ by $$\phi(\mu)(f) = \int_{X}d\mu(x)f(x)$$

Then $\phi$ is well-defined, one-to-one, and the weak convergence of the measures $\mu_{n}$ to $\mu$ is equivalent to the weak-* convergence of the sequence $\phi(\mu_n)$ to $\phi(\mu)$.

  • 2
    $\begingroup$ You can find a nice discussion about this problem on the following PDF: math.ucsd.edu/~bdriver/240-01-02/Lecture_Notes/current_versions/… What you are looking for starts at p. 17. After understand how to decompose the Linear functional and use Riesz-Markov theorem give a look at Appendix in Functional Analysis Conway's book. $\endgroup$ – Leandro Jan 8 '16 at 5:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.