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A standard result in probability theory asserts that a sequence of probablity measures $\mu_n$ on the Borel $\sigma$-algebra of $\bf R$ converges in law or weakly to a probability measure $\mu$ if and only if the characteristic functions $\widehat{\mu_n}$ converge pointwise to $\widehat{\mu}$. Convergence in law means that $\int f d\mu_n \rightarrow \int f d\mu$ for all $f$ bounded continuous.

When $f$ is for example $C^1$ with compact support, this follows from the formula $$ \int_{\bf R} f(x) d\mu(x) = {1\over 2\pi} \int_{\bf R} \hat{f}(t) \hat{\mu}(t) \, dt. $$ Unfortunately the integral on the right hand side is not well defined for all $f$ bounded continuous, so additional work is needed to get the full result.

Is there a well-known result in (Schwartz) distribution theory that can make sense of the formula for all probability measure $\mu$ and all $f$ bounded continuous so as to be able to get the convergence result directly? $$ \langle f , \mu \rangle = {1\over 2\pi} \langle \hat{f} , \hat{\mu} \rangle. $$ I guess that one can always build an adhoc pair of dual spaces for which this formula holds. Maybe convolution can help here since duality relations are sometimes particular cases of convolution relations.

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    $\begingroup$ If you're OK with $f \in C_b^\infty$ rather than $f \in C_b$, then there is a concept of integrable distributions: functionals on $C_b^\infty$, which should give what you are looking for. This class is discussed already in L. Schwartz's Théorie des distributions (Hermann, Paris, 1966). $\endgroup$
    – Mateusz Kwaśnicki
    Commented Feb 27, 2022 at 10:40
  • $\begingroup$ Your formulation of the theorem for convergence in law is missing a tightness/continuity-condition. The correct formulation can be found here: en.wikipedia.org/wiki/Lévy%27s_continuity_theorem $\endgroup$
    – Ben Deitmar
    Commented Feb 28, 2022 at 7:33
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    $\begingroup$ @Tardis My definition is correct. See here for the definition of the convergence in law (or convergence in distribution) en.wikipedia.org/wiki/… together with the portmanteau theorem. $\endgroup$
    – coudy
    Commented Feb 28, 2022 at 8:44
  • $\begingroup$ @coudy Yes, the definition is correct, but the convergence in distribution is only equivalent to the pointwise convergence of the characteristic functions when certain additional conditions are met. These conditions are not stated in the Wikipedia article about convergence in dirtibution (see en.wikipedia.org/wiki/…), but they are stated in the article about Levy's continuity theorem (see en.wikipedia.org/wiki/Lévy%27s_continuity_theorem). $\endgroup$
    – Ben Deitmar
    Commented Feb 28, 2022 at 10:15
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    $\begingroup$ @Tardis I am assuming here that the limit is the Fourier transform of a probability measure. Hence, no additional hypothesis is needed. $\endgroup$
    – coudy
    Commented Feb 28, 2022 at 12:55

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