If $\mu$ is a complex finite Borel measure on a separable real Hilbert space $H$ then $$x \mapsto \hat \mu (x) = \int \limits _H \Bbb e ^{\Bbb i \langle x, y \rangle } \Bbb d \mu _{(y)}$$ is continuous.

This slightly reminds me of showing that the convolution of a function in $L^p$ and another one from $L^{\frac {p+1} p}$ is continuous. In this latter case, the proof was done in steps, showing things for step functions, then for linear combinations of them and finally taking a limit, but I do not know whether this approach can be mimicked here.


An application of the Lebesgue dominated convergence theorem quickly proves the above. Question closed.

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    $\begingroup$ Show sequential continuity using the dominated convergence theorem. $\endgroup$ – PhoemueX Oct 2 '15 at 16:40
  • $\begingroup$ @PhoemueX: Pfff... disappointingly easy, I have no excuse for not having thought about it. $\endgroup$ – Alex M. Oct 2 '15 at 16:41

Since Hilbert spaces are first-countable, it will be enough to check continuity using sequences (as opposed to using nets). We shall base our proof on Lebesgue's dominated convergence theorem (for which using sequences and not nets is essential, as shown here and here).

Let $x_n \to x$ in $H$. Note that the function $y \mapsto \Bbb e ^{\Bbb i \langle x_n, y \rangle}$ converges pointwisely to the function $y \mapsto \Bbb e ^{\Bbb i \langle x, y \rangle}$ and that the absolute values of all these functions are precisely $1$. Since $\mu$ is a finite measure, the constant function $1$ is integrable, so Lebesgue's theorem applies and shows that $$\lim _n \hat \mu (x_n) = \lim _n \int \limits _H \Bbb e ^{\Bbb i \langle x_n, y \rangle} \Bbb d \mu = \int \limits _H \lim _n \Bbb e ^{\Bbb i \langle x_n, y \rangle} \Bbb d \mu = \int \limits _H \lim _n \Bbb e ^{\Bbb i \langle x, y \rangle} \Bbb d \mu = \hat \mu (x) ,$$ which means that $\hat \mu$ is continuous.

Note that We have used only the metric structure of $H$, so the above result stays valid for the larger class of Hausdorff (real or complex, not necessarily complete) F-spaces.


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