# Continuity of the Fourier transform of a measure

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.

Edit:

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

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

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.