Connection between the Fourier transform of a measure and the support of the measure 
Let $H$ be a separable real Hilbert space, $\mu$ a complex finite Borel measure on it with Fourier transform $\hat \mu$, and $V$ a finite dimensional subspace. Let $P : H \to V$ be the orthogonal projector onto $V$. Show that if $\hat \mu = \hat \mu \circ P$ (composition of functions), then $\text{supp} \ \mu \subseteq V$.

The trouble with this one is that I really do not know where to start from, even though intuitively I can understand what happens here. I tried to show that if $U \subseteq H \setminus V$ is an open subset, then $\int \limits _H 1_U \ \Bbb d \mu = 0$, but I failed to connect this with the $\Bbb e ^{\Bbb i \langle x, y \rangle }$ in $\hat \mu$. How to use it?
 A: Suppose for now $\mu$ real.
Let $|\mu|^{\perp}$ the pushforward of $|\mu|$ with respect to the projection $P^{\perp}:H\rightarrow V^{\perp}$. Using the hypothesis and a change of variables, we get that for every $y\in V^{\perp}$,
$$\widehat{|\mu|^{\perp}}(y) =
\int \limits_{V^{\perp}}e^{i(x,y)}d|\mu|^{\perp}(x) =
\int \limits_He^{i(P^\perp x,y)}d|\mu|(x) =
\int \limits_He^{i(x,P^\perp y)}d|\mu|(x) =
\widehat {|\mu|} (P^\perp y) = \\
\widehat {|\mu|} \circ P (P^\perp y) =
\widehat {|\mu|} (0) =
\int \limits_He^{i(x,0)}d|\mu|(x) =
\int \limits_He^{i(P^\perp x,0)}d|\mu|(x) =
\int \limits_{V^{\perp}}e^{i(x,0)}d|\mu|^{\perp}(x) =
\widehat {|\mu|^\perp} (0) ,$$
so $\widehat {|\mu|^\perp}$ is a constant function (equal to $\widehat {|\mu|} (0) = |\mu| (H)$). But the measure $|\mu|(H)\delta_0$ has exactly the same Fourier transform on $V^\perp$, so necessarily $|\mu|^{\perp}=|\mu|(H)\delta_0$ on $V^{\perp}$.
Now, given $U\subset H\backslash V$, we  have $U\subset  (P^{\perp})^{-1}P^{\perp}U$. Since $0\not\in P^{\perp}U$, $|\mu|$ is monotonic and $\mu\leq |\mu|$, we have
$$\mu (U) \le |\mu|(U) \le |\mu|((P^{\perp})^{-1}P^{\perp}U)=|\mu|^{\perp}(P^{\perp}U)=0 ,$$
so $\mu(U)=0$.
Now, if $\mu$ is complex, one can separate its real and imaginary part and apply
the previous case to both.
