Given any Borel measure $\mu$ on $\mathbb{R}$, define a map that sends any $f\in C\_c(\mathbb{R})$ to $$T\_\mu(f)(y)=\int \langle\exp(-i x \lambda),f(x)\rangle\exp(iy\lambda)d\mu(\lambda).$$
Here $\langle\cdot,\cdot\rangle$ denotes the inner product of functions with respect to the Lebesgue measure.


Is (a multiple of) Lebesgue measure the only Borel measure such that 


1. $T\_\mu$ is defined for all $f\in C\_c(\mathbb{R})$.
2. $T\_\mu(f)$ has exponential decay for all $f\in C\_c(\mathbb{R})$.


Any reference is welcome.



Restrictions on spectral measure
No. This becomes clear if you use more suggestive notation: you are sending $f$ to the (inverse) Fourier transform of $\widehat{f}\mu$. Exponential decay can be obtained from (a variant of) the Paley-Wiener Theorem: if $\widehat{g}$ is holomorphic on a strip $|\textrm{Im}\, z|<c$ and
$$
\sup\_{|y|<c}\int\_{-\infty}^{\infty}|\widehat{g}(x+iy)|\, dx <\infty ,
$$
then $g$ has exponential decay.


I want to apply this to $\widehat{g}=\widehat{f}h$, and then we'll take $d\mu=h\, dt$.
Notice that $\widehat{f}$ doesn't make any trouble because $f$ is compactly supported, and thus $|\widehat{f}(x+iy)|\lesssim e^{Ly}$ is uniformly bounded on the whole strip. So we can take something like
$h(t) = 1/(t^2+1)$.


There are of course many other examples, but they all have to be of this type because it is also true that, conversely, the Fourier transform of an exponentially decaying function is holomorphic on a strip. Thus we must take $d\mu=h\, dt$ absolutely continuous with a holomorphic density $h$.


