Problem involving Fourier-Stieltjes transform Suppose $\mu$ is a positive measure on $\mathbb{R}$ with compact support and satisfying $\|\mu\|=1$. Suppose that
$$(*)~~~~~~~\int_{\mathbb{R}} \big| u \widehat{\mu}(u)\big|\; du < \infty,$$
where $\widehat{\mu}$ denotes the Fourier-Stieltjes of $\mu$. The author of the paper that I'm reading then says :

(*), in turn, implies that, actually, $d\mu(t) = m(t) \; dt$ for some function $m \in C^1(\mathbb{R})$.

Why is that ?
Equation (*) means that $u\widehat{\mu} \in L^1(\mathbb{R})$. I know a result that goes like this :

If $f \in L^1(\mathbb{R})$ and $xf(x) \in L^1(\mathbb{R})$ then $f
> \in C^1(\mathbb{R})$.

Is it what is being used here ? If yes, how can we make sure that $\widehat{\mu}$ is integrable ?
 A: The measure $\mu$ can be reconstructed from its Poisson integral for $y > 0$:
\begin{align}
  M(x,y) & =\int_{-\infty}^{\infty} \frac{y}{(t-x)^2+y^2}d\mu(t) \\
   & = \frac{1}{2i} \int_{-\infty}^{\infty}\frac{1}{t-x-iy}-\frac{1}{t-x+iy}d\mu(t) \\
   & = \frac{1}{2}\int_{-\infty}^{\infty}\left(\int_{0}^{\infty}e^{-is(t-x-iy)}ds-\int_{-\infty}^{0}e^{-is(t-x+iy)}ds\right)d\mu(t) \\
   & = \frac{1}{2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-|s|y}e^{-is(t-x)}dsd\mu(t) \\
   & = \frac{1}{2}\int_{-\infty}^{\infty}e^{-|s|y}e^{isx}\int_{-\infty}^{\infty}e^{-ist}d\mu(t)ds \\
   & = \frac{\sqrt{2\pi}}{2}\int_{-\infty}^{\infty}e^{-|s|y}e^{isx}\hat{\mu}(s)ds.
\end{align}
This reconstruction is through the Herglotz inversion
\begin{align}
 \frac{1}{2}\{\mu[a,b]+\mu(a,b)\}
   & =\lim_{y\downarrow 0}\frac{1}{\pi}\int_{a}^{b}M(x,y)dx \\
   & =\lim_{y\downarrow 0}\frac{1}{\sqrt{2\pi}}\int_{a}^{b}\int_{-\infty}^{\infty}e^{-|s|y}e^{isx}\hat{\mu}(s)ds dx
\end{align}
Because of the assumption that $s\hat{\mu}(s)$ is in $L^1$, then $\hat{\mu}$ is in $L^1$ as well because $\hat{\mu}(s)$ is well-behaved near $s=0$ ($\hat{\mu}$ is continuous on $\mathbb{R}$ by bounded convergence.) Under this weaker assumption,
$$
    \frac{1}{2}\{\mu[a,b]+\mu(a,b)\}=\int_{a}^{b}\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{isx}\hat{\mu}(s)ds\right)dx.
$$
Because the right side is continuous in $a$ and $b$, then
$$
       \mu[a,b] = \mu(a,b) = \int_{a}^{b}\left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{isx}\hat{\mu}(s)ds\right)dx.
$$
Therefore, the measure $\mu$ is absolutely continuous with respect to Lebesgue measure, with a continuous density function
$$
            m(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{isx}\hat{\mu}(s)ds.
$$
This density function $m$ must be real and non-negative because $d\mu(x)=m(x)dx$ is a positive measure.
It appears to me that the above argument goes through without modification if $\mu$ is any finite positive Borel measure on $\mathbb{R}$ for which $\hat{\mu}(s) \in L^1(\mathbb{R})$. However, the final part of concluding that $m \in C^1(\mathbb{R})$ makes use of the stated condition:
$$
            \int_{-\infty}^{\infty}|u\hat{\mu}(u)|du < \infty.
$$
Then $m'$ exists, is continuous, and is given by
$$
             m'(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{isx}s\hat{\mu}(s)ds.
$$
