Fourier transform of smoothed measure converges uniformly globally (not just on compact sets)? Let $\mu$ be a probability measure on $\mathbb{R}$.  
Let $\psi_{n}(x) = n \psi(n x)$ where $n \geq 1$ and $\psi: \mathbb{R} \to \mathbb{R}$ is $C^{\infty}$, compactly supported, and $\int \psi(x) dx = 1$. 
Let $\mu_n = \psi_n \ast \mu$
I know that $$\widehat{\mu_{n}} \to \widehat{\mu} \quad \text{uniformly on compact sets},$$  where $\widehat{ \phantom{f} }$ denotes the Fourier transform. (See uniform convergence of characteristic functions or Weak convergence implies uniform convergence of characteristic functions on bounded sets.)
But I read somewhere that 
\begin{align}\label{1}\tag{1}
\widehat{\mu_{n}} \to \widehat{\mu} \quad \text{uniformly on $\mathbb{R}$}. 
\end{align}
And I haven't been able to prove it. 
Question: Is \eqref{1} true? Why or why not? 
Remark: It is clearly true if $\widehat{\mu}$ goes to zero at infinity.

Yes, it's true if $\hat\mu$ vanishes at infinity. So  try the first measure you can think of such that $\hat\mu$ does not vanish at infinty: If $\mu=\delta_0$ then $\hat\mu=1$, while $\hat\mu_n(\xi)=\hat\psi(\xi/n)$, which certainly does not tend to $1$ uniformly since $\hat\psi$ vanishes at infinity.
