Showing that the Fourier transform of a finite Borel measure $\mu$ with compact support is Lipschitz continuous function in $\mathbb{R}^n$ Let $\mu$ be a finite Borel measure on $\mathbb{R}^n$ with compact support, that is
$$\mathrm{supp}(\mu) = \{x\in X\mid \forall N_x\in \tau:(x\in N_x\implies \mu(N_x) > 0)\}$$
Here is a related article on the support.
Define the Fourier transform of $\mu$ as $\hat{\mu} = \int_{\mathbb{R}^n}e^{-2\pi i \xi \cdot x}d\mu x, \xi \in \mathbb{R}^n$. It is clear to me that $||\hat{\mu}||_\infty \leq \mu(\mathbb{R}^n)$, but I am having trouble proving that $\hat{\mu}$ is Lipschitz continuous. Namely, suppose that $\mathrm{supp}(\mu)\in B(0, R)$. My textbook says that then
$$|\hat{\mu}(x) - \hat{\mu}(y)| \leq R\mu(\mathbb{R}^n)|x - y|$$
for $x,y\in\mathbb{R}^n$. What is not clear to me is that how you can massage $|x - y|$ out of $e^{-2\pi i x\cdot \xi} - e^{-2\pi i y\cdot \xi}$ for a fixed $\xi\in\mathbb{R}^n$. In single variable case one way to obtain a Lipshitz upperbound would be to use triangle inequality and apply the mean value theorem for the cosine and sine term. But I don't think we can apply the same strategy directly in our case, for approximation in one of the iterated integrals by e.g.
$$|\cos(2\pi\left(x_k\xi_k + \text{rest}\right)) - \cos(2\pi\left(y_k\xi_k + \text{rest}\right))| \leq |\xi_k||x_k - y_k|$$
will not produce the desired inequality.

$|e^{ia}-e^{ib}|\leq |a-b|$ for all real numbers $a$ and $b$.
$|(-2\pi x.\xi_1)-(-2\pi x.\xi_2)|\leq 2\pi |x.\xi_1-x.\xi_2|\leq 2\pi \|x\|\|\xi_1-\xi_2\|$ and there is a bound fo $\|x\|$ on the support.
