Is the Gaussian density Lipschitz continuous? More precisely, define $\phi(x) = \frac1{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$. 
Does there exists a constant $L$ such that $$|\phi(x)-\phi(y)|\le L|x-y|,$$ for all $x,y\in \mathbb{R}$. If yes, what is the minimal $L$?
 A: $$\forall x\in\mathbb{R},\phi'(x)=-\frac{x}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} $$
Thus $\lim\limits_{x\rightarrow\pm\infty}\phi'(x)=0$ and since $\phi'$ is continuous, it is bounded on $\mathbb{R}$. According to the mean value theorem,
$$ \forall(x,y)\in\mathbb{R}^2,|\phi(x)-\phi(y)|\leqslant\|\phi'\|_{\infty}|x-y| $$
On the other hand, if $L>0$ is the minimal value such that
$$ \forall (x,y)\in\mathbb{R}^2,|\phi(x)-\phi(y)|\leqslant L|x-y| $$
then because of what said above, $L\leqslant\|\phi'\|_{\infty}$. Moreover
$$ \forall x\neq y,\left|\frac{\phi(x)-\phi(y)}{x-y}\right|\leqslant L $$
Taking the limit $y\rightarrow x$ gives $|\phi'(x)|\leqslant L$ and thus $\|\phi'\|_{\infty}\leqslant L$. Finally $L=\|\phi'\|_{\infty}$ and $\|\phi'\|_{\infty}$ is the minimal value. We can also find the value of $\|\phi'\|_{\infty}$ :
$$ \forall x\in\mathbb{R},\phi''(x)=\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}(x^2-1) $$
Studying the sign of this expression gives the variations of $\phi'$ so that $\|\phi'\|_{\infty}=|\phi'(1)|=\frac{e^{-\frac{1}{2}}}{\sqrt{2\pi}}$.
A: In fact, one generalize the accepted answer to show that every 1D gaussian density is $1/4$-Lipschitz, irrespective of the mean and standard deviation.

Indeed, consider a 1D Gaussian density centered at $\mu \in \mathbb R$, with standard deviation $\sigma$, i.e consider the function $\phi_{\mu,\sigma} : \mathbb R \to \mathbb R$ defined by
$$
\phi_{\mu,\sigma}(x) := \frac{\exp(-(x-\mu)^2/(2\sigma^2))}{\sqrt{2\pi\sigma^2}}.
$$
One computes the derivative as
$$
 \phi'_{\mu,\sigma}(x) = -\frac{x-\mu}{\sigma^2}\phi_{\mu,\sigma}(x).
$$
Using arguments analogous to @Tuvasbien, it is easy to argue that $\phi_{\mu,\sigma} \in L^{\infty}(\mathbb R)$, and the Lipschitz constant of $\phi_{\mu,\sigma}$ is $\|\phi'_{\mu,\sigma}\|_\infty$.
Now, the derivative of $\phi_{\mu,\sigma}'$ is given by
$$
\phi_{\mu,\sigma}''(x)=\sigma^{-2}(1-\sigma^{-2}(x-\mu)^2) \phi_{\mu,\sigma}(x),
$$
which vanishes iff $(x-\mu)^2/\sigma^2 = 1$.  I leave it to the OP to argue that the maximum of $x \mapsto |\phi_{\mu,\sigma}'(x)|$ is attained when $x = 1\pm \sigma^2 \mu$, from which we get
$$
Lip(\phi_{\mu,\sigma}) = \|\phi'_{\mu,\sigma}\|_\infty = \frac{\exp(-1/(2\sigma^2))}{\sqrt{2\pi \sigma^2}} \le \frac{e^{-1/2}}{\sqrt{2\pi}} \le \frac{1}{4}.
$$
