of the first part of either Theorem \[thm:families\] or Corollary \[cor:two-sided\], and any distribution $\eta$ additionally satisfying $\langle\eta,1\rangle=0$ will be good for the second part of the statements of therein contained.

\[ex:explicit\] We work out Example \[ex:uniform\] explicitly in the special case of $P={\mathbb{R}}$, $U=(-\frac12,\frac12)$, $\mu=\chi_{U}dx$ the uniform distribution on $U$, $k\in{\mathbb{N}}$, $\eta=(-1)^k\partial^k\delta_0$. Let $\psi\colon{\mathbb{R}}\to[0,1]$ be a compactly-supported, $C^\infty$ function that is equal to 1 in a neighborhood of 0, satisfies $\psi(x)=\psi(-x)$, $\int_{\mathbb{R}}\psi(x)dx=1$, and is decreasing on $[0,+\infty)$. Pick some $0<q<1$. For $t\in{\mathbb{R}}$ with $|t|$ small enough, we let $\mu_t$ be the measure that corresponds to the density $\rho(x,t)\,dx$ with $$\rho(x,t)=\chi_U(x)+(-1)^k(\operatorname{sgn}t)|t|^q\frac{d^k\psi}{dx^k}\left(|t|^{\frac{q-1}{k+1}}x\right), \quad t\neq 0,$$ and $\rho(x,0)=\chi_U(x)$. We need $|t|$ to be small enough that $\rho(x,t)$ will be nonnegative for all $x$; this defines $\mu_t$ only for small $|t|$, yet its smooth extension to all of $t\in{\mathbb{R}}\setminus\{0\}$ exists. We remark that at $t=0$ and $x=0$, the function $\rho$ is not differentiable with respect to $t$ because the slope is vertical.

Let us check that this family $(\mu_t)_{t\in{\mathbb{R}}}$ works, that is, that its derivative is indeed equal to $\eta$. Let $f\in C^\infty({\mathbb{R}})$; then we have $$\begin{aligned}
 \frac{d}{dt}\int f\,d\mu_t&=\frac{d}{dt}\int_{\mathbb{R}}f(x)  \rho(x,t)dx\\
  &=\frac{d}{dt}\int f(x)\left(\chi_U(x)+(-1)^k\operatorname{sgn}t|t|^q\frac{d^k\psi}{dx^k}(|t|^{\frac{q-1}{k+1}}x)\right)dx \\
  &=\frac{d}{dt} \operatorname{sgn}t|t|^\frac{q+k}{k+1}\int \frac{d^kf}{dx^k}(x)\psi(|t|^\frac{q-1}{k+1}x)dx
 \end{aligned}$$ after $k$ integrations by parts. In 