order to simplify the notation, let $h=d^kf/dx^k$. We take the derivative and we get $$\begin{aligned}
  &(\operatorname{sgn}t)|t|^\frac{q-1}{k+1}\left(\tfrac{q+k}{k+1}  \int h(x)\psi(|t|^\frac{q-1}{k+1}x)dx +\tfrac{q-1}{k+1}|t|^\frac{q-1}{k+1}\int xh(x)\psi'(t^\frac{q-1}{k+1}x)dx\right)\\
  &=(\operatorname{sgn}t)|t|^\frac{q-1}{k+1}\left(\tfrac{q+k}{k+1}  \int h(x)\psi(|t|^\frac{q-1}{k+1}x)dx-\tfrac{q-1}{k+1}\int (xh(x))'\psi(|t|^\frac{q-1}{k+1}x)dx\right)\\
  &=(\operatorname{sgn}t)|t|^\frac{q-1}{k+1}\left(\int h(x)\psi(|t|^\frac{q-1}{k+1}x)dx-\tfrac{q-1}{k+1}\int xh'(x)\psi(|t|^\frac{q-1}{k+1}x)dx\right).
 \end{aligned}$$ This is easily seen to tend to $h(0)-\frac{q-1}{k+1}0\,h'(0)=h(0)=\langle\eta,f\rangle$ as $t\to0$, which is what we wanted.

\[rmk:support\] The conditions in Corollary \[cor:two-sided\] force the support of $\eta$ to be contained inside the support of the measure $\mu$. However, take a look at the following example.

\[ex:singcont\] For the conditions of Theorem \[thm:families\] or Corollary \[cor:two-sided\] to be satisfied, it is *not* true that the support of the distribution $\eta$ must be contained inside the set of points that hold the measure of $\mu$. (Note that the support of $\mu$ is the closure of this set.) Immitating the construction of the Cantor denisty, it is easy to construct a singular-continuous measure supported on the interval $[0,1]$ whose measure will be contained in a set of Lebesgue measure zero; for instance, one can recursively subdivide the interval in halves and assign 1/3 of the measure to the rationals in the lower half and 2/3 to the rationals in the upper half, and iterating for the halves of each of these intervals, while declaring that the 