f\,d\nu_a=a\int f\,d\mu<\int f\,d\mu$, thus contradicting .

If $\mu$ is a probability and $\langle\eta,1\rangle=1$, we may divide each measure $\mu_t$ in the family by its total mass to get probability measures $\mu_t/\mu_t(P)$; the condition $\langle\eta,1\rangle=1$ ensures that this procedure does not break the differentiability of the normalized family at $t=0+$.

Let us recall that the *strong topology* on $\mathscr E'$ is the one induced by the seminorms $|\cdot|'_{K,m}$ defined in equation .

Recall that $\mathscr E'$ is a Montel space. A set $A$ in $\mathscr E'$ is, by definition, *bounded* if for each $f\in C^\infty(P)$ there is a constant $C_f>0$ such that $\langle\xi,f\rangle{\leqslant}C_f$ for all $\xi$ in $A$. Thus $X$ is bounded. Proposition 34.5 in [@treves] tells us that, in Montel spaces, the strong and weak\* topologies coincide in bounded sets, so this is the case in $X$. It is easy to see that, within $X$, the seminorm $|\cdot|'_{\overline U,n}$ dominates all other seminorms of the form $|\cdot|'_{K,k}$ with $K$ compact and $k>0$. Thus, within the set $X$ the seminorm $|\cdot|'_{\overline U,n}$ induces the weak\* topology. Its nondegeneracy is easy to check as well.

Examples and remarks {#sec:examples}
--------------------

\[ex:deltas\] We show that the derivative of a family of measures $(\mu_t)_t$ 