A question about formal power series and Amice transform Let's call $L$ the continuous dual space of $C(\mathbb{Z_p},\mathbb{Q_p})=\{f:\mathbb{Z_p}\rightarrow\mathbb{Q_p}\mid f\text{ continuous}\}$, so $L$ consists on the continuous linear functions from $C(\mathbb{Z_p},\mathbb{Q_p})$ to $\mathbb{Q_p}$. We can define the Amice transform from $L$ to $\mathbb{Z}_p[[T]]\otimes \mathbb{Q}_p$ : recalling that $\int_{\mathbb{Z}_p}f(x)\mu(x):=\mu(f)$ for $\mu\in L$, we have $$\mu\longmapsto A_{\mu}(T)=\sum_{n=0}^\infty \left( \int_{\mathbb{Z}_p}\binom{x}{n}\mu(x) \right)T^n.$$
Until this, everything is ok. Then every text write without problems that $$\sum_{n=0}^\infty \left( \int_{\mathbb{Z}_p}\binom{x}{n}\mu(x) \right)T^n=\int_{\mathbb{Z}_p}(1+T)^x\mu(x). $$
I have some difficulties to understand this. I know that what they do is to write $$(1+T)^x=\sum_{n=0}^\infty \binom{x}{n}T^n,$$ but then they bring the variable $T$ inside $\mu$ as a constant, and I have problems to give a formal meaning on this. How have I to treat $T$? 
 A: Let $C(Z_p,Q_p)$ be the space of continuous functions $x\in Z_p \mapsto f(x) \in Q_p$ and $\mu \in C(C(Z_p,Q_p),Q_p)$. 
Since ${m \choose n}$ is an integer and a polynomial in $m$ then ${x\choose n} \in C(Z_p,Z_p)$ and $\mu({x\choose n}) \in Q_p$.

Assume $\lim_{n \to \infty} \mu({x\choose n})= 0$. 

For some $k_\mu, \mu({x\choose n}) \in p^{-k_\mu} Z_p$.
For $t \in p Z_p$, the series $\sum_{n=0}^\infty {x \choose n} t^n$ converges in $ C(Z_p,Z_p)$, to $(1+t)^x \in C(Z_p,Z_p)$.
Let the formal series $$A(\mu)(T) = \sum_{n=0}^\infty \mu({x \choose n}) T^n \in p^{-k_\mu} Z_p[[T]] \subset Z_p[[T]] \otimes_{Z_p} Q_p$$
The defining series for $A(\mu)(t)$ converges as a function of $t\in Z_p$ and for $t \in p Z_p$, $A(\mu)(t)=\mu((1+t)^x)$. 
The point of this is that the polynomials ${x \choose n}$ are dense in $C(Z_p,Q_p)$, for $b_n \to 0,f = \sum_{n=0}^\infty b_n {x \choose n} \in C(Z_p,Q_p)$ and $\mu(f) = \sum_{n=0}^\infty b_n \mu({x \choose n})$.
If you want to avoid the assumption $\lim_{n \to \infty} \mu({x\choose n})= 0$ I think you should tell which particular $\mu$ you want to look at.
