Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 27 / 2 / iii / Solution

Take an arbitrary $f\in C({\mathbb{Z}}_p,{\mathbb{Q}}_p)$. Compactness of ${\mathbb{Z}}_p$ bounds its field absolute value, so some $p^{m}f$ takes values in ${\mathbb{Z}}_p$. Apply part (ii) to obtain a continuous $h$ with $h(x+1)-h(x)=p^{m}f(x)$. Then $g=p^{-m}h$ belongs to the same continuous-function space and satisfies $f=g(\cdot +1)-g$.

Translation invariance by one alone now gives $L(f)=L(g(\cdot +1))-L(g)=0$. Since $f$ was arbitrary,This proves translation-invariant linear forms on p-adic continuous functions vanish without assuming that the linear form itself is continuous. A proof using only density of polynomials would not suffice for an arbitrary, potentially discontinuous linear form; surjectivity of the forward difference operator on the entire function space is what is needed.