Tangent vectors on the algebra of trigonometric polynomials Let $G$ be a compact real Lie group and ${\sf Trig}(G)$ the algebra of trigonometric polynomials on $G$ (defined in the Hewitt-Ross, Abstract harmonic analysis, (27.7)), i.e. the algebra of functions $u:G\to {\mathbb C}$ which can be represented as complex linear combinations of functions of the form
$$
x\in G\mapsto \langle U(x)\xi,\eta\rangle\in{\mathbb C},
$$
where $U:G\to B(H)$ is an arbitrary irreducible unitary continuous representation of $G$ (in an arbitrary Hilbert space $H$), and $\xi,\eta\in H$.
Let us call a tangent vector on ${\sf Trig}(G)$ in a point $a\in G$ an arbitrary linear functional $f:{\sf Trig}(G)\to{\mathbb C}$ such that:
1) $f$ preserves involution:
$$
f(\overline{u})=\overline{f(u)},\qquad u\in {\sf Trig}(G),
$$  
2) $f$ satisfies the Leibniz identity in the point $a$:
$$
f(u\cdot v)=u(a)\cdot f(v)+f(u)\cdot v(a), \qquad u,v\in {\sf Trig}(G).
$$
A question: is it true that every such functional $f:{\sf Trig}(G)\to{\mathbb C}$ is a "true tangent vector" on $G$, i.e. it
can be represented in the form 
$$
f(u)=\lim_{t\to 0}\frac{u(\gamma(t))-u(\gamma(0))}{t},\qquad u\in {\sf Trig}(G),
$$
for some smooth curve $\gamma:{\mathbb R}\to G$ with $\gamma(0)=a$?
An equivalent statement: is it true that every such functional $f:{\sf Trig}(G)\to{\mathbb C}$ can be extended to a functional $h:C^\infty(G)\to{\mathbb C}$ with the same properties?
 A: Yes, of course. The functions you have are representation functions on $G$ (the representations are finite dimensional as well); they are thus algebraic functions on the complexification of $G$. It is well known that the "algebraic Lie algebra is the true tangent space for algebraic groups.  
A: There is some difficulty.
Added in edit: Only if $G$ is not compact, and infinite dimensional $H$ are also used. 


*

*$G\ni x\mapsto u(x)=\langle U(x)\xi,\eta\rangle$ is in $C^\infty(G)$ if and only if one of $\xi,\eta\in H$ is a smooth vector. See section 5 of here for a complete and transparent proof. 

*If you consider only smooth trigonometric functions, then this is true, since smooth trigonometric functions are dense in $C^\infty(G)$, and the usual proof on manifolds goes through if you replace bump functions by smooth trigonometric functions which approximate bump functions. 
The Taylor expansion to first order with remainder should go through in the algebra of smooth trigonometric functions.  
Is this enough information, or should I try to write down a proof?
