I'm trying to understand a proof of the following simple statement:
Let $A:C^\infty(\mathbb{S}^1,\mathbb{C})\to C^\infty(\mathbb{S}^1,\mathbb{C})$ be a continuous linear operator on the Frechet space $C^\infty(\mathbb{S}^1,\mathbb{C})$. If $A$ commutes with all rotations $R_s$, then $A$ commutes with the derivative operator $D:=d/dx$.
First note that the rotations $R_s$ act on $C^\infty(\mathbb{S}^1,\mathbb{C})$ via translations: $$R_s\cdot u(x):=u(x+s).$$ So if $A$ commutes with rotations $R_s$, then we have $$(Au)(x+s)=A(u(x+s)).$$ Now, the author provides a proof wherein he says "differentiate this equation with respect to $s$ at $0$ and use the continuity of $A$ to get $DAu=ADu.$" The LHS is fine, because the derivative operator just remains on the outside and acts on the function $(Au)(x)$, but for the RHS, $\textbf{I'm unsure how exactly we are using the continuity of $A$.}$ My thought is to do something like:
$$\frac{d}{ds}\Big|_{s=0}A(u(x+s))=\lim_{s\to0}\frac{A(u(x+s))-A(u(x))}{s}=A\left(\lim_{s\to0}\frac{(u(x+s))-(u(x))}{s}\right).$$ Now this doesn't feel right since, although $1/s$ seems constant with respect to the linear operator $A$, I then don't see why we even need to use the hypothesis that $A$ commutes with rotations. It seems this argument works for any continuous, linear operator $A$.
I'd appreciate any guidance!
