How to show that $v \mapsto \pi(f)v$ is differentiable?

Question

Let $G$ be a compact group. Let $(\pi, V)$ be a representation of $G$ and $f$ a smooth function on $G$. Define \begin{align} \pi(f)v = \int_G f(x)\pi(x) v dx. \end{align} We have \begin{align} & \pi(y)\pi(f)v \\ & = \int_G f(x)\pi(x) v dx \\ & = \int_G f(x) \pi(y) \pi(x) v dx \\ & = \int_G f(x) \pi(y x) v dx \\ & = \int_G f(y^{-1}x) \pi(x) v dx. \end{align} Why this implies that $v \mapsto \pi(f)v$ is differentiable? Thank you very much.