Show $T:G\rightarrow L^1(G)$ by $y\mapsto f(y\cdot)$ is continuous for fixed $f\in L^1(G)$

Question

Let $G$ be a locally compact group, and show $T_f:G\rightarrow L^1(G)$ by $y\mapsto f(y\cdot)$ is continuous for fixed $f\in L^1(G)$

Answer 1

We only have to show $T_f$ is continuous for $f$ in a dense subset of $L^1(G)$, which we pick to be of the one of continuous differentiable functions with compact support $C^1_c(G)$. Then, because $f\in C^1_c(G) \Rightarrow T_f (y)\in C^1_c(G)$, it suffices to show that a neighborhood $U$ of $T_f(1)$ is pulled back to a neighborhood $V$ of $1$. But since $f$ is continuous, for every $y \in V$ we can pick a neighborhood $V_y$ such that $\forall z\in V_y$ $$\sup_{z\in \sup(f)}\{|f(zx)-f(yx)|\}\le \epsilon\Rightarrow |f(z\cdot)-f(y\cdot)|_{L^1}\le \epsilon\cdot\mu(\sup(f)).$$ Picking $\epsilon$ accordingly, we can ensure that $T_z(f)\in U$. Hence $V$ is open and this completes the proof.