# Mapping $G$ into its group algebra as left multiplication. Continuous?

I am reading an appendix on Group algebras which contains the following Proposition which I am trying to prove:

Proposition: Let $G$ be a locally compact group, with $\zeta\in L^{p}(G)$ fixed. Define $L_{x}\in L^{p}(G)$ by $L_{x}(\zeta)[\lambda] = \zeta(x\lambda)$. The map $$G\mapsto L^{p}(G), x\mapsto L_{x}(\zeta)$$ is continuous.

My Proof Attempt:

It was not obvious to me that this map is well-defined; but I think I have verified it below. Can someone check my logic and confirm I am using the notation correctly? My measure theory background always leaves me feeling uneasy when abstract measure space show up.

Let $\mu$ denote the left Haar measure for $G$.

\begin{align*} \int_{G}|L_{x}(\zeta)|^{p}d\mu &= \int_{G}|\zeta(xy)|^{p}\mu(dy)\\ &= \int_{G}|\zeta(y)|^{p}\mu(x\cdot dy)\text{, (transformation property)}\\ &= \int_{G}|\zeta(y)|^{p}\mu(dy)\text{, (by Left }G\text{-invariance)}\\ &= \|\zeta\|_{p}^{p} <\infty\text{, because }\zeta\in L^{p}(G) \end{align*}

Therefore $L_{x}(\zeta)\in L^{p}(G)$, so the map is well-defined.

Now to show the map is continuous, it suffices by $G$-invariance of $\mu$ to show that whenever $x_{\alpha}\to 1$, we have $\|L_{x_{\alpha}}(\zeta) - \zeta\|_{p}\to 0$.

I'm stuck here though, as I have no idea how to estimate $\int_{G}|\zeta(x_{\alpha}y) - \zeta(y)|^{p}d\mu$. What is the trick?

• Can you do it for the characteristic function of an open set? Notice that is enough, for they span $L^p$. – Mariano Suárez-Álvarez May 12 '13 at 2:57
• Yes I think I can, thanks for the hint. – roo May 12 '13 at 3:24
• Alternatively, show that the action of $G$ on $C_c(G)$ is continuous with respect to the sup-norm since functions in $C_c(G)$ are uniformly continuous. Then use that $C_c(G)$ is dense in $L^p$ for $p \lt \infty$. A more detailed outline is here. – Martin May 14 '13 at 1:06
• Thank you very much for the reference. I'll try that approach. – roo May 14 '13 at 18:16