Left regular action of $L^1(G)$ on $L^2(G)$ is given by convolution.

Question

This seems basic and I couldn't find it by googling so I thought I'd post it along with an answer.

If $G$ is a locally compact topological group and $\pi:G\rightarrow U(H)$ is a unitary representation of $G$ on a Hilbert space $H$, we can talk about an action, $\pi'$, of $L^1(G, \mathbb{C})$ on $H$. Namely for $f\in L^1(G, \mathbb{C})$, and $v\in H$, define $$\pi'(f)v := \int_G f(y)\cdot(\pi(y)(v))dy$$ where $dy$ is Haar measure and the integral requires the theory of integration of Banach space valued functions. That is, we define the integral as a limit of integrals of step functions which form a Cauchy sequence in $L^1(G,H)$ and which converge pointwise almost everywhere to $y \mapsto f(y)\cdot(\pi(y)(v))$.

In particular if $\pi$ is the left-regular representation of $G$ on $L^2(G, \mathbb{C})$, and $g\in L^2(G, \mathbb{C})$, one gets $\pi'(f)(g)$ that looks a lot like the ordinary old convolution of $f$ and $g$. But how do you actually know they're the same? That is how do you know

$$x\mapsto \bigg((f\ast g)(x)=\int_Gf(y)g(y^{-1}x)dy\bigg)$$ is the same element of $L^2(G, \mathbb{C})$ as $$\int_G f(y)(L_yg)dy$$ since the former is a family of ordinary integrals parametrized by $x$ and the latter is a single integral of an $L^2(G, \mathbb{C})$ valued function?

More generally, let $X$ and $Y$ be measure spaces. Suppose $$\Phi:X\times Y\rightarrow \mathbb{C}$$ has the property that $$\Phi_x:Y\rightarrow \mathbb{C}, y\mapsto \Phi(x,y)$$

is in $L^1(Y, \mathbb{C})$ for every $x\in X$, and $$\Phi^y:X\rightarrow \mathbb{C}, x\mapsto \Phi(x,y)$$ is in $L^p(X)$ for all $y$ and some fixed $p\in [1,\infty)$. Suppose further that the map $$Y\mapsto L^p(X, \mathbb{C}), y \mapsto \Phi^y$$ lies in $L^1(Y,L^p(X, \mathbb{C}))$ in the sense of the theory of integration of functions taking values in a Banach space. Denote by $\int_Y \Phi^y dy\in L^p(X, \mathbb{C}) $ its integral in the sense of this theory. How do we know that $$ (\int_Y \Phi^y dy)(x) = \int_Y\Phi(x,y)dy~~~~~~~~~~~~~~(1)$$ holds for almost all $x\in X$?

Answer 1

We may take a sequence $(\sigma_n)_{n\in\mathbb{N}}\in L^1(Y,L^p(X, \mathbb{C}))$ of step functions (that is functions taking only finitely many values (in $L^p(X, \mathbb{C})$) converging in $L^1(Y,L^p(X, \mathbb{C}))$ to $(y\mapsto \Phi^y) $. Such a step function may be written $$\sigma = \sum_{j=1}^{k}\chi_{A_j}(y)f_j(x)~~~~~~~~~~~~(2)$$ for $f_1,...,f_k\in L^p(X, \mathbb{C})$ and $A_1,...,A_k$ sets with finite measure in $Y$. Now since $\sigma_i \rightarrow (y\mapsto \Phi^y) $, we have $$\lim_{i\rightarrow \infty}\int_Y\|\sigma_i^y(x) - \Phi^y(x) \|_{L^p(X)}dy=0.$$ Using the integral form of Minkowski's inequality, we get $$ \int_Y\|\sigma^y_i - \Phi^y \|_{L^p(X)}dy\geq \| \int_Y \sigma_i(x,y) - \Phi(x,y) dy \|_{L^p(X)}$$ so that $$\| \int_Y \sigma_i(x,y)dy - \int_Y\Phi(x,y) dy \|_{L^p(X)} \rightarrow 0.$$ Then passing to a subsequence if necessary we may assume $$\int_Y \sigma_i(x,y)dy\rightarrow \int_Y\Phi(x,y) dy ~~~~~(3)$$ for almost all $x\in X.$

Recall the $\sigma_i$ have the form (2) above. Since for such $\sigma$, $\int_Y \sigma dy = \int_Y \sigma^y dy$ is then by definition just $\sum \mu(A_j) f_j(x)\in L^p(X, \mathbb{C})$, it is clear $(1)$ holds for $\Phi=\sigma_i$ and all $x \in X$. Thus $$\int_Y \sigma_i(x,y)dy=(\int_Y \sigma_i^y dy)(x)\xrightarrow{L^p(X)} (\int_Y \Phi^y dy)(x)$$ where the convergence holds by definition of the $\sigma_i$. Thus passing to another subsequence if necessary we have $$\int_Y \sigma_i(x,y)dy\rightarrow (\int_Y \Phi^y dy)(x) ~~~~~~~ (4)$$ for almost all $x$.

Putting (3) and (4) together gives the result.