Background:
Given a Banach algebra $A$, we can turn $A^{*}$, the Banach space dual of $A$, into a Banach $A$-bimodule via the following module actions:
For $x\in A, f\in A^{*}$, $x.f:y\mapsto f(yx)$ and $f.x:y\mapsto f(xy)$.
Example:
Let $G$ be a locally compact group with left Haar measure $\mu$.
Take $A = L^{1}(G)$, and identify $A^{*}$ with $L^{\infty}(G)$ as usual.
Now for a fixed $f\in A$, $g\in A^{*}$, $g.f$ is an element of $A^{*}$ and thus identifies with an element of $L^{\infty}(G)$, but which one?
That is, up to almost everywhere, I should be able to evaluate $g.f$ at a point $x\in G$.
My Logic:
What I do know, from the definition of the action of $A^{*}$ on $A$ is that $$\int_{G}g.f(x)h(x)\mu(dx) = \int_{G}g(x)[f*h](x)\mu(dx)$$ for any $h\in L^{1}(G)$.
So I can do some manipulations with the right hand side to get
\begin{align*} \int_{G}g(x)[f*h](x)\mu(dx) &= \int_{G}g(x)\int_{G}f(y)h(y^{-1}x)\mu(dy)\mu(dx)\\ &= \int_{G}\int_{G}g(x)f(y)h(y^{-1}x)\mu(dy)\mu(dx)\\ &= \int_{G}\int_{G}g(x)f(y)h(y^{-1}x)\mu(dx)\mu(dy)\\ &= \int_{G}\int_{G}g(x)f(y)h(x)\mu(dx)\mu(dy)\\ &= \int_{G}\int_{G}g(yx)f(y)h(x)\mu(dy)\mu(dx)\\ &= \int_{G}\left[\int_{G}g(yx)f(y)\mu(dy)\right]h(x)\mu(dx)\\ \end{align*}
Therefore can I say that $g.f(x) = \int_{G}g(yx)f(y)\mu(dy)$ for almost every $x\in G$?