Question

Using the standard notation, let $\lambda_gf(x)=f(g^{-1}x)$ be the left regular representation and $\rho_gf(x)=f(xg)$ be the right regular one, acting on the space $V$ of complex-values functions on $G$.

Let $b:G\to \ell_2(G)$ be a cocycle for $\rho$; i.e., $b_g=\rho_g f-f\in \ell_2(G)$ for some $f\in V$ and every $g\in G$.

Question: Does $a_g=\lambda_g f-f$ belong to $\ell_2(G)$ for every $g\in G$? If not, what is the "smallest" subspace $W$ of $V$, $\ell_2(G)\subseteq W\subseteq V$ for which this happens?

The most interesting case seems to be when $G$ is amenable.

Edit: The metric interpretation is the following. Take all $f\in V$ whose discrete gradient with respect to the left-invariant metric is $\ell_2$-summable. Then $W$ is the space consisting of gradients of these $f$, with respect to the right-invariant metric.

Answer 1

I think, the following is an example. Let $G$ be the infinite dihedral group with involution $s\in G$. Next, let $h: {\mathbb R}\to {\mathbb R}$ be a positive $L^2$-function and $f(n):= \int_{-\infty}^n h(x)dx$ be a function on ${\mathbb Z}$. Now, extend $f$ to $G$ by $f(zs)=f(z)$, $z\in {\mathbb Z}$. Then $f$ determines an $\ell_2$-cocycle $b_g$ but $a_s(x)$ is not in $\ell_2$ (since it does not tend to zero as $|x|\to\infty$.