Suppose $M$ is an abelian von Neumann algebra, carrying a (point-ultraweakly) continuous action $G\curvearrowright M$ of a locally compact, second-countable group.

Let $y: G\to\cal U(M)$ be an ultraweakly continuous unitary cocycle, i.e., $y_{gh}=y_g\cdot (g.y_h)$ for all $g,h\in G$. Suppose that $y$ is uniformly close to the trivial cocycle, i.e., $$ \sup_{g\in G} \|y_g-1\|=:\varepsilon<1. $$

My Question:Is $y$ always a coboundary? This means $y_g=v\cdot(g.v)^*$ for some $v\in\cal U(M)$. Can one say more about the possible choices for $v$, for example $\|v-1\|\leq C(\varepsilon)$ for some universal constant, where $\lim_{\varepsilon\to 0} C(\varepsilon)=0$?

**Ideas:**
I suspect that this should hold for
$C(\varepsilon)=\varepsilon(1+\frac{1+\varepsilon}{1-\varepsilon})$.
(*The block of text below also gives a justification for amenable groups*)

This would be implied by the following statement: (I guess an addendum to the above question is whether what follows is always true.)

(Conjectured) Lemma:Let $S$ be the ultraweak closure of the convex hull of $\{y_g\}_{g\in G}$. Then there exists an element $s\in S$ with $s=y_g\cdot (g.s)$ for all $g\in G$.

First let us convince ourselves why this would solve the problem. Indeed, if such an element exists, then $s^*s=g.(s^*s)$ follows immediately for all $g\in G$, i.e., the element $s^*s$ is fixed by the action. Moreover, since $s$ is a limit of convex combinations of the $y_g$, it follows that $\|s-1\|\leq\varepsilon$. Therefore it is invertible and $v=s|s|^{-1}$ is a unitary that trivializes $y$ as desired. Furthermore $$ \|v-1\| = \|s|s|^{-1}-s+s-1\| \leq \|s\| \underbrace{ \||s|^{-1}-1\| }_{\leq \varepsilon/(1-\varepsilon)} + \|s-1\| \leq \frac{(1+\varepsilon)\varepsilon}{1-\varepsilon}+\varepsilon. $$

**Proof of the above lemma for amenable groups:**
(*Note: The following argument appears to work in any von Neumann algebra*)

Suppose $G$ is amenable. Let $\mu$ be the left-invariant Haar measure. Then there is a sequence of (non-zero) functions $f_n\in L^1(G)$ with $0\leq f_n\leq 1$ such that $\displaystyle 0=\lim_{n\to\infty} \max_{g\in K} \frac{\|f_n-g.f_n\|_1}{\|f_n\|_1}$ for all compact sets $K\subseteq G$. Set $$ s_n = \|f_n\|^{-1} \int_G f(g)y_g~d\mu(g) \in M. $$ Evidently $s_n\in S$, and a brief calculation (using the fact that $y$ is a cocycle) shows for every compact set $K\subseteq G$ that $$ \max_{g\in K} \| s_n-y_g\cdot(g\cdot s_n) \| \leq \max_{g\in K} \frac{\|f_n-g.f_n\|_1}{\|f_n\|_1} \stackrel{n\to\infty}{\longrightarrow} 0. $$ Thus we may obtain the desired element $s\in S$ as some cluster point of $(s_n)_n$ in the ultraweak topology.

**Problem:** Although I believe I can generalize the above argument further to the case where the $G$-action is amenable, I am not so happy with this approach.
I actually believe that amenability of $G$ might be a red herring for the specific setup of this question, especially due to $M$ being abelian.

For example, for some arbitrarily badly-behaved $G$, consider the case where the $G$-action on $M$ is trivial (so highly non-amenable). Then $y$ has a disintegration into a family of group homomorphisms $M\to\mathbb T$, but by assumption these take value in $\{z\mid |z-1|<1\}$, so in fact $y=1$ and the claim trivially holds. This toy example leads me to suspect that I am going at it the wrong way, and that an element $s\in S$ as required by the above lemma may always exist by some more clever Hahn-Banach trickery on compact convex sets.