Trivializing unitary cocycles in abelian von Neumann algebras that are uniformly close to the trivial one 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.
 A: With your uniformly closeness assumption, $b(g) := \sqrt{-1}\log y_g$ is a usual additive cocycle (i.e., $b(gh) = b(g) + g\cdot b(h)$) which is moreover real and bounded. I think any bounded cocycle into $L^\infty(X)$ is a coboundary. By exponentiating it, one gets the unitary element $v$. Here's a standard proof for the case $G$ being countable. The general locally compact second countable case will follow by considering a countable dense subgroup. 
We assume $b$ is real and put $\xi(x) := \sup_{h \in G} b(h)(x)$. It is not difficult to see $\xi$ is invariant under the the affine order-preserving action $\rho_g\colon \zeta \mapsto g\cdot\zeta + b(g)$. Indeed,
$$\begin{array}{lcl}(\rho_g\xi)(x) & = & \xi(g^{-1}x) + b(g)(x) \\ & = & \sup_h b(h)(g^{-1}x)+b(g)(x) \\ & = & \sup_h b(gh)(x) \\ & = & \xi(x). \end{array}$$
This means $b(g) = \xi - g\cdot\xi$. 
In case $G \curvearrowright X$ is a topological minimal action and $y_g \in {\mathcal U}(C(X))$, one can take $v$ to be continuous. For this, see the proof of Theorem 2.6 in [N. S. Ormes, Pacific J. Math. 195 (2000), no. 2, 453–476], which I replicate here. For every $\epsilon>0$, the subset $K := \{ x \in X : \xi(x) \le \epsilon\}$ is closed and $x\in gK \Leftrightarrow \xi(x) \le b(g)(x) + \epsilon$. By definition, $\bigcup_g gK = X$, and so $K$ has non-empty interior $K^\circ$ by the Baire category theorem. One has $X = \bigcup_g gK^\circ$ by minimality and $b(g) \le \xi \le b(g)+\epsilon$ on $gK^\circ$. This proves continuity of $\xi$.
