Quasimorphisms and Bounded Cohomology: Quantitative Version?

Question

Consider maps from a discrete group $\Gamma$ to the additive group $\mathbb{R}$. A function $f:\Gamma \to \mathbb{R}$ is called a quasimorphism if it is locally close to being a group homomorphism. More precisely, $f$ is a quasimorphism if there exists some constant $C$ such that for every $x,y \in \Gamma$, $$|f(xy)-f(x)-f(y)|<C$$ In other words, $f(xy)$ is at bounded distance from $f(x)+f(y)$.

A natural question is whether any quasimorphism is simply a homomorphism perturbed by a bounded function. In this regard, we can define the bounded cohomology of $\Gamma$ with coefficients in $\mathbb{R}$, and consider the comparison homomorphism $$c:H_b^2(\Gamma,\mathbb{R}) \to H^2(\Gamma,\mathbb{R})$$ and it is known that the kernel of this map is precisely the space of non-trivial quasimorphisms. More precisely $$ker(c) = QM(\Gamma)/(Hom(\Gamma,\mathbb{R}) \oplus C_b(\Gamma,\mathbb{R}))$$ where $QM(\Gamma)$ is the space of quasimorphisms, $C_b(\Gamma,\mathbb{R})$ is the space of all bounded functions, and $Hom(\Gamma,\mathbb{R})$ is the space of homomorphisms.

So when $ker(c)$ is trivial, every quasimorphism is trivial: it is obtained by perturbing a homomorphism by a bounded function.

Is there a quantitative version of this statement? That is, suppose $C$ is the uniform bound for the quasimorphism $f$ as above, and suppose $ker(c)$ is trivial. Then can we actually give a concrete bound for the distance of $f$ from the homomorphism in $Hom(\Gamma,\mathbb{R})$ in terms of $C$?

All I find in references is that $f$ can be written as a sum of a homomorphism and a bounded function, but I have not come across any actual bound on that function in terms of the original bound on $f(xy)-f(x)-f(y)$. This seems like a very natural question, so do help out with directions and references if available.

*UPDATE: So in one direction it seems to be easy. Suppose $f=\phi+g$ where $\phi \in Hom(\Gamma,\mathbb{R})$ and $g:\Gamma \to \mathbb{R}$ is a bounded function with $|g(x)|<D$ for every $x \in \Gamma$. Then $$|g(xy)-g(x)-g(y)|<3D$$ by the triangle inequality, and $$|f(xy)-f(x)-f(y)|\leq |\phi(xy)-\phi(x)-\phi(y)|+|g(xy)-g(x)-g(y)|$$ and so $$|f(xy)-f(x)-f(y)|< 3D$$ So if $f$ is globally $D$-close to a homomorphism, then it is locally $3D$-close to being a homomorphism, or a $3D$-quasimorphism. But I am interested more in the other direction.

Answer 1

Suppose that $f : \Gamma \to \mathbf{R}$ is a trivial quasi-isomorphism and write $f = \phi + g$ where $\phi \in Hom(\Gamma,\mathbf{R})$ and $g : \Gamma \to \mathbf{R}$ is a bounded function as you did. Here we prove that if $$|f(x) + f(y) - f(xy)| \le C,$$ for all $x,y \in \Gamma$, then $g$ is bounded by $C$.

Suppose that $g$ is bounded by $D$ and that the bound is optimal, namely $D = \sup_{x \in \Gamma} |g(x)|$. For simplicity, let us assume for the moment that there exists $x \in \Gamma$ such that $|g(x)| = D$. First of all, as $\phi$ is a group homomorphism, we have $$|g(x) + g(y) - g(xy)|= |f(x) + f(y) - f(xy)|.$$

Since

$$2D - |g(x^2)| \le\left|2|g(x)| - |g(x^2)|\right| \le|2g(x) - g(x^2)| = |2f(x) - f(x^2)| \le C,$$
we have $|g(x^2)| \ge 2D - C$. On the other hand $D \ge |g(x^2)|$ by assumption, so $C \ge D$. Therefore $g$ is bounded by $C$.

In the general case, that is if we don't assume that $|g(x)| = D$ for some $x \in \Gamma$, then for every $\varepsilon > 0$ we can still find $x \in \Gamma$ such that $|g(x)| \ge D - \varepsilon$. A similar argument shows that $C + 2\varepsilon \ge D$ for every $\varepsilon > 0$. Therefore $C$ is a bound of $g$.