Let $G$ be a locally compact Hausdorff (LCH) topological group with left Haar measure $\mu$. Given a compact unit neighborhood $U$, consider the function
$$
\Phi: \quad G \to (0,\infty), \quad x \mapsto \mu(U x U).
$$
My question is:



> 
> Can one give a natural characterization of the groups for which this function is bounded?
> 
> 
> 


One conjecture (see below): This happens exactly for IN groups, which are groups for which there exists a compact unit neighborhood $U$ satisfying $x U x^{-1} = U$ for all $x \in G$.


Thoughts/observations:


1. The question is independent of the choice of $U$. Indeed, if $U,V$ are both compact unit neighborhoods, then $U \subset \bigcup\_{i=1}^n x\_i V$ and $U \subset \bigcup\_{j=1}^m V y\_j$ for suitable $x\_i,y\_j \in G$, and this easily allows to bound $\mu(U x U)$ in terms of $\mu(V x V)$.
2. If $G$ is not unimodular, then $\Phi$ is *not* bounded, since $\Phi(x) \geq \mu(U x) = \Delta(x) \cdot \mu(U)$, so that $\Phi$ is bounded from below (up to a constant) by the modular function, which is unbounded for non-unimodular groups.
3. If $G$ is an IN group, then $\Phi$ is bounded. Indeed, by the first observation from above we can choose $U$ to satisfy $x U x^{-1} = U$ for all $x$, and then $\Phi(x) = \mu(U x U) = \mu(x U U) = \mu(U U)$ for all $x \in G$.


What I have not been able to show is that if $\Phi$ is bounded, then $G$ needs to be IN. Of course, it could be that this simply does not hold.



For what LCH groups is the Haar measure $\mu(U x U)$ bounded?
Your conjecture is correct.


Suppose that we have a compact unit neighbourhood $U$ such that $\mu(UxU) \ll 1$ for all $x$. As you have already noted, we can take $\mu$ to be unimodular, and the choice of neighbourhood is not relevant, so we may assume without loss of generality that $U$ is symmetric: $U^{-1} = U$. (This makes $U$ what is called an [approximate group](https://en.wikipedia.org/wiki/Approximate_group) in arithmetic combinatorics, and the intuition that $U$ should behave like a subgroup of $G$ is underlying the arguments below.) We allow implied constants in asymptotic notation to depend on $U$, thus for instance $\mu(U) \asymp 1$.


Note that the conjugate $x U x^{-1}$ of $U$ is commensurate with $U$ in the sense that $\mu( U \cdot x U x^{-1} ) = \mu( U x U ) \ll 1$. (In the language of arithmetic combinatorics, $U$ stays close to its conjugates $xUx^{-1}$ in [Ruzsa distance](https://en.wikipedia.org/wiki/Ruzsa_triangle_inequality).) In the spirit of the isomorphism theorems (or the [Ruzsa covering lemma](https://en.wikipedia.org/wiki/Freiman%27s_theorem#Ruzsa_covering_lemma) in arithmetic combinatorics), one now expects $U$ and $xUx^{-1}$ to have large intersection, and this can be accomplished (at the cost of enlarging $U$ to $U^2$) by the following convolution argument (cf. the double counting argument that shows that $|H \cdot K| = |H| |K| / |H \cap K|$ for finite subgroups $H,K$ of $G$). Observe that the convolution $1\_U \* 1\_{xUx^{-1}}$ has an $L^1(G,\mu)$ norm of $\mu(U) \mu(x U x^{-1}) \asymp 1$ and is supported on $U x U x^{-1}$, which has measure $O(1)$. Thus there must exist a point $y \in G$ where $1\_U \* 1\_{xUx^{-1}}(y) \gg 1$, thus
$$ \mu( yU \cap x U x^{-1} ) \gg 1$$
which implies
$$ \mu( (yU \cap x U x^{-1})^{-1} \cdot (yU \cap x U x^{-1}) ) \gg 1$$
and hence
$$ \mu( U^2 \cap x U^2 x^{-1} ) \gg 1.$$
To put it another way, the inner products of the functions $1\_{x U^2 x^{-1}}$ with $1\_{U^2}$ are uniformly bounded from below.


We can now use an "ergodic" argument to extract an invariant object (in the spirit of the [Alaoglu--Birkhoff ergodic theorem](https://mathscinet.ams.org/mathscinet-getitem?mr=2026)).
Let $S$ be the closed convex hull in $L^2(G,\mu)$ of the functions $1\_{x U^2 x^{-1}}$. By the [Hilbert projection theorem](https://en.wikipedia.org/wiki/Hilbert_projection_theorem), this set has a unique element $f$ of minimal norm. All elements of $S$ have inner product with $1\_{U^2}$ uniformly bounded from below, so $f$ does also; in particular, $f$ is non-trivial. Since $S$ is conjugation-invariant, symmetric, bounded and consists of non-negative functions, $f$ must be non-negative, symmetric, and conjugation-invariant.


At this point one could already extract a conjugation-invariant set of positive finite measure by taking level sets of $f$, but this is not quite regular enough for the conjecture, so we take a convolution to achieve an additional smoothing (in the spirit of the [Steinhaus theorem](https://en.wikipedia.org/wiki/Steinhaus_theorem)). The convolution $f\*f$ is then in $C\_0(G)$ (this follows from a standard limiting argument, approximating $f$ in $L^2(G,\mu)$ by $C\_c(G)$ functions and using Young's inequality), conjugation-invariant, and strictly positive at the origin; taking level sets, we obtain a non-trivial conjugation-invariant compact unit neighbourhood, as required.


