Let $f$ be a nice function (smooth if you want), supported on $B_{2R}\subset\mathbb{R}^{n}$. Denote by $\beta_{R}$ the (normalized) ball average of $f$ (wrt the Haar measure on $\mathbb{R}^{n}$, for a ball of radius $R$). Assume $H<R$. One should think about $H$ as being some power $R$, something like $H=R^{1-\epsilon}$.
The following inequality is easy to verify $$\lvert \beta_{R}(f) - \beta_{R}\star\beta_{H}(f)\rvert \ll_{f} \frac{H}{R}$$ where the dependence on $f$ is by some Lipschitz norm. The inequality follows at once as $f(x)-f(x+h)$ get cancelled when integrating over $B_{R}$, except for a tubular neighborhood of width $O(H)$ around the boundary of the ball, $\partial B_{R}$.
On the other hand, in the Fourier side, this double convolution have a major impact, it becomes a double multiplier (and some asymptotics would actually show the double multiplier is integrable, while the single one is not (as the FT is $L^{2}$ normalized)).
My question is - how can one see the above inequality from the Fourier side? I am willing to take lesser inequality (in terms of powers say, and maybe dependence on some higher Sobolev norm rather than just Lipschitz norm).