If f : G → ℂ has mean zero (i.e. ∑ g, f g = 0), then so does convOp μ f.
Iterating the convolution operator convOp μ preserves the mean-zero property of f.
For a probability measure μ, the $K$-fold convolution convPow μ K minus the constant c
equals the $K$-fold iterate of convOp μ applied to δ_{1_G} - c.
The set of operator-norm ratios ‖convOp μ f‖₂ / ‖f‖₂ over nonzero mean-zero functions
f : G → ℂ is bounded above (so the supremum σ₁(T_μ) is well-defined).
Operator-norm bound: for any mean-zero f : G → ℂ with nonzero $L^2$ norm,
‖convOp μ f‖₂² ≤ σ₁(T_μ)² · ‖f‖₂².
Iterated operator-norm bound: for any mean-zero f, the $K$-fold convolution satisfies
‖(T_μ)^K f‖₂² ≤ σ₁(T_μ)^{2K} · ‖f‖₂².
The function δ_{1_G} - 1/|G| (point mass at the identity minus the uniform
distribution) has total sum zero.
The squared $L^2$ norm of δ_{1_G} - 1/|G| is at most $1$ (in fact equals $1 - 1/|G|$).
$L^2$ mixing for random walks on finite groups: if μ is a probability measure on a
finite group G with second-largest singular value σ₁(T_μ), then after K steps the
distribution convPow μ K is close to uniform in $L^2$:
$$\|T_\mu^K \delta_{g_0} - 1/|G|\|_{L^2}^2 \le \sigma_1(T_\mu)^{2K}.$$