There was asking to prove that the best coercitivity constant $c_p$ for $\psi$ is $2^{-1/p}$. In fact this is not true.
For a given simple function $ f=\sum\limits_{k=1}^n a_k\chi_{A_k} $ denote $x_k=\mu(A_k)$. Consider special case $a_1=-1$, $a_2=0$, $a_3=1$ and $x_1=\varepsilon$, $x_2=1-2\varepsilon$, $x_3=\varepsilon$ where $\varepsilon\in(0,2^{-1})$. Then $ c_p\leq\Vert\psi(f)\Vert_p/\Vert f\Vert_p=(\varepsilon+2^{1-p}(1-2\varepsilon))^{1/p} $ Since left hand side is independent of $\varepsilon$ we conclude $ c_p\leq\min_{\varepsilon\in(0,2^{-1})}(\varepsilon+2^{1-p}(1-2\varepsilon))^{1/p}=2^{(1-\max(2,p))/p} $ But even the bound $ b_p=2^{(1-\max(2,p))/p} $ is not rough. Numeric test showed that for $p=3$, $a_1=0.079$, $a_2=0.079$, $a_3=-1$ with $x_1=0.879$, $x_2=0.99$, $x_3=0.022$ gives $ c_3< 0.612176<0.629960\approx b_3 $
Here is a Mathematica code to check this
FNorm[a_, x_, n_, p_] := (Sum[Abs[a[[k]]]^p x[[k]], {k, 1, n}])^(1/p); FImageNorm[a_, x_, n_, p_] := (Sum[ Abs[(a[[k]] + a[[l]])/2]^p x[[k]] x[[l]], {l, 1, n}, {k, 1, n}])^(1/p); FOpNorm[a_, x_, n_, p_] := FImageNorm[a, x, n, p]/FNorm[a, x, n, p] OpNorm = 1; A = {}; X = {}; p = 3; With[{n = 3, R = 1, M = 100000}, For[i = 0, i < M, i++, a = RandomReal[{-R, R}, n]; x = RandomVariate[GammaDistribution[1, 1], n]; x = x/Total[x]; norm = FOpNorm[a, x, n, p]; If[norm < OpNorm, {OpNorm, A, X} = {norm, a, x}, Continue[]]; ] ] Print[{{OpNorm, 2.^((1-Max[2,p])/p)}, A, X}]