Comparison inequality between a function and its double s-polar (s<0)
Establish that for every integer n ≥ 1 and s ∈ (−1/n, 0), for any even s-concave function g: R^n → [0,∞) with 0 < ∫_{R^n} g(x) dx < ∞, the integral of g dominates that of its double s-polar up to the factor (1+ns), namely prove ∫_{R^n} g(x) dx ≥ (1+ns) · ∫_{R^n} L_s L_s g(x) dx, with equality for g = 1_{[-1,1]^n}. Here L_s is the s-polar defined by L_s g(y) = inf_{x: g(x)>0} [(1 − s⟨x,y⟩)_+^{1/s} / g(x)].
References
The second conjecture compares the integral of g with the integral of its double s-polar. Conjecture Let s∈(−\frac{1}{n},0) and let g:R{n}→R_{+} be even s-concave such that 0<\int g<+∞. Then, \begin{equation} \int_{R{n}} g(x) dx \ge (1+ns) \int_{R{n}} L_s L_{s} g(x) dx, \end{equation} with equality for g=1_{[-1,1]n}.
                — On Mahler's conjecture for even s-concave functions in dimensions 1 and 2
                
                (2412.12372 - Fradelizi et al., 16 Dec 2024) in Conjecture (labelled “comp-bipolar”), Section 2