Let φ be a closed concave function, and M = max x ∈ R d φ ( x ). Let D r := { φ ≥ r } be the level set. Then given r ≤ s ≤ M, λ d ( D s ) ≥ ( M − s M − r ) d λ d ( D r )where λ d is the Lebesgue measure on R d .

Question

Let   φ be a closed concave function, and   M  =      max          x      ∈                        R                d              φ  (  x  ). Let       D    r    :=  {  φ  ≥  r  } be the level set. Then given   r  ≤  s  ≤  M,      λ    d    (      D    s    )  ≥            (                  M      −      s              M      −      r                          )              d        λ    d    (      D    r    )where       λ    d   is the Lebesgue measure on             R        d  .

smeachtaczm · Accepted Answer

Hint. First think   d  =  2 (or   3). Consider a triangle (cone) with base       D    r   and height   M  −  r. Show that for any horizontal slice at height   s  −  r, the legth (area) of that slice equals             M      −      s              M      −      r            D    r  .

Let varphi be a closed concave function, and M=max_(x in R^d) varphi(x). Let D_r:={varphi>=r} be the level set. Then given r<=s<=M, lambda_d(D_s)>=(M−s/M−r)^d lambda_d(D_r) where lambda_d is the Lebesgue measure on R^d.

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