Harmonic Analysis and Convex Geometry, Fall 2013, November 8.

Instructor: Dmitry Ryabogin

Assignment 9.

1. Problem 1.

Given any subset E of ℝ^{d} and α ≥ 0, we define the exterior α-dimensional Hausdorff measure of E by m_{α}^{∗}(E) = lim_{δ→0^{+}} H_{α}^{δ}(E), where

H_{α}^{δ}(E) := inf{ ∑_{k=1}^{∞}( diam F_{k})^{α} : E ⊂ ⋃_{k=1}^{∞} F_{k}, diam F_{k} ≤ δ ∀k },

and diam S = sup_{x,y∈S} |x − y| stands for the diameter of S.

a) Let 0 < δ_{1} < δ_{2}. Prove that H_{α}^{δ_{1}} ≥ H_{α}^{δ_{2}}. Conclude that the above limit exists (might be infinite).

b) Prove that H_{α}^{δ}(E) ≤ m_{α}^{∗}(E).

c) Prove that E_{1} ⊂ E_{2} implies m_{α}^{∗}(E_{1}) ≤ m_{α}^{∗}(E_{2}).

d) Prove that m_{α}^{∗}( ⋃_{j=1}^{∞} E_{j}) ≤ ∑_{j=1}^{∞} m_{α}^{∗}(E_{j}) for any countable family of sets in ℝ^{d}.

Hint: For each j choose a cover {F_{j,k}}_{k=1}^{∞} of E_{j} by sets of diameter less than δ such that

∑_{k=1}^{∞}( diam F_{j,k})^{α} ≤ H_{α}^{δ}(E_{j}) + \frac{ϵ}{2^{j}} .

Then use the fact that ⋃_{j,k} F_{j,k} is a cover of E.

e) If the distance d(E_{1}, E_{2}) > 0, then m_{α}^{∗}(E_{1} ∪ E_{2}) = m_{α}^{∗}(E_{1}) + m_{α}^{∗}(E_{2}).

Hint: To show that m_{α}^{∗}(E_{1} ∪ E_{2}) ≥ m_{α}^{∗}(E_{1}) + m_{α}^{∗}(E_{2}) fix 0 < ϵ < d(E_{1}, E_{2}). Given any cover of E_{1} ∪ E_{2} with sets F_{j} of diameter less than 0 < δ < ϵ, put F′_{j} = E_{1} ∩ F_{j}, F′′_{2} = E_{2} ∩ F_{j} and use

∑_{j}( diam F′_{j})^{α} + ∑_{l}( diam F′′_{l})^{α} ≤ ∑_{k=1}^{∞}( diam F_{k})^{α}.

f) Conclude (do not prove, google if necessary) that since m_{α}^{∗} satisfies all the properties of the metric Caratheodory exterior measure, it is a countably additive measure when restricted to the Borel sets. This measure (we will denote it m_{α}) is called the αdimensional Hausdorff measure. g) Prove that the α-dimensional Hausdorff measure is invariant under translations and rotations. Moreover, ∀λ > 0, m_{α}(λE) = λm_{α}(E).

h) Prove that the quantity m_{0}^{∗} counts the number of points in E ⊂ ℝ, while m_{1}(E) = m(E) for all Borel sets E ⊂ ℝ (here m denotes the usual Lebesgue measure).

i) Prove that m_{d}(E) ≤ 2^{d}c_{d}m(E).

Hint: Let E ⊂ ⋃_{j} F_{j} be a covering with ∑_{j}( diam F_{j})^{d} ≤ m_{d}(E) + ϵ. Find a closed

ball B_{j} centered at a point of F_{j} so that B_{j} ⊃ F_{j} and diam B_{j} = 2 diamF_{j}. To get the result, use that

m(E) ≤ ∑_{j} m(B_{j}) = ∑_{j} c_{d}( diam B_{j})^{d} = 2^{d}c_{d} ∑_{j}( diam F_{j})^{d} ≤ 2^{d}c_{d}(m_{d}(E) + ϵ).

2. Problem 2*. Let K be a convex body in ℝ^{d}, d ≥ 2. The width w_{K}(θ) of K in the direction θ ∈ S^{d−1} is defined as h_{K}(θ)+h_{K}(−θ). The body K is called of constant width if w_{K} is constant (independent of the direction θ). Prove that in the two-dimensional case among the bodies of the given width the Reuleaux triangle is the one of the smallest area. What is the body of the given constant width of the smallest volume/surface area in ℝ^{d}, d ≥ 3?