Real Analysis.

Instructor: Dmitry Ryabogin

Assignment VI.

1. Problem 1. a) Let K be a convex origin-symmetric set in R^{n}. Define the polar body K^{∗} of K as K^{∗} := {x ∈ R^{n} : x · y ≤ 1 ∀y ∈ K}. Prove that (K^{∗})^{∗} = K. Hint. To show that (K^{∗})^{∗} ⊆ K assume the contrary: there exists x ∈ (K^{∗})^{∗} \ K. Prove that there exists a hyperplane H_{α,u} := {y ∈ R^{n} : y · u = α}, such that K lies in a half-space {y ∈ R^{n} : y · u ≤ α}, and x ∈ {y ∈ R^{n} : y · u > α}. What can you say about z = u/α?

b) Let f be a differentiable function in R^{n}, homogeneous of degree m, that is g(tx) = t^{m}g(x) for every t > 0. Prove Euler’s formula: mg(x) = ∇g(x) · x.

c) Assume that a convex body K is such that its support function h_{K} is differentiable in R^{n} \ {0}. Prove that h_{K}(x) = ∇h_{K}(x) · x. Conclude that ∇h_{K}(x) belongs to a tangent plane H := {y ∈ R^{n} : y · x = h_{K}(x)} to K at the point ∇h_{K}(x).

d) Use convexity of K to prove that for every x, y ∈ K^{∗}, ∇h_{K}(x) · x ≥ ∇h_{K}(x) · y.

e) Use h_{K}(x) = ∇h_{K}(x) · x of c) and a) to prove ∇h_{K}(x) ∈ (K^{∗})^{∗} = K. Conclude that ∇h_{K}(x) ∈ K ∩ H.

2. Problem 2. a) Let K be a convex origin-symmetric body in R^{n}, ρ_{K}(x) := max{λ > 0 : λx ∈ K} be the distance function of K and let A ∈ GL(n) (an invertible linear transformation from R^{n} to R^{n}). Prove that ρ_{AK}(x) = ρ_{K}(A^{−1}x).

b) Let A, K be as in a). Prove that h_{AK}(x) = h_{K}(A^{t}x), where A^{t} is a conjugate operator, defined as Ax · y = x · A^{t}y.

c) Use ρ_{K^{∗}}(u) = 1/h_{K}(u), u ∈ S^{n−1}, to show that h_{(AK)^{∗}}(u) = h_{(A^{−1})^{t}K^{∗}}(u). Conclude that (AK)^{∗} = (A^{−1})^{t}K^{∗}.

d) Conclude that for any A ∈ GL(n), vol_{n}(AK)vol_{n}((AK)^{∗}) = vol_{n}(K)vol_{n}(K^{∗}). e)************************************************************************ Prove that for any K,

vol_{n}(K)vol_{n}(K^{∗}) ≥ vol_{n}(B_{1}^{n})vol_{n}(B_{∞}^{n}) = \frac{4^{n}}{n!}

. Hint. I have no clue.

3. Problem 3. a) Let K, L be two convex origin-symmetric bodies in R^{n}. Prove that

(K ∩ L)^{∗} = ∪_{θ∈[0,1]}((1 − θ)K^{∗} + θL^{∗}).

b) Prove that (B_{n}^{2})^{∗} = B_{n}^{2}.

c) Let K := {x ∈ R^{n} : ∑_{k=1}^{n} |x_{k}|^{2}/a_{k}^{2} ≤ 1, a_{k} > 0} be an ellipsoid. Find K^{∗}.

d) Let 1 ≤ p < ∞, and let B_{n}^{p} := {x ∈ R^{n} : ∑_{k=1}^{n} |x_{k}|^{p} ≤ 1}, B_{n}^{∞} := {x ∈ R^{n} :

sup_{k=1,...,n} |x_{k}| ≤ 1}. Prove that (B_{n}^{p})^{∗} = B_{n}^{q}, where 1/p + 1/q = 1. Hint. You might want to use the Lagrange multipliers. Then define

F_{x}(y) := x · y − λ(∑_{k=1}^{n} y_{k}^{p} − 1), λ > 0.

Then (we might assume that y_{i}, x_{i} > 0, we also assume that y is precisely the point on the boundary where the max is achieved)

\frac{∂F}{∂y_{i}} = 0, \frac{∂F}{∂λ} = 0, or x_{i} = λpy_{i}^{p−1}, ∑_{k=1}^{n} y_{k}^{p} = 1.

Rising the first equation to the power p/(p − 1) and summing up in i, we see that the second condition implies

∑_{i=1}^{n} x_{i}^{p/(p−1)} = (λp)^{p/(p−1)}, ( ∑_{i=1}^{n} x_{i}^{p/(p−1)})^{(p−1)/p} = λp.

But,

∑_{i=1}^{n} y_{i}x_{i} = λp ∑_{i=1}^{n} y_{i}^{p} = λp,

and

h_{B_{n}^{p}}(x) = x · y = ( ∑_{i=1}^{n} x_{i}^{p/(p−1)})^{(p−1)/p}, x ∈ S^{n−1}.

Observe that we have a maximum:

\frac{∂^{2}F}{∂y_{i}∂y_{j}} = 0,  j ≠ i,  \frac{∂^{2}F}{∂y_{i}^{2}} = −μp(p − 1)y_{i}^{p−2} < 0.