Fractional Brownian Motion

Solutions

Solutions to exercises from Chapter 1

Solution to Exercise 1.1. Let us establish existence and uniqueness of the point of minimum.

1. 1) Existence of the point of minimum follows from section 1.4, because the particular form of the kernel z(t, s) was never used in the proof.
2. 2) Similarly, uniqueness can be obtained in the same way as it was done in section 1.6. However, we put a short proof of uniqueness here.

Note that f(a) ≥ 0 for all a ∈ L₂([0,1]). Hence, the functionals f(a) and (f(a))² reach their minima on the same set of points.

Suppose that f(a) reaches minimum at points a₁ and a₀ with essentially different functions a₁ and a₀, which means that ||a₁ — a₀|| > 0; here, all norms are in L₂([0, 1]). Then, and (min f)².

If is easy to show that f²(a) is a convex function of a, and z(1, •)||² is a strictly convex function of a. The convexity of f²(a) follows from the facts that square function is convex and increasing in the range of f(a) and the functional f(a) is convex due to Lemma 1.4. The strict convexity of follows from Remark 1.8.

From the convexity and strong convexity, ...