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 aL2([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 ||a1a0|| > 0; here, all norms are in L2([0, 1]). Then, images and images (min f)2.

If is easy to show that f2(a) is a convex function of a, and images 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 images follows from Remark 1.8.

From the convexity and strong convexity, ...

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