## Miersemann E. 's Calculus of variations PDF

By Miersemann E.

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Extra resources for Calculus of variations

Sample text

NECESSARY CONDITIONS and that gj ∈ C 1 (Rn ) for each j ∈ I ∪ E. Let x ∈ V be a local minimizer of f in V and let I0 ⊆ I be the subset of I where the inequality constraints are active, that is, I0 = {j ∈ I; gj (x) = 0}. Let w ∈ T (V, x) and xk , tk are associated sequences to w, then for k ≥ k0 , k0 sufficiently large, we have for each j ∈ I0 0 ≥ gj (xk ) − gj (x) = gj (x), xk − x + o(||xk − x||). It follows that for each j ∈ I0 gj (x), w ≤ 0 for all w ∈ T (V, x). If j ∈ E, we obtain from 0 = gj (xk ) − gj (x) = gj (x), xk − x + o(||xk − x||) and if j ∈ E, then gj (x), w = 0 for all w ∈ T (V, x).

Exercise. In the following we use for (fx1 (x), . . , fxn ) the abbreviations f (x), ∇f (x) or Df (x). 4. Suppose that V ⊂ X is convex. Then f is convex on V if and only if f (y) − f (x) ≥ f (x), y − x for all x, y ∈ V. Proof. (i) Assume f is convex. Then for 0 ≤ λ ≤ 1 we have f (λy + (1 − λ)x) ≤ λf (y) + (1 − λ)f (x) f (x + λ(y − x)) ≤ f (x) + λ(f (y) − f (x)) f (x) + λ f (x), y − x + o(λ) ≤ f (x) + λ(f (y) − f (x)), which implies that f (x), y − x ≤ f (y) − f (x). (ii) Set for x, y ∈ V and 0 < λ < 1 x1 := (1 − λ)y + λx and h := y − x1 .

Suppose that w ∈ T (V, x) and that tk , xk → x are associated sequences, that is, xk ∈ V , tk > 0 and wk := tk (xk − x) → w. Then f (x)wk , wk + ||wk ||2 η(||xk − x||) ≥ 0, which implies that f (x)w, w ≥ 0. 4 that f is convex on V . 1. 1 Exercises 1. Assume V ⊂ Rn is a convex set. Show that Y = V − V := {x − y : x, y ∈ V } is a convex set in Rn , 0 ∈ Y and if y ∈ Y then −y ∈ Y . 2. 3. 3. Show that T (V, x) is a cone with vertex at zero. 4. Assume V ⊆ Rn . Show that T (V, x) = {0} if and only if x ∈ V is not isolated.