Page 192 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 1: Preliminaries 179

We hence obtain, appealing to the two above relations, that

¯Z ¯ Z
¯ £ ¤ ¯ £ ¯ ¯¤
¯
¯ ϕ + uϕ dx ¯ ≤ ||ϕ − ψ| + |u| ϕ − ψ ¯ dx
u x i |u x i
¯ x i ¯ x i x i
Ω Ω
£ ° ° ¤
°
°
≤ kuk W 1,p kψ − ϕk L p 0 + ϕ x i − ψ x i L p 0 ≤ kuk W 1,p .
Since is arbitrary, we have indeed obtained that
Z Z
ϕdx = − uϕ dx, i =1, ..., n .
u x i
x i
Ω Ω
7.1.4 Convex analysis
Exercise 1.5.1. (i) We firstshowthatif f is convex then, for every x, y ∈ R,
0
f (x) ≥ f (y)+ f (y)(x − y) .
Apply the inequality of convexity

1
[f (y + λ (x − y)) − f (y)] ≤ f (x) − f (y)
λ
and let λ → 0 to get the result.
We next show the converse. Let λ ∈ [0, 1] and apply the above inequality to
find

f (x) ≥ f (λx +(1 − λ) y)+ (1 − λ) f (λx +(1 − λ) y)(x − y)
0
f (y) ≥ f (λx +(1 − λ) y) − λf (λx +(1 − λ) y)(x − y) .
0
Multiplying the first inequality by λ, the second one by (1 − λ) and summing
the two of them, we have indeed obtained the desired convexity inequality

f (λx +(1 − λ) y) ≤ λf (x)+ (1 − λ) f (y) .
(ii) We next show that f is convex if and only if, for every x, y ∈ R,

0
[f (x) − f (y)] (x − y) ≥ 0 . (7.10)
0
Once this will be established, we will get the claimed result, namely that if
2
f ∈ C (R) then f is convex if and only if f (x) ≥ 0 for every x ∈ R.We start
00
by assuming that f is convex and hence apply (i) to get
f (x) ≥ f (y)+ f (y)(x − y)
0
f (y) ≥ f (x)+ f (x)(y − x) .
0