Page 196 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 1: Preliminaries 183

Indeed, f being convex, we have

0 0
f (0) ≥ f (ξ) − ξf (ξ) ⇒ lim f (ξ)= ±∞ .
ξ→±∞
Moreover, since f 00 > 0, wehavethat f is strictly increasing and therefore
0
invertible and we hence obtain
ξ = f 0−1 (v) .
1
The hypotheses on f clearly imply that ξ = f 0−1 is C (R).
Step 3. We now conclude that
f ∗0 = f 0−1 . (7.13)

Indeed, we have from (7.12) that
0
0
∗0
f (v)= ξ (v)+ ξ (v) v − f (ξ (v)) ξ (v)
0
0
0
= ξ (v)+ ξ (v)[v − f (ξ (v))] = ξ (v) .
1
00
∗
Wethereforehavethat f is C (R) and (7.13). Furthermore, since f > 0,we
2
∗
deduce that f is as regular as f (so, in particular f is C ).
∗
Exercise 1.5.7. (i) Let h> 0. Use the convexity of f and (1.14) to write
p p
0
hf (x) ≤ f (x + h) − f (x) ≤ α 1 (1 + |x + h| )+ α 1 (1 + |x| )
p p
−hf (x) ≤ f (x − h) − f (x) ≤ α 1 (1 + |x − h| )+ α 1 (1 + |x| ) .
0
We can therefore find eα 1 > 0,so that
p p
e α 1 (1 + |x| + |h| )
0
|f (x)| ≤ .
h
Choosing h =1 + |x|,wecan surely find α 2 > 0 so that (1.15) is satisfied, i.e.,
³ ´
p−1
0
|f (x)| ≤ α 2 1+ |x| , ∀x ∈ R .
The inequality (1.16) is then a consequence of (1.15) and of the mean value
theorem.
(ii) Note that the convexity of f is essential in the above argument. Indeed,
2
taking, for example, f (x)= x+sin x ,we find that f satisfies (1.14) with p =1,
but it does not verify (1.15).
(iii) Of course if f satisfies (1.15), we have by straight integration
0
x
Z
f (x)= f (0) + f (s) ds
0
0
that f verifies (1.14), even if f is not convex.