Page 33 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

20 Preliminaries

p
(iii) If 1 ≤ p< ∞ and if u ν u in L , then there exists a constant γ> 0
so that ku ν k L p ≤ γ,moreover kuk L p ≤ lim inf ν→∞ ku ν k L p.The result remains
∗
valid if p = ∞ and if u ν u in L .
∞
(iv) If 1 <p < ∞ and if there exists a constant γ> 0 so that ku ν k L p ≤ γ,
p p
then there exist a subsequence {u ν i } and u ∈ L so that u ν i u in L . The
∗
u in L .
∞
result remains valid if p = ∞ and we then have u ν i
p
(v) Let 1 ≤ p ≤∞ and u ν → u in L ,thenthere exist a subsequence {u ν i }
p
and h ∈ L such that u ν i → u a.e. and |u ν i | ≤ h a.e.
Remark 1.21 (i) Comparing (ii) and (iii) of the theorem, we see that the weak
convergenceensures thelower semicontinuity of thenorm, whilestrongconver-
gence guarantees its continuity.
n
(ii) The most interesting part of the theorem is (iv). We know that in R ,
Bolzano-Weierstrass Theorem ascertains that from any bounded sequence we can
p
extract a convergent subsequence. This is false in L spaces (and more generally
in infinite dimensional spaces); but it is true if we replace strong convergence by
weak convergence.
(iii) The result (iv) is, however, false if p =1; this is a consequence of
1
the fact that L is not a reflexive space. To deduce, up to the extraction of a
L 1 ≤ γ, we need
subsequence, weak convergence, it is not sufficient to have ku ν k
a condition known as “equiintegrability” (cf. the bibliography). This fact is the
reason that explains the difficulty of the minimal surface problem that we will
discuss in Chapter 5.

We now turn to Riemann-Lebesgue theorem that allows to easily construct
weakly convergent sequences that do not converge strongly. This theorem is
particularly useful when dealing with Fourier series (there u (x)=sin x or
cos x).

Theorem 1.22 (Riemann-Lebesgue Theorem). Let 1 ≤ p ≤∞, Ω =
Q
n p n
i=1 (a i ,b i ) and u ∈ L (Ω).Let u be extended by periodicity from Ω to R
and define
1 Z
u ν (x)= u (νx) and u = u (x) dx
meas Ω Ω
∗
p
then u ν u in L if 1 ≤ p< ∞ and, if p = ∞, u ν u in L .
∞
Proof. To make the argument simpler we will assume in the proof that
n
Ω =(0, 1) and 1 <p ≤∞. For the proof of the general case (Ω ⊂ R or p =1)