Page 163 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

150 Minimal surfaces

We therefore only need to show that if ku 0 k C 2,α ≤ , we have indeed (5.20).
Note that for every ν ≥ 2,wehave
⎧
⎨ ∆ (u ν+1 − u ν )= N (u ν ) − N (u ν−1 ) in Ω
⎩
u ν+1 − u ν =0 on ∂Ω .
From Step 1 and Step 2, we find that, for every ν ≥ 2,

ku ν+1 − u ν k 2,α ≤ C kN (u ν ) − N (u ν−1 )k 0,α
C C
(5.22)
2 2
≤ C (ku ν k C 2,α + ku ν−1 k C 2,α) ku ν − u ν−1 k C 2,α .
Similarly for ν =1,wehave

2 3
ku 2 − u 1 k C 2,α ≤ C kN (u 1 )k C 0,α ≤ C ku 1 k C 2,α . (5.23)
From now on, since all norms will be C 2,α norms, we will denote them simply
by k·k. From (5.22), we deduce that it is enough to show

2
2
C (ku ν k + ku ν−1 k) ≤ K, ν ≥ 2 (5.24)
to obtain (5.20) and thus the theorem. To prove (5.24), it is sufficient to show
that
µ 4 2 ¶
C
ku ν k ≤ C 1+ ,ν ≥ 2 . (5.25)
1 − K
The inequality (5.22) will then follow from the choice of in (5.21). We will
prove (5.25) by induction. Observe that by Step 1 and from (5.18), we have
µ µ 4 2 ¶¶
C
ku 1 k ≤ C ku 0 k ≤ C ≤ C 1+ . (5.26)
1 − K
We now show (5.25) for ν =2. We have, from (5.23) and from (5.26), that

³ ´
2 2 ¡ 4 2 ¢
ku 2 k ≤ ku 1 k + ku 2 − u 1 k ≤ ku 1 k 1+ C ku 1 k ≤ C 1+ C
and hence since K ∈ (0, 1), we deduce the inequality (5.25). Suppose now, from
the hypothesis of induction, that (5.25) is valid up to the order ν and let us
prove thatitholds true for ν +1. We trivially have that
ν
X
ku ν+1 k ≤ ku 1 k + ku j+1 − u j k . (5.27)
j=1