hLu, ue i ≥eckuk2 for all u∈X.
Proof. We decompose u =v +y into v ∈ V and y ∈ Y. The positivity of L on the subspace Y and LX⋆
←X ≤C for some C >0 lead to hLu, ui=hLy, yi+hLy, vi+hLv, yi+hLv, vi
≥ckyk2−C(2kykkvk+kvk2)≥mkyk2−Mkvk2,
where the last step is due to LemmaA.4.1. The proof is finished by applying the squared triangle inequality
kuk2 ≤(kyk+kvk)2 ≤2kyk2+ 2kvk2 to the positivity estimate
hLu, ue i=hLu, ui+λkvk2≥mkyk2+ (λ−M)kvk2, where we have to choose λ > M =C+ C2
2c.
A.6 Lipschitz Inverse
Lemma A.6.1. Let X,·
X
and Y,·
Y
be Banach spaces withx0 ∈X and denote by L: X → Y a linear homeomorphism. If there exist positive constants δ, c1, c2 >0 and a mapping F: Bδ(x0)⊆X →Y such that
(i) F(x1)−F(x2)Y ≤c0
x1−x2
X,
(ii) c0 < c1 ≤ 1 L−1
X←Y
,
(iii) Lx0+F(x0)Y ≤δ(c1−c0), then the equation
(L+F)(x) = 0
has a unique solution x⋆ ∈ Bδ(x0), and the stability estimate x1−x2X ≤ 1
c1−c0
(L+F)(x1)−(L+F)(x2)Y
holds for all x1, x2 ∈ Bδ(x0).
140 Appendix A. Auxiliaries Proof. By defining T(x) =−L−1F(x), we rewrite the equationLx+F(x) = 0 as an equivalent fixed point problem T(x) =x. From the inequality
T(x1)−T(x2)X ≤L−1X
←Y
F(x1)−F(x2)Y ≤ c0
c1
x1−x2
X
for x1, x2 ∈ Bδ(x0) and T(x1)−x0
X ≤T(x1)−T(x0)X +T(x0)−x0
X
≤ c0 c1
x1−x0
X +L−1
X←Y
F(x0) +Lx0
Y
≤ c0 c1δ+ 1
c1δ(c1−c0) =δ,
we conclude thatT is a contraction on the closed ballBδ(x0). Hence, the existence of a unique solution follows from the contraction mapping principle. Moreover, the stability estimate is a consequence of
x1−x2
X ≤(I−T)x1−(I−T)x2
X +T x1−T x2
X
≤L−1X
←Y
(L+F)x1−(L+F)x2
Y +c0 c1
x1−x2X
for x1, x2 ∈ Bδ(x0).
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