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Optimal Higher Order Time Discretizations 79
Proof. Let R ∈ D k be such that [ψ m (R)](x ∗ )=4forsomex ∗ ∈ ]0,α m (R)[.
(A similar argument works if [ψ m (R)](x ∗ ) = 0.) For any ε > 0, ψ m (R +
ε) =ψ m (R) +εx m+1 > 4 in a small neighborhood of x ∗ . This implies that
α m (R + ε) α m (R) − ε.
Now ε>0 is arbitrary small, so that lim inf α m (R n ) ≥ α m (R). The continuity
of α m at R follows, since α m is upper semi-continuous by Lemma 1. ⊓⊔
Lemma 3. The set of solutions of the optimization problem (20) is a convex
subset of D k .
Proof. Let us first prove that any local maximum of α m belongs to D k . Indeed,
it is easy to see that, if R/∈ D k , the function
t ∈ R ↦→ α m (R + t)
is continuous and strictly monotone in the neighborhood of the origin. This
shows that R cannot be a local maximum of α m .
Let R 1 and R 2 be two solutions of (20):
By definition of α m
α m (R 1 )=α m (R 2 )=α m.k ≡
sup α m (R).
R∈P k−1
∀x ≤ α m,k , 0 ≤ [ψ m (R 1 )] (x) ≤ 4 and 0 ≤ [ψ m (R 2 )] (x) ≤ 4.