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Optimal Higher Order Time Discretizations 77
α m,k = α m (R m,k ). (21)
Clearly, since P k−1 ⊂ P k , α m,k increases with k. We have also α m,k > 0, since
P m (0) = 1 (m ≥ 1).
For what follows, it is useful to introduce the following affine map:
ψ m : P → P
(22)
∣ R → ψ m (R) =Q m + x m+1 R,
where we recall that Q m (x) =xP m (x). Note that ψ m maps P k−1 into P m+k .
Lemma 1. The function R ∈ P k−1 ↦→ α m (R) ∈ R ∗ + has the following properties:
(i) It goes to 0 at infinity:
lim α m(R) =0.
‖R‖→+∞
(ii) It is upper semi-continuous:
R n → R in P k−1 =⇒ α m (R) ≥ lim sup α m (R n ).
Proof. Let r j (R) denote the coefficient of x j in R ∈ P k−1 and consider R n ∈
P k−1 such that
‖R n ‖ ∞ ≡ sup |r j (R n )|−→+∞.
0≤j≤k−1
Referring to the fact that P k−1 is finite dimensional, one can find a subsequence
(still denoted R n for simplification) and a fixed non-zero polynomial
ϕ ∈ P k−1 such that, as soon as ϕ(x) ≠0,
R n (x) ∼‖R n ‖ ∞ ϕ(x)
(n → +∞).
For such positive values of x, [ψ m (R n )](x) /∈ [0, 4] for sufficiently large n
which means that α m (R n ) < x =⇒ lim sup α m (R n ) < x. Since ϕ is a
non-zero polynomial, one can find arbitrarily small values of such x so that
lim sup α m (R n ) ≤ 0. As α m (R n ) is a sequence of positive real numbers, this
means that α m (R n ) tends to 0.
On the other hand, let R n ∈ P k−1 be a sequence converging to R. Letε
be any arbitrarily small positive number. By the uniform convergence of R n
to R in the interval I R (ε) =[0,α(R)+ε] wehave:
lim ‖ψ m(R n ) − 2‖ L
n→+∞ ∞ (I R(ε)) = ‖ψ m (R) − 2‖ L ∞ (I R(ε)) > 2.
Thus, there exists an integer N ε such that:
n ≥ N ε =⇒ ‖ψ m (R n ) − 2‖ L ∞ (I R(ε)) > 2 =⇒ α m (R n )