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ϵ, satisfiesP u∈C0∞(ω) andP u= 0 in [p.3 ↑3 — p.4 ↓15] Proof of Proposition 1.1 should be replaced by: Let ω be the open set in Definition 1.1

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シェア "ϵ, satisfiesP u∈C0∞(ω) andP u= 0 in [p.3 ↑3 — p.4 ↓15] Proof of Proposition 1.1 should be replaced by: Let ω be the open set in Definition 1.1"

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Hyperbolic Systems With Analytic Coefficients:正誤表

[p.2 8] with|τ|< ϵ, satisfiesP u= 0 in =with|τ|< ϵ, satisfiesP u∈C0(ω) andP u= 0 in

[p.3 3 — p.4 15] Proof of Proposition 1.1 should be replaced by: Let ω be the open set in Definition 1.1. Take an open setV such thatKVω. Then for any f C0(Vϵ) there exists a unique u ∈H(ω) satisfying P u =f in ω and vanishing in x0 ≤ −ϵ. Denote by T the map T : C0(Vϵ)3f 7→u∈ H(ω). Note that H(ω) is a Fr´echet space equipped with countable semi- norms k · kHp(ω), p = 0,1, . . .. Assume that C0(Vϵ) 3 fj f in C0(Vϵ) and T fj = uj u in H(ω). SinceP uj = fj it is clear that P u = f and u= 0 in x0 ≤ −ϵ. From the uniqueness of the solution one has T f = uand hence the graph of T is closed. From the Banach’s closed graph theorem it follows thatT is a continuous map. Therefore for anyp∈Nthe inverse image of{u∈H(ω)| kukHp(ω) <1}, which is a neighborhood of 0 in H(ω), is a neighborhood of 0 inC0(Vϵ), that is there existδ >0 andq∈Nsuch that

f ∈C0(Vϵ), kfkHq(V)< δ=⇒ kT fkHp(ω)<1.

For any f C0(Vϵ) the Hq(V) norm of δf /kfkHq(V) is less than 1 then from the uniqueness of the solution we conclude that for anyf ∈C0(Vϵ) and u∈H(ω) satisfying P u=f inω and vanishing inx0≤ −ϵ satisfies

kukHp(ω)≤δ1kfkHq(V).

[p.21 12] polynomial inx=polynomial iny [p.27 10]Proof of Proposition 1.6. =⇒Proof.

[p.29 9]Proof of Proposition 1.6. =⇒Proof.

[p.72 5]hρ(sY −tX) = (−1)rhρ (−sY+tX) = (−1)rhρ (Y)∏

(−s−λj(tX)) = hρ(sY −tX) = (−1)rhρ(−sY +tX) = (−1)rhρ(Y)∏

(−s−λj(tX)) [p.125 5]C|x|2lt(x)2(Qql1)

l1+l2l

φ(x)

ε(x) |∂tQ+1+l1xl2f|2dxdt

=⇒C|x|2l|r|2(qkl)t(x)2(Qql1)

l1+l2l

t(x)

ε(x) |∂tQ+1+l1lx2f|2dt [p.125 4]C|t−ε|2(Qk)t

ε|∂tQ+1xlf|2dxdt

=⇒Delete

[p.125 3] for q+l+ 1≤Q,k+l ≤q and =for|t| ≤t(x),q+l+ 1≤Q, k+l≤qand

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[p.130 7〜11] should be replaced by

|∂tkxlrqFq1|2

≤C

k1+k2k

|r|2(qlk1)|x|2lt(x)2(Qqlk21)+1

l1+l2l

t ε

|∂tQ+1+l1xl2f|2dt

≤C|x|2l|r|2(qlk)t(x)2(Qql1)+1

l1+l2l

t ε

|∂tQ+1+l1lx2f|2dt

hence we conclude the proof.

[p.132 10] Since tpu= 0 on t = sν(x) and t = σν+1(x), |x| = δ(T−t) are space-like curves =Sincetpu= 0 ont=sν(x)

[p.164 1]P(x) =⇒P(ξ)

[p.197 10]intervals=⇒neighborhoods

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