By Hokkanen, Morosanu

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**Extra resources for Functional Methods For Differential Equaqtions**

**Example text**

12). 11), (P g, en )HE = 0 for all n ∈ IN∗ . Since {e1 , e2 , . } is a complete orthonormal basis of HE , P g = 0. Hence (g, v)H = 0 for all v ∈ HE . Since HE is dense in H, g = 0. It remains to prove (iii). 11) en + em = ∗ HE = sup v∈HE , v=0 (en + em , v)H (Qen + Qem , v)HE = sup = v HE v HE v∈HE , v=0 (µn en + µm em , v)HE = v HE v∈HE , v=0 sup µ2n + 2µn µm (en , em )HE + µ2m . By the parallelogram identity, 2(en , em )HE∗ = en + em 2 ∗ HE − en 2 ∗ HE − em 2 ∗ HE = 2(en , em )HE . λ n λm ∗ ∗ .

2) with (u0 , f ) := (u0n , fn ). 8) to conclude the proof. 3 (H. Br´ezis). 2) has a unique strong solution u such that t → t 2 u (t) belongs to L2 (0, T ; H), t → ψ u(t) is integrable on [0, T ] and absolutely continuous on [δ, T ], for all δ ∈ (0, T ). If, in addition, u0 ∈ D(ψ), then u ∈ L2 (0, T ; H), t → ψ u(t) is absolutely continuous on [0, T ], and ψ u(t) ≤ ψ(u0 ) + 1 2 T f (s) 2 H ds for all t ∈ [0, T ]. 3 Let C be a nonempty closed subset of H. A continuous semigroup of contractions on C is a family of operators S(t): C → C, t ≥ 0, satisfying: (A1) S(0)x = x for all x ∈ C; (A2) S(t + s)x = S(t)S(s)x for all x ∈ C, t, s ≥ 0; (A3) for every x ∈ C, the mapping t → S(t)x is continuous on [0, ∞); (A4) S(t)x − S(t)y H ≤ x−y H for all x, y ∈ C, t ≥ 0.