Is A−1 an infinitesimal generator?∗ - Applied Mathematics

Proof. We again concentrate on the integral term in (11). We have that, using property c.,∫ ∞0‖τh ac (tτ)T (t)x 0 ‖dt≤≤∫ ∞0∫ ∞Mτ|h ac (tτ)|Me −ωt ‖x 0 ‖dt0M 2= MM 2 ‖x 0 ‖τ 1 4τ(tτ) 3/4 e−ωt ‖x 0 ‖dt∫ ∞= MM 2 ‖x 0 ‖τ 1 π √ 24ω 1 4 Γ( 3 4 ).Combining this inequality with equation (11), we see that (15) holds.01t 3/4 e−ωt dtExample 3.5. In our example we take X = C 0 (0, 1), i.e., the space of all continuous functionson [0, 1), which have limit zero at one. This is a Banach space under the supremum norm.On this state space we choose as semigroup{f(t + η) t + η ∈ [0, 1)(T (t)f) (η) =(16)0 t + η ≥ 1It is not hard to see that this is a strongly continuous semigroup on X with ‖T (t)‖ ≤ 1, andT (t) = 0 for t ≥ 1. So it is a contraction semigroup with growth bound minus infinity. Inparticular, the semigroup is exponentially stable.From Lemma 3.2 we have that( )∫ 1−ηe A−1τ f (η) = f(η) − τh ac (tτ)f(t + η)dt= f(η) −0∫ (1−η)τ0( ) ˜th ac (˜t)fτ + η d˜t. (17)By constructing a suitable sequence of functions f’s, we show that ‖e A−1τ ‖ behaves like 4√ τ.Therefore we fix τ.Let φ ε be a continuous function on R which is positive, bounded by one, φ ε (0) = 0, andφ ε (t) = 1 for |t| ≥ ε. Furthermore, define ψ τ be a continuous function on [0, ∞) which ispositive, bounded by one, ψ τ (t) = 0 for t ≥ τ, and ψ τ (t) = 1 for t < τ − 1.Let t k , k = 1, . . . , K be the zeros of h ac in [0, τ].[ K]∏g ε (t) = φ ε (t − t k ) sign(h ac (t))ψ τ (t), t ≥ 0.k=1It is not hard to see that g ε is continuous on [0, ∞) with g ε (t) = 0 for t ≥ τ and sup t≥0 |g ε (t)| =1. Hence f(t) := −g ε (t/τ) is an element of X with norm one.Next choose in equation (17) η = 0 and f(t) = −g ε (τt), then( )∫ τ( ) ˜te A−1τ f (0) = f(0) − h ac (˜t)f0 τ + 0 d˜t∫≥ f(0) +|h ac (˜t)|d˜t.DRAFT{˜t∈[0,τ−1]||˜t−t k |≥ɛ, for k=1,...K}8