Kenneth Kuttler

19.14. GENERAL THEORY OF CONTINUOUS SEMIGROUPS 579

(λ I−A)−1 on X . The estimate 19.14.79 follows easily from the definition of R(λ ) when-ever λ > α as follows.

∥R(λ )x∥=∥∥∥(λ I−A)−1 x

∥∥∥= ∥∥∥∥∫ ∞

0e−λ tS (t)xdt

∥∥∥∥≤∫

∞

0e−λ tMeαtdt ∥x∥ ≤ M

|λ −α|∥x∥

Why is D(A) dense? I will show that ∥λR(λ )x− x∥ → 0 as λ → ∞, and it was shownabove that R(λ )x and therefore λR(λ )x ∈D(A) so this will show that D(A) is dense in X .For λ > α where ∥S (t)∥ ≤Meαt ,

∥λR(λ )x− x∥=∥∥∥∥∫ ∞

0λe−λ tS (t)xdt−

∫∞

0λe−λ txdt

∥∥∥∥≤∫

∞

∥∥∥λe−λ t (S (t)x− x)∥∥∥dt

=∫ h

∥∥∥λe−λ t (S (t)x− x)∥∥∥dt +

∫∞

∥∥∥λe−λ t (S (t)x− x)∥∥∥dt

≤∫ h

∥∥∥λe−λ t (S (t)x− x)∥∥∥dt +

∫∞

hλe−(λ−α)tdt (M+1)∥x∥

Now since S (t)x− x→ 0, it follows that for h sufficiently small

≤ ε

∫ h

0λe−λ tdt +

λ −αe−(λ−α)h (M+1)∥x∥

≤ ε

λ −αe−(λ−α)h (M+1)∥x∥< ε

whenever λ is large enough. Thus D(A) is dense as claimed.Why is A a closed operator? Suppose xn → x where xn ∈ D(A) and that Axn → ξ . I

need to show that this implies that x ∈ D(A) and that Ax = ξ . Thus xn→ x and for λ > α,

(λ I−A)xn→ λx−ξ . However, 19.14.79 shows that (λ I−A)−1 is continuous and so

xn→ (λ I−A)−1 (λx−ξ ) = x

It follows that x ∈ D(A) . Then doing (λ I−A) to both sides of the equation, λx− ξ =λx−Ax and so Ax = ξ showing that A is a closed operator as claimed.

Definition 19.14.6 The linear mapping for λ > α where ∥S (t)∥ ≤Meαt given by

(λ I−A)−1 = R(λ )

is called the resolvent.

The following corollary is also very interesting.

19.14. GENERAL THEORY OF CONTINUOUS SEMIGROUPS 579(AI—A)~| on X. The estimate 19.14.79 follows easily from the definition of R (A) when-ever A > a as follows.|R(A)x|| =|](a7—a)' a = |I e*'S (t) xdt0< fe MeMat |x| < Lx0=||A —a|Why is D(A) dense? I will show that ||AR(A)x—x|| + 0 as A > oo, and it was shownabove that R(A).x and therefore 1R (A) x € D(A) so this will show that D(A) is dense in X.For A > @ where ||S(t)|| < Me™,|AR(A)x—xl| = | [ret stoysar— [ae Maar0 0< [ae (S(t)x—x)|| ae= [ Jae™ (S(t)x—)|]ar+ [ Jae™ (S(¢)x—a)||atfers sacs [Pace rarars iNow since S (t)x —x — 0, it follows that for / sufficiently small€ fh oa A —(A-a@)h<os — M+1<5 [de Mart Se OO" (M+ 1)€ A< _ste ON (M+ 1) [xl] <whenever A is large enough. Thus D(A) is dense as claimed.Why is A a closed operator? Suppose x, — x where x, € D(A) and that Ax, — €. Ineed to show that this implies that x € D(A) and that Ax = €. Thus x, > x and ford >a,(AI —A) x, — Ax —&. However, 19.14.79 shows that (AJ —A)~' is continuous and soXy 3 (AI—A)' (Ax—&) =xIt follows that x € D(A). Then doing (AJ—A) to both sides of the equation, Ax — § =Ax — Ax and so Ax = & showing that A is a closed operator as claimed. JDefinition 19.14.6 The linear mapping for A > a where ||S(t)|| <Me™ given by(AI—A) | =R(A)is called the resolvent.The following corollary is also very interesting.