ISEM Team-Lecce. holds for some c > 0 and all y Y. Assume T (t)y Y for all t 0 and T ( )y C(R +, Y ) for all y Y. For t 0 we define the operators

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1 ISEM Tem-Lecce EXERCISE 3.. Let A generte the C -semigroup T ( ) on Bnch spce X. Let J : X E be n isomorphism to nother Bnch spce E, Y X be Bnch subspce which is equipped with norm Y such tht X c Y holds for some c > nd ll Y. Assume T (t)y Y for ll t nd T ( ) C(R +, Y ) for ll Y. For t we define the opertors S(t) = JT (t)j on E, T Y (t) = T (t) on Y. Show tht S( ) nd T Y ( ) re C -semigroups on E nd Y, respectivel, nd compute their genertors. How does the result simplif if Y (nd thus Y is closed in X)? is equivlent to X Let us consider S(t) = JT (t)j on E nd let (B, D(B)) be its genertor. Obviousl from the linerit nd the continuit of J nd J we know tht S(t) is C -semigroup. Let us show tht D(B) = J(D(A)) nd B = JAJ. Let x be in J(D(A)); we hve: S(t)x x JT (t)j x JJ x = = ( T (t)j x J ) x = J = JAJ x The lst equlit proves tht J(D(A)) D(B) nd tht B = JAJ on J(D(A)). A similr rgument pplied to J proves tht J (D(B)) D(A) or, equivlentl, D(B) J(D(A)). Let us consider now T Y (t) = T (t) Y on Y. B hpothesis, to show tht T Y (t) is C -semigroup we onl hve to prove tht T Y (t) B(Y ) for ll t. Since T Y (t) is defined on the whole of Y, from the closed grph theorem, we cn equivlentl show tht T Y (t) is closed. Let us fix t > nd consider the sequences x n x nd T Y (t)x n, which converge in Y with respect to Y. It follows from the hpothesis tht the two sequences converge with respect to X s well. Since T Y (t) is continous, we hve T Y (t)x = nd so T Y (t) is closed. Let (C, D(C)) be its genertor. Let us define in Y the liner opertor (A, D(A )) b D(A ) = D(A) Y A (Y ), A = A on D(A ). D(A ) is the lrgest set in which the restriction A of A in Y is well defined. Trivill, C A, i.e., A is n extension of C. If x D(C) Y, the following it exists in Y with respect to Y : T Y (t)x x T (t)x x = = Cx.

2 Hence, it follows from the hpothesis tht the sme it reltion holds with respect to X. It follows tht x D(A) nd Ax = Cx Y. For λ R, λ > mx ω (A), ω (C)}, b Proposition 3.4 we hve λ ρ(a) ρ(c) nd, for ll Y : R(λ, C) = So, for ll x D(A ), we hve: e λt T Y (t) dt = e λt T (t) dt = R(λ, A). x = R(λ, A)(λI A)x = R(λ, C)(λI A)x D(C), which proves tht D(A ) D(C) nd then A = C. If Y is equivlent to X then Y is closed in X. For ll x Y D(A), we hve T Y (t)x x Ax = Y = Y nd so Y D(A) A (Y ) = A (Y ). D(C) = D(A) Y nd C = A Y. It follows from the first prt tht INTERMEZZO. Let J R be n intervl. Given u n W p (J), if u n u nd u n f in L p (J) then u W p (J) nd u = f. Proof. If u n u nd u n f in L p (J), then u n u nd u n f with respect to in ll compct intervls in J. So, for ll x, J we hve u n (x) u n () = u n(t)dt f(t)dt. We know tht u n u.e. up to subsequences, so b the bove reltion the entire sequence u n u.e.; finll, we obtin u(x) u() = which proves tht u W p (J) nd u = f. f(t)dt, EXERCISE 3.2. Let X = L p (R), p [, + ) nd T (t)f = f( + t) for t R. From Exercise 2.2 we know tht T ( ) is C -group. Show tht its genertor is given b Au = u on D(A) = W p (R). Further show tht σ(a) = ir. Let us denote b (A, D(A)) the genertor of T ( ) nd define the liner opertor (B, D(B)) b D(B) = W p (R) Bu = u 2

3 We hve immeditel tht (B, D(B)) is closed opertor. In fct, s proved in the previous intermezzo, if u n u in L p (R) nd Bu n = u n f in L p (R), then u W p (R) nd Bu = u = f. In order to show tht A = B, we firstl prove tht A B; if f D(A) nd g = Af, then t (T (t)f f) converges to g in Lp (R) s t +. So for ll, b R h + f(s + h) f(s) ds = h B Lebesgue differentition theorem, we hve [ f(s + h) f(s) ds = h + h h + h +h = f(b) f().e. b g(s) ds. f(s) ds h +h ] f(s) ds = So we hve proved f D(B) nd Af = f = Bf. Now, since T ( ) is contrction group, b Proposition 3.4 we hve λ ρ(a) for ll λ >.Conversel, for λ >, solving in L p (R) the differentil eqution u λu = g for g L p (R), we find tht (λi B) is bijective nd R(λ, B)g(t) = e λ(s t) g(s) ds. Therefore λ ρ(a) ρ(b) for ll λ > nd b Lemm 3.6 we hve A = B. Let us prove tht σ(a) = ir. Appling Proposition 3.4 to the contrction C -semigroups T (t) nd S(t) defined b S(t)f := f( t) we obtin tht σ(a) ir. To verif the reverse inclusion, let λ R nd tke the following functions ϕ n Cc (R) Wp (R) = D(A): if s 2n if s n ϕ n (s) := n n (2n s) (2n + s) if n s 2n if 2n s n We notice tht ϕ n, ϕ n p + nd ϕ n p = (2n p ) p for p > nd is bounded for p =. Now define, for ll n N, the function f n = ϕ n e iλ. Immeditel, f n W p (R) = D(A) nd f n p = ϕ n p + s n. Moreover, we hve: Af n = f n = iλf n + ϕ ne iλ, hence iλf n Af n p = ϕ n p s n for p > nd is bounded for p =. This implies tht iλ A cnnot hve bounded inverse, otherwise we would hve f n p. So ir σ(a) nd finll σ(a) = ir. EXERCISE 3.3. Let X = L p (, ) nd p [, + ). For t nd s (, ) set f(s + t), s + t < T (t)f(s) =, s + t 3

4 From Exercise 2.3 we know tht T ( ) is C -semigroup. Show tht its genertor is given b Au = u on D(A) = u W p (, ) u() = }. Let (B, D(B)) be the genertor of T (t). In order to prove tht B = A, let us first show tht B A. As in the previous exercise, we find tht if f D(B) then f Wp (, ) nd Bf = f. Furthermore, T (t)f D(B) Wp (, ) for ll t, so the left trnsltion of f is continuous nd this fct implies tht f() = necessril. We hve just proved tht D(B) D(A) nd A = B on D(B). Since T (t) is contrction semigroup, for ll λ > we hve tht λ ρ(b) nd: R(λ, B)g(t) = Solving the Cuch problem for g L p (, ): u λu = g we obtin: t R(λ, A)g(t) = e λ(s t) g(s) ds, g L p (, ). u() = t e λ(s t) g(s) ds So, if λ >, then λ ρ(a) ρ(b) nd, b lemm 3.6, A = B. EXERCISE 3.4. A C -semigroup T ( ) is clled qusicontrctive if there is number ω R such tht T (t) e ωt holds for ll t. Chrcterize the genertors of such C -semigroups s in Theorem 3.8 (without using Theorem 3.). Find n equivlent norm on X = C (R) such tht the left trnsltion group is not qusicontrctive for this norm. For ever ω R we m define the semigroup S(t) := e ωt T (t). Trivill, T (t) is qusicontrctive with T (t) e ωt if nd onl if S(t) is contrctive. Furthermore, denoting with (B, D(B)) nd (A, D(A)) the genertors of S(t) nd T (t) respectivel, we hve: nd λ ρ(b) λ + ω ρ(a) R(λ, A) = R(λ ω, B) Hence, using the Hille-Yosid Theorem, we cn chrcterize qusicontrctive semigroup T (t) b the equivlence of the following sttements: (i) T (t) is qusicontrctive with T (t) e ωt for ll t (ii) The genertor A of T (t) is closed, densel defined nd the following estimte holds: λ > ω λ ρ(a) nd (λ ω)r(λ, A) 4

5 (iii) The genertor A of T (t) is closed, densel defined nd the following estimte holds: λ C, Re(λ) > ω, λ ρ(a) nd R(λ, A) Re(λ) ω. Let T (t) be qusicontrctive semigroup on C (R). Now, to find n equivlent norm on C (R) such tht T (t) is not qusicontrctive, we define the following function: 2 if s w(s) := + s if < s < if s. For ll f C (R), let us consider the norm given b: f w := sup f(s)w(s). s R This norm is obviousl equivlent to, s for ll f C (R) we hve: f f w 2 f. Let now h be the function in C (R) defined b: h(s) := if s s if s It is immeditel seen tht h w =. Indeed ( h w = mx sup [,] ( s ), sup( s ) [,] Now, let us estimte T (t) for ll < t <. We hve: T (t) w = ) =. sup T (t)f w = sup f( + t) w h( + t) w = f w= f w= = sup h(s + t)w(s) h( t + t)w( t) = w( t) = + t s R Since T () = nd the right derivte of + t is infinite in t =, for ever ω > there is right neighborhood U of such tht T (t) > e ωt for ll t U \ }. Finll, for ever ω >, nd hence for ever ω R, there exists t > such tht T (t) > e ωt. So, the semigroup T (t) is not qusicontrctive. 5