2 Unbounded operators We will gather some information on operators in Banach and Hilbert spaces. Throughout this chapter let X0, X1; and X2 be Banach spaces and H0, H1, and H2 be Hilbert spaces over the eld K 2 fR; Cg. 2.1 Operators in Banach spaces The main dierence of continuous (or bounded) linear operators, that is, ( ) kBxk L(X0;X1) := B : X0 ! X1 ; B linear, kBk := sup < 1 x2X0nf0g kxk (with the usual abbreviation L(X0) := L(X0;X0)) and the set of unbounded linear operators is that the latter are dened only on a subset of X0. In order to dene unbounded linear operators, we will rst take a more general point of view and introduce (linear) relations. This perspective will turn out to be the natural setting later on. Denition. A subset A ⊆ X0 × X1 is called a relation in X0 and X1. We dene the domain, range and kernel of A as follows dom(A) := fx 2 X0 ; 9y 2 X1 :(x; y) 2 Ag ; ran(A) := fy 2 X1 ; 9x 2 X0 :(x; y) 2 Ag ; ker(A) := fx 2 X0 ;(x; 0) 2 Ag : The image, A[M], of a set M ⊆ X0 under A is given by A[M] := fy 2 X1 ; 9x 2 M :(x; y) 2 Ag : A relation A is called bounded, if for all bounded M ⊆ X0 the set A[M] ⊆ X1 is bounded. For a given relation A we dene the inverse relation −1 A := f(y; x) 2 X1 × X0 ;(x; y) 2 Ag : A relation A is called linear, if for all (x; y); (u; v) 2 A, λ 2 K we have (x+λu, y+λv) 2 A. A linear relation A is called linear operator or just operator from X0 to X1, if A[f0g] = fy 2 X1 ; (0; y) 2 Ag = f0g: In this case, we also write A: dom(A) ⊆ X0 ! X1 to denote a linear operator from X0 to X1. Moreover, we shall write Ax = y instead of (x; y) 2 A in this case. A linear operator A, which is not bounded, is called unbounded. 12 2 Unbounded operators For completeness, we also dene the sum, composition, and scalar multiples of relations. Denition. Let A ⊆ X0 × X1, B ⊆ X0 × X1 and C ⊆ X1 × X2 be relations, λ 2 K. Then we dene A + B := f(x; y + w) 2 X0 × X1 ;(x; y) 2 A; (x; w) 2 Bg ; λA := f(x; λy) 2 X0 × X1 ;(x; y) 2 Ag ; CA := f(x; z) 2 X0 × X2 ; 9y 2 X1 :(x; y) 2 A; (y; z) 2 Cg : For a relation A ⊆ X0 × X1 we will use the abbreviation −A := −1A (so that the minus sign only acts on the second component). We now proceed with topological notions for relations. Denition. Let A ⊆ X0 × X1 be a relation. A is called densely dened, if dom(A) is dense in X0. We call A closed, if A is a closed subset of the direct sum of the Banach spaces X0 and X1. If A is a linear operator then we will call A closable, whenever A ⊆ X0 × X1 is a linear operator. Proposition 2.1.1. Let A ⊆ X0 × X1 be a relation, C 2 L(X2;X0) and B 2 L(X0;X1). Then the following statements hold. (a) A is closed if and only if A−1 is closed; (b) A is closed if and only if A + B is closed; (c) if A is closed, then AC is closed. Proof. The rst statement follows upon realising that X0×X1 3 (x; y) 7! (y; x) 2 X1×X0 is an isomorphism. For the second statement, it suces to show that the closedness of A implies the same for . Let be a sequence in convergent in A+B ((xn; yn))n A+B X0 ×X1 to some . Since , it follows that in is convergent (x; y) B 2 L(X0;X1) ((xn; yn − Bxn))n A to (x; y − Bx) in X0 × X1. Since A is closed, (x; y − Bx) 2 A. Thus, (x; y) 2 A + B. For the third statement of the present proposition, let be a sequence in ((wn; yn))n AC convergent in to some . Since is continuous, converges to . X2 × X1 (w; y) C (Cwn)n Cw Hence, (Cwn; yn) ! (Cw; y) in X0 × X1 and since (Cwn; yn) 2 A and A is closed, it follows that (Cw; y) 2 A. Equivalently, (w; y) 2 AC, which yields closedness of AC. We shall gather some other elementary facts about closed operators in the following. We will make use of the following notion. Denition. Let A: dom(A) ⊆ X0 ! X1 be a linear operator. Then the graph norm of q is dened by 2 2. A dom(A) 3 x 7! kxkA := kxk + kAxk Lemma 2.1.2. Let A: dom(A) ⊆ X0 ! X1 be a linear operator. Then A is closed if and only if dom(A) equipped with the graph norm is a Banach space if and only if for all in convergent in such that is convergent in we have (xn)n dom(A) X0 (Axn)n X1 limn!1 xn 2 dom(A) and A limn!1 xn = limn!1 Axn. 13 2 Unbounded operators Proof. For the rst equivalence, it suces to observe that dom(A) 3 x 7! (x; Ax) 2 A, where dom(A) is endowed with the graph norm, is an isomorphism. The last statement is an easy reformulation of the denition of closedness of A ⊆ X0 × X1. Unless explicitly stated otherwise (e.g. in the form dom(A) ⊆ X0, where we regard dom(A) as a subspace of X0), for closed operators A we always consider dom(A) as a Banach space on its own right; that is, we shall regard it as being endowed with the graph norm. Lemma 2.1.3. Let A: dom(A) ⊆ X0 ! X1 be a closed linear operator. Then A is bounded if and only if dom(A) ⊆ X0 is closed. Proof. First of all note that boundedness of A is equivalent to the fact that the graph norm and the X0-norm on dom(A) are equivalent. Hence, the closedness and bounded- ness of A implies that dom(A) ⊆ X0 is closed. On the other hand, the embedding ι: (dom(A); k·k ) ,! (dom(A); k·k ) A X0 is continuous and bijective. Since the range is closed, the open mapping theorem implies −1 that ι is continuous. This yields the equivalence of the graph norm and the X0-norm and, thus, the boundedness of A. For unbounded operators, obtaining a precise description of the domain may be dicult. However, there may be a subset of the domain which essentially (or approximately) describes the operator. This gives rise to the following notion of a core. Denition. Let A ⊆ X0 × X1. A set D ⊆ dom(A) is called a core for A provided A \ (D × X1) = A. Proposition 2.1.4. Let A 2 L(X0;X1), and D ⊆ X0 a dense linear subspace. Then D is a core for A. Corollary 2.1.5. Let A: dom(A) ⊆ X0 ! X1 be a densely dened, bounded linear operator. Then there exists a unique B 2 L(X0;X1) with B ⊇ A. In particular, we have B = A and kAxk kBk = sup : x2dom(A);x6=0 kxk The proofs of Proposition 2.1.4 and Corollary 2.1.5 are asked for in Exercise 2.2. 2.2 Operators in Hilbert spaces Let us now focus on operators on Hilbert spaces. In this setting, we can additionally make use of scalar products h·; ·i, which in this course are considered to be linear in the second argument (and anti-linear in the rst, in the case when K = C). For a linear operator A: dom(A) ⊆ H0 ! H1 the graph norm of A is induced by the scalar product (x; y) 7! hx; yi + hAx; Ayi ; 14 2 Unbounded operators known as the graph scalar product of A. If A is closed then dom(A) (equipped with the graph norm) is a Hilbert space. Of course, no presentation of operators in Hilbert spaces would be complete without the central notion of the adjoint operator. We wish to pose the adjoint within the relational framework just established. The denition is as follows. ∗ Denition. For a relation A ⊆ H0 × H1 we dene the adjoint relation A by ∗ −1? A := − A ⊆ H1 × H0; where the orthogonal complement is computed in the direct sum of the Hilbert spaces H1 and H0; that is, the set H1 × H0 endowed with the scalar product (x; y); (u; v) 7! hx; ui + hy; vi. Remark 2.2.1. Let A ⊆ H0 × H1. Then we have ∗ A = f(u; v) 2 H1 × H0 ; 8(x; y) 2 A : hu; yi = hv; xig : In particular, if A is a linear operator, we have ∗ A = f(u; v) 2 H1 × H0 ; 8x 2 dom(A): hu; Axi = hv; xig : ∗ Lemma 2.2.2. Let A ⊆ H0 × H1 be a relation. Then A is a linear relation. Moreover, we have −1 ? ? −1 −1 A∗ = − A? = (−A)−1 = − A−1 = (−A)? = − A? : The proof of this lemma is left as Exercise 2.3. ∗ −1 Remark 2.2.3. Let A ⊆ H0 × H1. Since A is the orthogonal complement of −A , it follows immediately that A∗ is closed. Moreover, A∗ = A∗ since A? = A?. Lemma 2.2.4. Let A ⊆ H0 × H1 be a linear relation. Then A∗∗ := (A∗)∗ = A: Proof. We compute using Lemma 2.2.2 !? ? −1−1 ? A∗∗ = (− (A∗))−1 = − − A? = A? = A: Theorem 2.2.5.