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A Note on the Definition of an Unbounded Normal Operator

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A Note on the Definition of an Unbounded Normal Operator A NOTE ON THE DEFINITION OF AN UNBOUNDED NORMAL OPERATOR MATTHEW ZIEMKE 1. Introduction Let H be a complex Hilbert space and let T : D(T )(⊆ H) !H be a generally unbounded linear operator. In order to begin the discussion about whether or not the operator T is normal, we need to require that T have a dense domain. This ensures that the adjoint operator T ∗ is defined. In the literature, one sees two definitions for what it means for a densely defined operator T to be normal. The first requires that T be closed and that T ∗T = TT ∗ while the second requires that D(T ) = D(T ∗) and kT xk = kT ∗xk for all x 2 D(T ) = D(T ∗). There is no harm in having these competing definitions as they are, in fact, equivalent. While the equivalence of these statements seems to be widely known, the proof is not obvious, and my efforts to find a detailed and self-contained proof of this fact have left me empty-handed. I attempt to give such a proof here. I also want to note that while we say a bounded linear operator T is normal if and only if T ∗T = TT ∗, the lone requirement of T ∗T = TT ∗ for an unbounded densely defined operator does not, in general, imply T is normal. I used [1] and [2] to write this note. 2. Proof of the equivalence We begin with a definition. Definition 2.1. Let T : D(T )(⊆ H) !H be a densely defined operator. We define the T -inner product on D(T ) × D(T ), denoted as h·; ·iT , by the equation hx; yiT = hx; yi + hT x; T yi; for all x; y 2 D(T ): It is easy to see that the T -inner product is, in fact, an inner product on D(T ). Further, 1=2 we denote the T -norm on D(T ) as k · kT and it is given by kxkT = hx; xiT . Note that 1=2 1=2 2 21=2 kxkT = hx; xiT = (hx; xi + hT x; T xi) = kxk + kT xk : 1 ZIEMKE The following lemma is fairly elementary and widely known, but for the sake of having the notes self-contained we give the statement and proof. Lemma 2.2. Let T : D(T )(⊆ H) !H be a densely defined operator. The inner product space (D(T ); k · kT ) is a Hilbert space if and only if T is closed. Proof. First, assume T is closed. Let (xn)n≥1 ⊆ D(T ) be a Cauchy sequence with respect to the T -norm. Then, for n; m 2 N, we have that 2 2 2 kxn − xmk + kT xn − T xmk = kxn − xnkT ! 0 as m; n ! 1 and so (xn)n≥1 and (T xn)n≥1 are Cauchy with respect to the norm k · k. So, there exists k·k k·k x 2 H and y 2 H such that xn −! x and T xn −! y. Since T is closed, x 2 D(T ) and T xn ! T x. Further, 2 21=2 kxn − xmkT = kxn − xk + kT xn − T xk ! 0 k·kT and so xn −−! x and therefore (D(T ); k · kT ) is a Hilbert space. k·k Now, suppose (D(T ); k·kT ) is a Hilbert space. Let (xn)n≥1 ⊆ D(T ) such that xn −! x 2 H k·k and T xn −! y 2 H. If n; m 2 N then 2 21=2 kxn − xmkT = kxn − xmk + kT xn − T xmk ! 0 as n; m ! 1 and so (xn)n≥1 is Cauchy with respect to k · kT . Since (D(T ); k · kT ) is complete, there exists k·kT z 2 D(T ) such that xn −−! z. Then 2 2 2 kxn − zk + kT xn − T zk = kxn − zkT ! 0 k·k k·k and so xn −! z and T xn −! T z. Hence x = z 2 D(T ) and T x = T z = y. Therefore T is closed. We are now ready to prove the equivalence of the two definitions. Theorem 2.3. Let T : D(T )(⊆ H) !H be a densely defined operator. The following are equivalent: (i) T is closed and T ∗T = TT ∗. (ii) D(T ) = D(T ∗) and kT xk = kT ∗xk for all x 2 D(T ) = D(T ∗). (ii) ) (i). Suppose D(T ) = D(T ∗) and kT xk = kT ∗xk for all x 2 D(T ) = D(T ∗). We k·k first want to show that T is closed. To this end, let (xn)n≥1 ⊆ D(T ) such that xn −! x 2 H 2 ZIEMKE k·k and T xn −! y 2 H. Let n; m 2 N. Then ∗ ∗ kT xn − T xmk = kT xn − T xmk ! 0 as n; m ! 1 ∗ ∗ since (T xn)n≥1 is Cauchy. So (T xn)n≥1 is Cauchy and, since T is closed, we have that ∗ ∗ k·k ∗ x 2 D(T ) = D(T ) and T xn −! T x. Then ∗ ∗ kT xn − T xk = kT xn − T xk ! 0 k·k and so T xn −! T x but T xn ! y. Therefore T x = y and so T is closed. Let x; y 2 D(T ) = D(T ∗). By the polarization identity 3 3 1 X 1 X (1) hT x; T yi = ikkT x + ikT yk2 = ikkT ∗x + ikT ∗yk2 = hT ∗x; T ∗yi: 4 4 k=0 k=0 Now, we would like to show D(T ∗T ) = D(TT ∗). To this end, let x 2 D(T ∗T ). Then x 2 D(T ) = D(T ∗) and T x 2 D(T ∗). So there exists C > 0 such that jhT x; T yij ≤ Ckyk; for all y 2 D(T ): Then Equation (1) imples jhT ∗x; T ∗yij ≤ Ckyk; for all y 2 D(T ) = D(T ∗): Thus T ∗x 2 D(T ∗∗). Since T is closed, we have that T = T ∗∗ and so T ∗x 2 D(T ). Hence x 2 D(TT ∗) and so D(T ∗T ) ⊆ D(TT ∗). Similarly, if x 2 D(TT ∗) then x 2 D(T ∗) = D(T ) and T ∗x 2 D(T ) = D(T ∗∗). So there exists C > 0 such that jhT ∗x; T ∗yij ≤ Ckyk; for all y 2 D(T ) = D(T ∗) and so, by Equation (1), we have that jhT x; T yij ≤ Ckyk; for all y 2 D(T ) = D(T ∗) and thus T x 2 D(T ∗). Therefore x 2 D(T ∗T ) and so D(TT ∗) = D(T ∗T ). Now, let x 2 D(T ∗T ) = D(TT ∗) and let y 2 D(T ) = D(T ∗). Then, by Equation (1), we have that hT x; T yi = hT ∗x; T ∗yi, that is, hT ∗T x; yi = hTT ∗x; yi, and so h(T ∗T − TT ∗)x; yi = 0 for all y 2 D(T ). Since D(T ) is dense, we have that (T ∗T − TT ∗)x = 0 and so T ∗T = TT ∗. [(i) ) (ii)]. Suppose T ∗T = TT ∗ and T is closed. Then D(T ∗T ) = D(TT ∗). Let x 2 D(T ∗T ) = D(TT ∗). Then kT xk2 = hT x; T xi = hT ∗T x; xi = hTT ∗x; xi = hT ∗x; T ∗xi = kT ∗xk2 3 ZIEMKE and so we have that kT xk = kT ∗xk whenever x 2 D(T ∗T ) = D(TT ∗). First, we would like to show D(T ∗T ) = D(TT ∗) is dense in D(T ) with respect to the ∗ ∗ T -norm. Let S : D(T T )(⊆ (D(T ); k · kT )) !H where S = T T + 1. Further, define J : D(T )(⊆ H) ! (D(T ); k · kT ) by Jx = x. Since D(T ) is dense in H, the map J is densely k·k defined. We claim that J is closed. Indeed, if (xn)n≥1 ⊆ D(T ) such that xn −! x 2 H and k·kT k·kT Jxn −−! y 2 D(T ) then, since Jxn = xn, we have that xn −−! y. Then 2 2 2 2 kxn − xk ≤ kxn − yk + kT (xn − y)k = kxn − ykT ! 0 k·k k·k and so xn −! y but xn −! x and so x = y and therefore Jx = y. Hence J is closed. Since J is densely defined and closed, J ∗ is densely defined. Next, we claim that S = J ∗. Note that D(S) = D(T ∗T ). If y 2 D(S) = D(T ∗T ) then, for any x 2 D(T ), hx; yi + hT x; T yi = hx; yi + hx; T ∗T yi = hx; (T ∗T + 1)yi = hx; Syi ∗ ∗ and so S ⊆ J . Let y 2 D(J ). Then there exists z 2 H such that hJx; yiT = hx; zi for all x 2 D(J) = D(T ). Let x 2 D(T ). Then hx; zi = hJx; yiT = hx; yiT = hx; yi + hT x; T yi and so hT x; T yi = hx; z −yi and so T y 2 D(T ∗), hence y 2 D(T ∗T ). Therefore we have that ∗ ∗ ∗ ∗ J = S. Since J is dense, we then have that D(T T ) = D(J ) is dense in (D(T ); k · kT ). ∗ ∗ k·kT Let x 2 D(T ). Then there exists (xn)n≥1 ⊆ D(T T ) = D(TT ) such that xn −−! x. Also, since 2 2 2 2 kxn − xk ≤ kxn − xk + kT (xn − x)k = kxn − xkT ! 0 k·k we have that xn −! x. Further, 2 2 2 2 kT xn − T xk ≤ kxn − xk + kT (xn − x)k = kxn − xkT ! 0 k·k so T xn −! T x. Let m; n 2 N. Then ∗ ∗ ∗ kT xn − T xmk = kT (xn − xm)k = kT (xn − xm)k = kT xn − T xmk ! 0: ∗ ∗ k·k Hence (T xn)n≥1 is k · k-Cauchy and so there exists z 2 H so that T xn −! z.
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