Unbounded Operators

Chapter 10

Unbounded Operators

10.1 GENERAL ASPECTS OF UNBOUNDED LINEAR OPERATORS Let us return to the general definition of linear operator given at the beginning of the previous chapter, without any assumption about continuity of the operator. For simplicity we will assume a Hilbert space setting, although much of what is stated below remains true for mappings between Banach spaces. We have the following important definition. Definition 10.1. If H is a Hilbert space and T: D(T) ⊂ H → H is a linear operator then we say T is closed if whenever un ∈ D(T), un → u and Tun → v then u ∈ D(T) and Tu = v. We emphasize that this definition is strictly weaker than continuity of T, since for a closed operator it is quite possible that un → u but the image sequence {Tun } is divergent. This could not happen for a bounded linear operator. It is immediate that any T ∈ B(H) must be closed. A common alternate way to define a closed operator employs the concept of the graph of T. Definition 10.2. If T: D(T) ⊂ H → H is a linear operator then we define the graph of T to be G(T) = {(u, v) ∈ H × H: v = Tu}

(10.1.1)

The definition of G(T) (and for that matter the definition of closedness) makes sense even if T is not linear, but we will only use it in the linear case. It is easy to check that H × H is a Hilbert space with the inner product (u1 , v1 ), (u2 , v2 ) = u1 , u2  + v1 , v2 

(10.1.2)

In particular, (un , vn ) → (u, v) in H × H if and only if un → u and vn → v in H. One may now verify (Exercise 10.2).

Techniques of Functional Analysis for Differential and Integral Equations http://dx.doi.org/10.1016/B978-0-12-811426-1.00010-6 © 2017 Elsevier Inc. All rights reserved.

159

160 Techniques of Functional Analysis for Differential and Integral Equations

Proposition 10.1. T: D(T) ⊂ H → H is a closed linear operator if and only if G(T) is a closed subspace of H × H. We emphasize that closedness of T does not mean that D(T) is closed in H—this is false in general. In fact we have the so-called Closed Graph Theorem. Theorem 10.1. If T is a closed linear operator and D(T) is a closed subspace of H, then T must be continuous on D(T). We refer to Theorem 2.15 of [33] or Theorem 2.9 of [5] for a proof. In particular if T is closed and unbounded then D(T) cannot be all of H. By far the most common types of unbounded operator which we will be interested in are differential operators. For use in the next example, let us recall that a function f defined on a closed interval [a, b] is absolutely continuous on [a, b] (f ∈ AC([a, b])) if for any  > 0 there exists δ  > 0 such that if {(ak , bk )}nk=1  is a disjoint collection of intervals in [a, b], and nk=1 |bk − ak | < δ then nk=1 |f (bk ) − f (ak )| < . Clearly an absolutely continuous function is continuous. Theorem 10.2. The following are equivalent. 1. f is absolutely continuous on [a, b]. 2. f is differentiable a.e. on [a, b], f  ∈ L1 (a, b) and  x f  (y) dy ∀x ∈ [a, b] f (x) = f (a) +

(10.1.3)

a

3. f ∈ W 1,1 (a, b) Furthermore, when f satisfies these equivalent conditions, the distributional derivative of f coincides with its pointwise a.e. derivative. Here, the equivalence of 1 and 2 is an important theorem of analysis, see for example, Theorem 11, Section 6.5 of [30], Theorem 7.29 of [40], or Theorem 7.20 of [32], while the equivalence of 2 and 3, follows from Theorem 8.2 and the definition of the Sobolev space W 1,1 . The final statement is also a direct consequence of Theorem 8.2. Example 10.1. Let H = L2 (0, 1) and Tu = u on the domain D(T) = {u ∈ H 1 (0, 1): u(0) = 0}

(10.1.4)

Here T is unbounded, as in Example 9.10, D(T) is a dense subspace of H, since for example, it contains D(0, 1), but it is not all of H. We claim that T is closed. To see this, suppose un ∈ D(T), un → u in H and vn = un → v in H. By our assumptions, Eq. (10.1.3) is valid, so  x vn (y) dy (10.1.5) un (x) = 0

Unbounded Operators Chapter | 10 161

We can then find a subsequence nk → ∞ and a subset  ⊂ (0, 1) such that unk (x) → u(x) for x ∈  and the complement of  has measure zero. For any x we also have that vnk → v in L2 (0, x), so that passing to the limit in Eq. (10.1.5) through the subsequence nk we obtain  x u(x) = v(s) ds ∀x ∈  (10.1.6) 0

If we denote the right-hand side by w then from Theorem 10.2 we get that w ∈ D(T), with w = v in the sense of distributions. Since u = w a.e., u and w coincide as elements of L2 (0, 1) and so we get the necessary conclusion that u ∈ D(T) with u = v. 2 The proper definition of D(T) was essential in this example. If we had defined instead D(T) = {u ∈ C1 ([0, 1]) : u(0) = 0} then we would not have been able to reach the conclusion that u ∈ D(T). An operator which is not closed may still be closeable, meaning that it has a closed extension. Let us define this concept carefully. Definition 10.3. If S, T are linear operators on H, we say that S is an extension of T if D(T) ⊂ D(S) and Tu = Su for u ∈ D(T). In this case we write T ⊂ S. T is closeable if it has a closed extension. If T is not closed, then its graph G(T) is not closed, but it always has a closure G(T) in the topology of H × H, which is then a natural candidate for the graph of a closed operator which extends T. This procedure may fail, however, because it may happen that (u, v1 ), (u, v2 ) ∈ G(T) with v1 = v2 so that G(T) would not correspond to a single-valued operator. If we know somehow that this cannot happen, then G(T) will be the graph of some linear operator S (you should check that G(T) is a subspace of H × H) which is obviously closed and extends T, thus T will be closeable. It is useful to have a clearer criterion for the closability of a linear operator T. Note that if (u, v1 ), (u, v2 ) are both in G(T), with v1 = v2 , then (0, v) ∈ G(T) for v = v1 − v2 = 0. This means that there must exist un → 0, un ∈ D(T) such that vn = Tun → v = 0. If we can show that no such sequence un can exist, then evidently no such pair of points can exist in G(T), so that T will be closeable. The converse statement is also valid and is easy to check. Thus we have established the following. Proposition 10.2. A linear operator on H is closeable if and only if un ∈ D(T), un → 0, Tun → v implies v = 0. Example 10.2. Let Tu = u on L2 (0, 1) with domain D(T) = {u ∈ C1 ([0, 1] : u(0) = 0}. We have previously observed that T is not closed, but we can check that the above criterion holds, so that T is closeable. Let un ∈ D(T) and un → 0, un → v in L2 (0, 1). As before,  x un (s) ds (10.1.7) un (x) = 0

162 Techniques of Functional Analysis for Differential and Integral Equations

Picking a subsequence nk → ∞ for which unk → 0 a.e., we get  x v(s) ds = 0 a.e.

(10.1.8)

0

The left-hand side is absolutely continuous so equality must hold for every x ∈ [0, 1] and by Theorem 10.2 we conclude that v = 0 a.e. 2 An operator which is closeable may in general have many different closed extensions. However, there always exists a minimal extension in this case, denoted T, the closure of T, defined by G(T) = G(T). It can be alternatively characterized as follows: T is the unique linear operator on H with the properties that (i) T ⊂ T and (ii) if T ⊂ S and S is closed then T ⊂ S. If T: D(T) ⊂ H → H and S: D(S) ⊂ H → H are closed linear operators then the sum S + T is defined and linear on D(S + T) = D(S) ∩ D(T), but need not be closed, in general. Choose, for example, any closed and densely defined linear operator T with D(T) = H and S = −T. Then the sum S + T is the zero operator, on the dense domain D(S ∩ T) = D(T) = H, which is not a closed operator. In this example S + T is closeable, but even that need not be true, see Exercise 10.13. One can show, however, that if T is closed and S is bounded, then S + T is closed. Likewise the product ST is defined on D(ST) = {x ∈ D(T) : Tx ∈ D(S)} and need not be closed even if S, T are. If S ∈ B(H) and T is closed then TS will be closed, but ST need not be (see Exercise 10.11). Finally consider the inverse operator T −1 : R(T) → D(T), which is well defined if T is one-to-one. Proposition 10.3. If T is one-to-one and closed then T −1 is also closed. Proof. Let un ∈ D(T −1 ), un → u, and T −1 un → v. Then if vn = T −1 un we have vn ∈ D(T), vn → v, and Tvn = un → u. Since T is closed it follows that v ∈ D(T) and Tv = u, or equivalently u ∈ R(T) = D(T −1 ) and T −1 u = v as needed.

10.2 THE ADJOINT OF AN UNBOUNDED LINEAR OPERATOR To some extent it is possible to define an adjoint operator, even in the unbounded case, and obtain some results about the solvability of the operator equation Tu = f analogous to those proved earlier in the case of bounded T. For the rest of this section we assume that T: D(T) ⊂ H → H is linear and densely defined. We will say that (v, v ∗ ) is an admissible pair for T ∗ if Tu, v = u, v ∗  ∀u ∈ D(T)

(10.2.9)

We then define D(T ∗ ) = {v ∈ H: there exists v ∗ ∈ H such that (v, v ∗ ) is an admissible pair for T ∗} (10.2.10)

Unbounded Operators Chapter | 10 163

and T ∗v = v∗

v ∈ D(T ∗ )

(10.2.11)

For this to be an appropriate definition, we should check that for any v there is at most one v ∗ for which (v, v ∗ ) is admissible. Indeed if there were two such elements, then the difference v1∗ − v2∗ would satisfy u, v1∗ − v2∗  = 0 for all u ∈ D(T). Since we assume D(T) is dense, it follows that v1∗ = v2∗ . Note that for v ∈ D(T ∗ ), if we define φv (u) = Tu, v for u ∈ D(T), then φv is bounded on D(T), since |φv (u)| = |u, v ∗ | = |u, T ∗ v| ≤ u T ∗ v

(10.2.12)

The converse statement is also true (see Exercise 10.5) so that it is equivalent to define D(T ∗ ) as the set of all v ∈ H such that u → Tu, v is a bounded linear functional on D(T). The domain D(T ∗ ) always contains at least the zero element, since (0, 0) is always an admissible pair. There are known examples for which D(T ∗ ) contains no other points (see Exercise 10.4). Here is a useful characterization of T ∗ in terms of its graph G(T ∗ ) ⊂ H × H. Proposition 10.4. If T is a densely defined linear operator on H then G(T ∗ ) = (V(G(T)))⊥

(10.2.13)

where V is the unitary operator on H × H defined by V(x, y) = (−y, x)

x, y ∈ H

(10.2.14)

We leave the proof as an exercise. Proposition 10.5. If T is a densely defined linear operator on H then T ∗ is a closed linear operator on H. We emphasize that it is not assumed here that T is closed. The conclusion that T ∗ must be closed also follows directly from Eq. (10.2.13), but we give a more direct proof below. Proof. First we verify the linearity of T ∗ . If v1 , v2 ∈ D(T ∗ ) and c1 , c2 are scalars, then there exist unique elements v1∗ , v2∗ such that Tu, v1  = u, v1∗  Tu, v2  = u, v2∗ 

for all u ∈ D(T)

(10.2.15)

Then Tu, c1 v1 + c2 v2  = c1 Tu, v1  + c2 Tu, v2  = c1 u, v1∗  + c2 u, v2∗  (10.2.16) = u, c1 v1∗ + c2 v2∗  for all u ∈ D(T), and so (c1 v1 + c2 v2 , c1 v1∗ + c2 v2∗ ) is an admissible pair for T ∗ . In particular c1 v1 + c2 v2 ∈ D(T ∗ ) and

164 Techniques of Functional Analysis for Differential and Integral Equations

T ∗ (c1 v1 + c2 v2 ) = c1 v1∗ + c2 v2∗ = c1 T ∗ v1 + c2 T ∗ v2

(10.2.17)

To see that T ∗ is closed, let vn ∈ D(T ∗ ), vn → v, and T ∗ vn → w. If u ∈ D(T) then we must have Tu, vn  = u, T ∗ vn 

(10.2.18)

Letting n → ∞ yields Tu, v = u, w. Thus (v, w) is an admissible pair for T ∗ implying that v ∈ D(T ∗ ) and T ∗ v = w, as needed. With a small modification of the proof, we obtain that Proposition 9.3 remains valid. Theorem 10.3. If T: D(T) ⊂ H → H is a densely defined linear operator then N(T ∗ ) = R(T)⊥ . Proof. Let f ∈ R(T) and v ∈ N(T ∗ ). We have f = Tu for some u ∈ D(T) and f , v = Tu, v = u, T ∗ v = 0 so N(T ∗ )

R(T)⊥ . To get the reverse inclusion, let v

(10.2.19) R(T)⊥ , so that Tu, v

⊂ ∈ = 0 = u, 0 for any u ∈ D(T). This means that (v, 0) is an admissible pair for T ∗ , so v ∈ D(T ∗ ) and T ∗ v = 0. Thus R(T)⊥ ⊂ N(T ∗ ) as needed. Example 10.3. Let us revisit the densely defined differential operator in Example 10.1. We seek here to find the adjoint operator T ∗ , and emphasize that one must determine D(T ∗ ) as part of the answer. It is typical in computing adjoints of unbounded operators that precisely identifying the domain of the adjoint is more difficult than finding a formula for the adjoint. Let v ∈ D(T ∗ ) and T ∗ v = g, so that Tu, v = u, g for all u ∈ D(T). That is to say,  1  1  u (x)v(x) dx = u(x)g(x) dx ∀u ∈ D(T) (10.2.20) 0

0

Let 

1

G(x) = −

g(y) dy

(10.2.21)

x

so that G(1) = 0 and G (x) = g(x) a.e., since g is integrable. Integration by parts then gives  1  1  1 u(x)g(x) dx = u(x)G (x) dx = − u (x)G(x) dx (10.2.22) 0

0

0

since the boundary term vanishes. Thus we have  1 u (x)(v(x) + G(x)) dx = 0 0

(10.2.23)

Unbounded Operators Chapter | 10 165

x Now in Eq. (10.2.23) choose u(x) = 0 v(y)+G(y) dy, which is legitimate since u ∈ D(T). The result is that  1 |v(x) + G(x)|2 dx = 0 (10.2.24) 0

1 which can only occur if v(x) = −G(x) = x g(y) dy a.e., implying that T ∗ v = g = −v  . The above representation for v also shows that v  ∈ L2 (0, 1) and v(1) = 0, that is D(T ∗ ) ⊂ {v ∈ L2 (0, 1): v  ∈ L2 (0, 1), v(1) = 0}

(10.2.25)

We claim that the reverse inclusion is also correct: If v belongs to the set on the right and u ∈ D(T) then  1  1  Tu, v = u (x)v(x) dx = − u(x)v  (x) dx = u, −v   (10.2.26) 0

0

Thus (v, −v  ) is an admissible pair for T ∗ , from which we conclude that v ∈ D(T ∗ ) and T ∗ v = −v  as needed. We clearly have N(T ∗ ) = {0} and R(T) = H. In summary we have established that T ∗ v = −v  with domain D(T ∗ ) = {v ∈ H 1 (0, 1): v(1) = 0}

(10.2.27)

2 We remark that if we had originally defined T on the smaller domain {u ∈ C1 ([0, 1]): u(0) = 0} we would have obtained exactly the same result for T ∗ as ∗ mentioned above. This is a special case of the general fact that T ∗ = T (see Exercise 10.14). For this unclosed version of T we still have N(T ∗ ) = {0} but concerning the range can only state that R(T) = H. Definition 10.4. If T = T ∗ we say T is self-adjoint. In this definition it is crucial that equality of the operators T and T ∗ must include the fact that their domains are identical. Example 10.4. If in the previous example we defined Tu = iu on the same domain we would find that T ∗ v = iv  on the domain (10.2.27). Even though the expressions for T, T ∗ are the same, T is not self-adjoint since the two domains are different. 2 A closely related property is that of symmetry. Definition 10.5. We say that T is symmetric if Tu, v = u, Tv for all u, v ∈ D(T). Example 10.5. Let Tu = iu be the unbounded operator on H = L2 (0, 1) with domain D(T) = {u ∈ H 1 (0, 1): u(0) = u(1) = 0}

(10.2.28)

166 Techniques of Functional Analysis for Differential and Integral Equations

One sees immediately that T is symmetric; however, it is still not self-adjoint since D(T ∗ ) = D(T) again, see Exercise 10.6. 2 If T is symmetric and u ∈ D(T), then (v, Tv) is an admissible pair for T ∗ , thus D(T) ⊂ D(T ∗ ) and T ∗ v = Tv for v ∈ D(T). In other words, T ∗ is always an extension of T whenever T is symmetric. We see, in particular, that any selfadjoint operator is closed and any symmetric operator is closeable. Proposition 10.6. If T is densely defined and one-to-one, and if also R(T) is dense, then T ∗ is also one-to-one and (T ∗ )−1 = (T −1 )∗ . Proof. By our assumptions, S = (T −1 )∗ exists. We are done if we show ST ∗ u = u for all u ∈ D(T ∗ ) and T ∗ Sv = v for all v ∈ D(S). First let u ∈ D(T ∗ ) and v ∈ D(T −1 ). Then v, u = TT −1 v, u = T −1 v, T ∗ u

(10.2.29)

This means that (T ∗ u, u) is an admissible pair for (T −1 )∗ and ST ∗ u = (T −1 )∗ T ∗ u = u as needed. Next, if u ∈ D(T) and v ∈ D(S) then u, v = T −1 Tu, v = Tu, Sv

(10.2.30)

Therefore (Sv, v) is admissible for T ∗ , so that Sv ∈ D(T ∗ ) and T ∗ Sv = v. Theorem 10.4. If T, T ∗ are both densely defined then T ⊂ T ∗∗ , and in particular T is closeable. Proof. If we assume that T ∗ is densely defined, then T ∗∗ exists and is closed. If u ∈ D(T) and v ∈ D(T ∗ ) then T ∗ v, u = v, Tu, which is to say that (u, Tu) is an admissible pair for T ∗∗ . Thus u ∈ D(T ∗∗ ) and T ∗∗ u = Tu, or equivalently T ⊂ T ∗∗ . Thus T has a closed extension, namely T ∗∗ . There is an interesting converse statement, which we state but will not prove here, see [2] section 46, or [33] Theorem 13.12. Theorem 10.5. If T is densely defined and closeable then T ∗ must be densely defined, and T = T ∗∗ . In particular if T is closed and densely defined then T = T ∗∗ .

10.3 EXTENSIONS OF SYMMETRIC OPERATORS It has been observed previously that if T is a densely defined symmetric operator then the adjoint T ∗ is always an extension of T. It is an interesting question whether such a T always possesses a self-adjoint extension—the extension would necessarily be different from T ∗ at least if T is closed, since then if T ∗ is self-adjoint so is T, by Theorem 10.5. We say that a linear operator T is positive if Tu, u ≥ 0 for all u ∈ D(T).

Unbounded Operators Chapter | 10 167

Theorem 10.6. If T is a densely defined, positive, symmetric operator on a Hilbert space H then T has a positive self-adjoint extension. Proof. Define u, ve = u, v + Tu, v

u, v ∈ D(T)

(10.3.31)

with corresponding norm denoted by ue . It may be easily verified that all of the inner product axioms are satisfied by ·, ·e on D(T), and u ≤ ue . Let He be the dense closed subspace of H obtained as the closure of D(T) in the  · e norm, and regard it as equipped with the ·, ·e inner product. For any z ∈ H the functional ψz (u) = u, z belongs to the dual space of He since |ψz (u)| ≤ u z ≤ ue z, in particular ψz e ≤ z as a linear functional on He . Thus by the Riesz Representation Theorem there exists a unique element z ∈ He ⊂ H such that ψz (u) = u, ze

u ∈ He

(10.3.32)

with z ≤ ze ≤ z. It may be checked that : H → H is linear, and regarded as an operator on H we claim it is also self-adjoint. To see this observe that for any u, z ∈ H we have u, z = ψz (u) = u, ze = z, ue = ψu (z) = z, u = u, z (10.3.33) Choosing u = z we also see that  is positive, namely z, z = z, ze ≥ 0

(10.3.34)

Next  is one-to-one, since if z = 0 and u ∈ He it follows that 0 = u, ze = u, z

∀u ∈ He

(10.3.35)

He

and since is dense in H the conclusion follows. The range of  is also dense in He , hence in H, because otherwise there must exist u ∈ He such that 0 = u, ze = u, z for all z ∈ H. From the above considerations and Proposition 10.6 we conclude that S = −1 exists and is a densely defined self-adjoint operator on H. We will complete the proof by showing that the self-adjoint operator S − I is a positive extension of T. For z, w ∈ D(T) we have z, we = (I + T)z, w = ψ(I+T)z (w) = w, (I + T)ze = (I + T)z, we (10.3.36) and so (I + T)z = z

∀z ∈ D(T)

(10.3.37)

by the assumed density of D(T). In particular D(T) ⊂ R() = D(S) and (I + T) z = −1 z = Sz for z ∈ D(T), as needed.

168 Techniques of Functional Analysis for Differential and Integral Equations

Finally we note that z, z = z, ze ≥ ||z||2 for any z ∈ H. Thus for x = z ∈ R() = D(S) if follows that x, Sx ≥ ||x||2 , which amounts to the positivity of S − I. A positive symmetric operator may have more than one self-adjoint extension, but the specific one constructed in the above proof is usually known as the Friedrichs extension. To clarify what all of the objects in the proof are, it may be helpful to think of the case that Tu = −u on the domain D(T) = C2 ( ) ∩ C0 ( ). In this case ue = uH 1 ( ) , H e = H01 ( ) (except endowed with the usual H 1 norm) and the Friedrichs extension will turn out to be the Dirichlet Laplacian discussed in detail in Section 13.4. The condition of positivity for T may be weakened, see Exercise 10.16.

10.4 EXERCISES 10.1. Let T, S be densely defined linear operators on a Hilbert space. If T ⊂ S, show that S∗ ⊂ T ∗ . 10.2. Verify that H × H is a Hilbert space with the inner product given by Eq. (10.1.2), and prove Proposition 10.1. 10.3. Prove the null space of a closed operator is closed. 10.4. Let φ ∈ H = L2 (R) be any nonzero function and define the linear operator   ∞ u(x) dx φ Tu = −∞

L1 (R) ∩ L2 (R).

on the domain D(T) = (a) Show that T is unbounded and densely defined. (b) Show that T ∗ is not densely defined, more specifically show that T ∗ is the zero operator with domain {φ}⊥ . (Since D(T ∗ ) is not dense, it then follows from Theorem 10.5 that T is not closeable.) 10.5. If T: D(T) ⊂ H → H is a densely defined linear operator, v ∈ H and the map u → Tu, v is bounded on D(T), show that there exists v ∗ ∈ H such that (v, v ∗ ) is an admissible pair for T ∗ . 10.6. Let H = L2 (0, 1) and T1 u = T2 u = iu with domains D(T1 ) = {u ∈ H 1 (0, 1): u(0) = u(1)} D(T2 ) = {u ∈ H 1 (0, 1): u(0) = u(1) = 0} Show that T1 is self-adjoint, and that T2 is closed and symmetric but not self-adjoint. What is T2∗ ? 10.7. If T is symmetric and R(T) = H show that T is self-adjoint. (Suggestion: It is enough to show that D(T ∗ ) ⊂ D(T).) 10.8. Show that if T is self-adjoint and one-to-one then T −1 is also selfadjoint. (Hint: All you really need to do is show that T −1 is densely defined.)

Unbounded Operators Chapter | 10 169

10.9. If T is self-adjoint, S is symmetric and T ⊂ S, show that T = S. (Thus a self-adjoint operator has no proper symmetric extension). 10.10. Let T, S be densely defined linear operators on H and assume that D(T + S) = D(T) ∩ D(S) is also dense. Show that T ∗ + S∗ ⊂ (T + S)∗ . Give an example showing that T ∗ + S∗ and (T + S)∗ may be unequal. 10.11. Assume that T is closed and S is bounded (a) Show that S + T is closed. (b) Show that TS is closed, but that ST is not closed, in general. 10.12. Prove Proposition 10.4. 10.13. Let H = 2 and define ∞  nxn , 4x2 , 9x3 , . . . (10.4.38) Sx = n=1

Tx = {0, −4x2 , −9x3 , . . .}

(10.4.39)

on D(S) = D(T) = {x ∈ 2 : n4 |xn |2 < ∞}. Show that S, T are closed, but S + T is not closeable. (Hint: For example, en /n → 0 but (S + T)en /n → e1 .) 10.14. If T is closable, show that T and T have the same adjoint. 10.15. Suppose that T is densely defined and symmetric with dense range. Prove that N(T) = {0}. 10.16. We say that a linear operator on a Hilbert space H is bounded below, if there exists a constant c0 > 0 such that Tu, u ≥ −c0 u2

∀u ∈ D(T)

Show that Theorem 10.6 remains valid if the condition that T be positive is replaced by the assumption that T is bounded below. (Hint: T + co I is positive.)