In this section, H will be a Hilbert space, real or complex, and T will denote an operator
which satisfies the following definition. A useful theorem about the existence of square
roots of certain operators is presented. This proof is very elementary. I found it in
[18].
Definition 24.6.1Let T ∈ℒ
(H, H )
satisfy T = T^{∗}(Hermitian) and for allx ∈ H,
(Tx,x) ≥ 0 (24.30)
(24.30)
Such an operator is referred to as positive and self adjoint. It is probably better to refer tosuch an operator as “nonnegative” since the possibility that Tx = 0 for some x≠0 is not beingexcluded. Instead of “self adjoint” you can also use the term, Hermitian.To save onnotation, write
With the above definition here is a fundamental result about positive self adjoint
operators.
Proposition 24.6.2Let S,T be positive and self adjoint such that ST = TS. ThenST is also positive and self adjoint.
Proof:It is obvious that ST is self adjoint. The only problem is to show that ST is
positive. To show this, first suppose S ≤ I. The idea is to write
∑n
S = Sn+1 + S2k
k=0
where S_{0} = S and the operators S_{k} are self adjoint. This is a useful idea because it is then
obvious that the sum is positive. If we want such a representation as above, then it follows
that S_{0}≡ S and
2
Sn+1 ≡ Sn − Sn.
Thus it is obvious that the S_{k} are all self adjoint. Also, the following claim holds.
Claim: I ≥ S_{n}≥ 0.
Proof of the claim:This is true if n = 0. Assume true for n. Then from the
definition,
2
Sn+1 = S2n (I − Sn)+ (I − Sn) Sn
and it is obvious from the definition that the sum of positive operators is positive. Therefore,
it suffices to show the two terms in the above are both positive. It is clear from the definition
that each S_{n} is Hermitian (self adjoint) because they are just polynomials in S. Also each
must commute with T for the same reason. Therefore,
( )
S2n(I − Sn)x,x = ((I − Sn )Snx,Snx) ≥ 0
and also
( )
(I − Sn )2 Snx,x = (Sn (I − Sn)x,(I − Sn)x) ≥ 0
This proves the claim.
Now each S_{k} commutes with T because this is true of S_{0} and succeding S_{k} are
polynomials in terms of S_{0}. Therefore,
( ( n ) )
(STx,x) = S + ∑ S2 Tx,x
n+1 k=0 k
n∑
= (Sn+1Tx,x) + (S2kTx,x)
k=0
n∑
= (Tx,Sn+1x) + (T Skx,Skx) (24.31)
k=0
All this was based on the assumption that S ≤ I. The next task is to remove this
assumption. Let ST = TS where T and S are positive self adjoint operators. Then consider
S∕
||S||
. This is still a positive self adjoint operator and it commutes with T just like S does.
Therefore, from the first part,
( )
0 ≤ -S--Tx,x = -1-(ST x,x).■
||S || ||S||
The proposition is like the familiar statement about real numbers which says that
when you multiply two nonnegative real numbers the result is a nonnegative real
number. The next lemma is a generalization of the familiar fact that if you have an
increasing sequence of real numbers which is bounded above, then the sequence
converges.
Lemma 24.6.3Let
{Tn}
be a sequence of self adjoint operators on a Hilbert space, Hand let T_{n}≤ T_{n+1}for all n. Also suppose there exists K, a self adjoint operator such that forall n,T_{n}≤ K. Suppose also that each operator commutes with all the others and that Kcommutes with all the T_{n}. Then there exists a self adjoint continuous operator, T such thatfor all x ∈ H,
Tnx → T x,
T ≤ K, and T commutes with all the T_{n}and with K.
Proof:Consider K − T_{n}≡ S_{n}. Then the
{Sn}
are decreasing, that is,
{(Snx,x)}
is a decreasing sequence and from the hypotheses, S_{n}≥ 0 so the above sequence is
bounded below by 0. Therefore, lim_{n→∞}
= ((S2− 2S S + S2 )x,x) = ((S2 − S S )x,x) + ((S2 − S S )x,x)
n n m m m m n n n m
(( 2 ) ) (( 2 2) )
≤ Sm − SmSn x,x ≤ Sm − Sn x,x
= ((Sm − Sn )(Sm +Sn )x,x) ≤ 2((Sm − Sn)Kx, x)
≤ 2 ((Sm − Sn)Kx, Kx )1∕2((Sm − Sn)x,x)1∕2
The last step follows from an application of the Cauchy Schwarz inequality along with the
fact S_{m}−S_{n}≥ 0. The last expression converges to 0 because lim_{n→∞}
(Snx,x)
exists for each
x. It follows
{Tnx}
is a Cauchy sequence. Let Tx be the thing to which it converges. T is
obviously linear and
and so TK = KT. Similarly, T commutes with all T_{n}.
In order to show T is continuous, apply the uniform boundedness principle, Theorem
23.1.8. The convergence of
{Tnx}
implies there exists a uniform bound on the norms,
||Tn ||
and so
|(Tnx,y)| ≤ C |x||y|.
Now take the limit as n →∞ to conclude
|(T x,y)| ≤ C |x||y|
which shows
||T||
≤ C. ■
With this preparation, here is the theorem about square roots.
Theorem 24.6.4Let T ∈ℒ
(H,H )
be a positive self adjoint linear operator.Then there exists a unique square root, A with the following properties. A^{2} = T,A ispositive and self adjoint, A commutes with every operator which commutes with T.
Proof: First suppose T ≤ I. Then define
A ≡ 0,A = A + 1 (T − A2 ).
0 n+1 n 2 n
From this it follows that every A_{n} is a polynomial in T. Therefore, A_{n} commutes with T and
with every operator which commutes with T.
Claim 1: A_{n}≤ I.
Proof of Claim 1: This is true if n = 0. Suppose it is true for n. Then by the assumption
that T ≤ I,
I − An+1 = I − An + 1 (A2 − T )
2 n
≥ I − An + 1 (A2 − I)
2 n
= I − An − 1 (I − An)(I + An )
(2 )
= (I − A ) I − 1 (I + A )
n 2 n
1
= (I − An)(I − An)2 ≥ 0.
Claim 2: A_{n}≤ A_{n+1}
Proof of Claim 2: From the definition of A_{n}, this is true if n = 0 because
A1 = T ≥ 0 = A0.
Suppose true for n. Then from Claim 1,
[ ]
A − A = A + 1 (T − A2 )− A + 1 (T − A2 )
n+2 n+1 n+1 2 n+1 n 2 n
1 ( 2 2 )
= An+1 − An + 2 An − A n+1
( 1 )
= (An+1 − An) I − 2 (An + An+1 )
( )
≥ (An+1 − An) I − 1 (2I) = 0.
2
Claim 3: A_{n}≥ 0
Proof of Claim 3: This is true if n = 0. Suppose it is true for n.
is a sequence of positive self adjoint operators which are bounded above by I
such that each of these operators commutes with every operator which commutes with
T. By Lemma 24.6.3, there exists a bounded linear operator, A such that for all
x,
Anx → Ax
Then A commutes with every operator which commutes with T because each A_{n} has this
property. Also A is a positive operator because each A_{n} is. From passing to the limit in the
definition of A_{n},
1 ( )
Ax = Ax + 2 T x− A2x
and so Tx = A^{2}x. This proves the theorem in the case that T ≤ I.
B. Then A has all the
right properties and A^{2} =
||T||
B^{2} =
||T||
(T ∕||T ||)
= T. This proves the existence part of the
theorem.
Next suppose both A and B are square roots of T having all the properties stated in the
theorem. Then AB = BA because both A and B commute with every operator which
commutes with T.
(A (A − B )x,(A − B )x),(B (A − B)x,(A − B)x ) ≥ 0 (24.32)
(24.32)
Therefore, on adding these,
((A2 − AB + BA − B2)x,(A − B)x)
(( 2 2) )
= A − B x,(A − B )x
= ((T − T)x,(A − B)x) = 0.
It follows both expressions in 24.32 equal 0 since both are nonnegative and when they are
added the result is 0. Now applying the existence part of the theorem to A, there exists a
positive square root of A which is self adjoint. Thus
(√ -- √-- )
A (A − B )x, A (A − B )x = 0
so
√--
A
(A− B )
x = 0 which implies A
(A− B )
x = 0. Similarly, B
(A− B )
x = 0. Subtracting
these and taking the inner product with x,
( )
0 = ((A (A − B )− B (A − B )) x,x ) = (A − B )2x,x = |(A − B)x|2
and so Ax = Bx which shows A = B since x was arbitrary. ■
class=”left” align=”middle”(U)24.7. ORDINARY DIFFERENTIAL EQUATIONS IN
BANACH SPACE