Analysis

24.6 Square Roots

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].

With the above definition here is a fundamental result about positive self adjoint operators.

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 S₀ = 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 S₀ ≡ 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 S₀ and succeding S_k are polynomials in terms of S₀. Therefore,

Consider S_n+1x.From the claim,

∑n 2 ∑n 2 (Sx, x) = (Sn+1x,x)+ |Skx| ≥ |Skx| k=0 k=0

and so lim_n→∞S_nx = 0. Hence from 24.31,

∑n lim inf (ST x,x) = (ST x,x) = lim inf (T Skx,Skx) ≥ 0. n→ ∞ n→ ∞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∕

. 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.

Proof: Consider K − T_n ≡ S_n. Then the

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→∞

exists. By Proposition 24.6.2, if n > m,

2 Sm − SnSm = Sm(Sm − Sn) ≥ 0

and similarly,

S S − S2= S (S − S ) ≥ 0. n m n n m n

Therefore, since S_n is self adjoint,

( ) |Tnx − Tmx|2 = |Snx− Smx |2 = (Sn − Sm )2x,x

= ((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→∞

exists for each x. It follows is a Cauchy sequence. Let Tx be the thing to which it converges. T is obviously linear and

(T x,x ) = lni→m∞ (Tnx,x) ≤ (Kx, x).

Also

(KT x,y) = nli→m∞ (KTnx, y) = nli→m∞ (TnKx, y) = (TKx, y)

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

implies there exists a uniform bound on the norms, and so

|(Tnx,y)| ≤ C |x||y|.

Now take the limit as n →∞ to conclude

|(T x,y)| ≤ C |x||y|

which shows

≤ C. ■

With this preparation, here is the theorem about square roots.

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,

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,

Claim 3: A_n ≥ 0

Proof of Claim 3: This is true if n = 0. Suppose it is true for n.

because A_n − A_n² = A_n ≥ 0 by Proposition 24.6.2.

Now

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 = A²x. This proves the theorem in the case that T ≤ I.

In the general case, consider T∕

. Then

( ) -T--x,x = --1-(Tx,x) ≤ |x|2 = (Ix,x) ||T || ||T ||

and so T∕

≤ I. Therefore, it has a square root, B. Let A = B. Then A has all the right properties and A² = B² = = 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,

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

x = 0 which implies Ax = 0. Similarly, Bx = 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