Kenneth Kuttler

13.12. ROOTS OF POSITIVE LINEAR MAPS 365

= (An+1−An)− (An+1−An)1m

(Am−1

n+1 +Am−2n+1 An + · · ·+Am−1

≥ (An+1−An)− (An+1−An) I = 0

since each An,An+1 ≤ I, so this proves the claim.Claim 3: An ≥ 0Proof of Claim 3: This is true if n = 0. Suppose it is true for n.

(An+1x,x) = (Anx,x)+1m(T x,x)− 1

m(Am

n x,x)

≥ (Anx,x)+1m(T x,x)− 1

m(Anx,x)≥ 0

because by Proposition 13.12.2, An−Amn = An

(I−Am−1

n)≥ 0 because An ≤ I.

Thus (Anx,x) is increasing and bounded above so it converges. Now let n > k. UsingProposition 13.12.2 AnAk ≥ A2

k and also

(An−Ak)(An +Ak)≤ 2(An−Ak) .

Thus the following holds.

∥Anx−Akx∥2 =((An−Ak)

2 x,x)=(A2

nx,x)−2(AnAkx,x)+

(A2

kx,x)

≤(A2

nx,x)−2(A2

kx,x)+(A2

kx,x)= ((An−Ak)(An +Ak)x,x)

≤ 2 [(Anx,x)− (Akx,x)]

which converges to 0 as k,n→ ∞. Therefore, limn→∞ Anx exists since {Anx} is a Cauchysequence. Let this limit be Ax. Then clearly A is linear. Also, since each An ≥ 0 and selfadjoint, the Cauchy Schwarz inequality implies

|(Ax,y)|= limn→∞|(Anx,y)| ≤ lim sup

n→∞

∣∣∣(Anx,x)1/2 (Any,y)1/2∣∣∣≤ ∥x∥∥y∥

so A is also continuous. Now (Ax,x) = limn→∞ (Anx,x)≥ 0 so A is positive and it is clearlyalso self adjoint since each An is. From passing to the limit in the definition of An,

Ax = Ax+1m(T x−Amx)

and so T x = Amx. This proves the theorem in the case that T ≤ I. Then if T > I, considerT/∥T∥. T/∥T∥ ≤ I and so there is B such that Bm = T/∥T∥ . Let A = ∥T∥1/m B. Thisproves the existence of the mth root. It is clear that A commutes with every continuouslinear operator that commutes with T because this is true of each of the iterates. In fact,each of these is just a polynomial in T . It remains to verify uniqueness.

Next suppose both A and B are mth roots of T having all the properties stated in the the-orem. Then AB = BA because both A and B commute with every operator which commuteswith T . Then from Proposition 13.12.2,((

Am−1 +Am−2B+ ...+Bm−1)(A−B)x,(A−B)x)≥ 0 (13.23)

Therefore, ((Am−Bm)x,(A−B)x) = (0,(A−B)x) = 0.

13.12. ROOTS OF POSITIVE LINEAR MAPS 365(Am EAM PA, +e + An!)1= (Anti —An) _ (An+i —An) n+1 n+1m> (Anti —An) _ (An+1 —An)I =0since each Ay, An+1 < TJ, so this proves the claim.Claim 3: A, > 0Proof of Claim 3: This is true if 7 = 0. Suppose it is true for 7.1 1(AntixX,x) = (Anx,x)+ mn” (Tx,x) — on (At’x,x)1 1> (A —(T ——(A >0> (Anx,x)+ m (Tx,x) m (Apx,x) >because by Proposition 13.12.2, An —A” = Ay, (I-A!) > 0 because Ay < I.Thus (A,x,x) is increasing and bounded above so it converges. Now let n > k. UsingProposition 13.12.2 AnAx > Az and also(An — Ax) (An +Ax) <2 (An — Ax) .Thus the following holds.\|Anx — Agx||* = (( h — Ax)? x,2) = (Ajx,x) —2 (AnAgx,x) + (Agx,x)< (Apx,x) —2 (Azx,x) + (Azx,x) = ((An — Ag) (An + Ag) X,x)<2 [(Anx,x) — (Agx,x)]which converges to 0 as k,n —> ce. Therefore, limy_,.0Anx exists since {A,x} is a Cauchysequence. Let this limit be Ax. Then clearly A is linear. Also, since each A, > 0 and selfadjoint, the Cauchy Schwarz inequality implies|(Ax,y)] = lim |(Anx,y)| < lim sup |(Anx,x)!/? (Any,y)!/} < all Illn—-»0oso A is also continuous. Now (Ax, x) = limp. (Anx,x) > 0 so A is positive and it is clearlyalso self adjoint since each A, is. From passing to the limit in the definition of Aj,1Ax = Ax+ —(Tx—A"x)mand so Tx = Ax. This proves the theorem in the case that T < J. Then if T > J, considerT/\|T\|. T/||T\| <1 and so there is B such that B” = T/||T||. Let A = |\T||!/"B. Thisproves the existence of the m'” root. It is clear that A commutes with every continuouslinear operator that commutes with T because this is true of each of the iterates. In fact,each of these is just a polynomial in T. It remains to verify uniqueness.Next suppose both A and B are m' roots of T having all the properties stated in the the-orem. Then AB = BA because both A and B commute with every operator which commuteswith 7. Then from Proposition 13.12.2,((A"-! +A" ?B+...+B""') (A—B)x,(A—B)x) >0 (13.23)Therefore, ((A” — B”).x, (A —B)x) = (0,(A—B)x) =0.