Mathematical Foundations of Quantum Mechanics 91From now on,B(H):=B(H,H) denotes the set of bounded operatorsA:H→H
over the complex Hilbert spaceH. This set is a complex Hilbert space defining the
linear combination of operatorsαA+βB∈B(H)forα, β∈CandA, B∈B(H)
by (αA+βB)x:=αAx+βBxfor everyx∈H.
B(H) acquires the structure of aunital Banach algebra. The complex vector
space structure is the standard one of operators, the associative algebra product is
the composition of operators with unit given byI, and the norm being the above
defined operator norm,
||A||:= sup
0
=x∈H||Ax||
||x||.This definition of||A||can be given also for an operatorA:D(A)→H,ifAis
bounded andD(A)⊂HbutD(A)=H. It immediately arises that
||Ax||≤||A||||x|| ifx∈D(A).As we already know,||·||is a norm overB(H). Furthermore it satisfies
||AB||≤||A||||B|| A, B∈B(H).It is also evident that||I||= 1. ActuallyB(H) is a Banach space so that:B(H)
is a unital Banach algebra. In fact, a fundamental result is the following theorem.
Theorem 2.2.12.IfHis a Hilbert space,B(H)is a Banach space with respect
to the norm of operators.
Proof. The only non-trivial property is completeness ofB(H). Let us prove it.
Consider a Cauchy sequence{Tn}n∈N⊂B(H). We want to prove that there exists
T∈B(H) with||T−Tn||→0asn→+∞. Let us defineTx= limn→+∞Txfor
everyx∈H. The limit exists becouse{Tnx}n∈Nis Cauchy from||Tnx−Tmx||≤
||Tn−Tm||||x||.LinearityofTis easy to prove from linearity of everyTn.Next
observe that||Tx−Tmx||=||limnTnx−Tmx||= limn||Tnx−Tmx|| ≤ ||x||is
mis sufficiently large. Assuming thatT∈B(H), the found inequality, dividing
by ||x||and taking the sup overxwith||x||=0provesthat||T−Tm|| ≤
and thus ||T−Tm|| →0form→+∞as wanted. This concludes the proof
becauseT∈B(H):||Tx||≤ ||Tx−Tmx||+||Tmx|| ≤ ||x||+||Tm||||x||and thus
||T||≤( +||Tm||)<+∞.
Remark 2.2.13.The result, with the same proof, is valid for the above defined
complex vectorspaceB(X,Y), provided the normed spaceYis||·||Y-complete. In