90 From Classical Mechanics to Quantum Field Theory. A Tutorial
Proposition 2.2.10. An operator A:X →Y,withX,Y normed spaces, is
continuous if and only if it is bounded.
Proof. It is evident thatAbounded is continuous because, forx, x′∈X,||Ax−
Ax′||Y ≤b||x−x′||X.Conversely,ifAis continuous then it is continuous for
x=0,so||Ax||y ≤ for >0if||x||X <δforδ>0 sufficiently small. If
||x||=δ/2 we therefore have||Ax||Y< and thus, dividing byδ/2, we also find
||Ax′||Y < 2 /δ,where||x′||X= 1. Multiplying forλ>0,||Aλx′||Y < 2 λ /δ
which can be re-written||Ax||Y < 2 δ||x||for everyx∈X,provingthatAis
bounded.
For bounded operators, it is possible to define theoperator norm,
||A||:= sup
0
=x∈X||Ax||Y
||x||X(
=sup
x∈X,||x||X=1||Ax||Y)
.
It is easy to prove that this is a true norm on the complex vector spacesB(X,Y)
of bounded operatorsT:X→Y, withX,Ycomplex normed spaces.
An important elementary technical result is stated in the following proposition.
Proposition 2.2.11.LetA:S→Ybe a bounded operator defined on the subspace
S⊂X,whereX,Y ar enormed spaces withYcomplete. IfSis dense inX,then
Acan be continously extended to a bounded opertaorA 1 :X→Y with the same
norm asAand this extension is unique.
Proof. Uniqueness is obvious from continuity: ifSxn→x∈XandA 1 ,A′ 1 are
continuous extensions,A 1 x−A′ 1 x= limn→+∞A 1 xn−A′ 1 xn= limn→+∞0=0.Let
us construct a linear contiuous extension. Ifx∈Xtere is a sequenceSxn→x∈
XsinceSis dense.{Axn}n∈Nis Cauchy because{xn}n∈Nis and||Axn−Axm||Y≤
||A||||xn−xm||X. So the limitA 1 x:= limn→+∞Axnexists becauseYis complete.
The limit does not depend on the sequence: ifSx′n→x,then||Axn−Ax′n||≤
||A||||xn−x′n||→0, soA 1 is well defined. It is immediately proven thatA 1 is linear
form linearity ofAand thereforeA 1 is an operator extendingAon the wholeX.By
construction||A 1 x||Y= limn→+∞||Axn||Y ≤limn→+∞||A||||xn||X≤||A||||x||X,
so||A 1 ||≤||A||, in particularA 1 is bounded. On the other hand,
||A 1 ||=sup{||A 1 x||||x||−^1 |x∈X\{ 0 }}≥sup{||A 1 x||||x||−^1 |x∈S\{ 0 }}
=sup{||Ax||||x||−^1 |x∈X\{ 0 }}=||A||,so that||A 1 ||≥||A||and therefore||A 1 ||=||A||.