MA2223: SOLUTIONS TO PROBLEM SHEET 3
1. Let `2 be the real vector space of all square summable sequences
x = (xn ) of real numbers with norm
kxk2 =
∞
X
x2i
! 12
i=1
Show that the unilateral shift operator
T : `2 → `2 ,
(x1 , x2 , x3 , . . .) 7→ (0, x1 , x2 , x3 , . . .)
is continuous.
Solution: First note that T is a linear operator. If x = (x1 , x2 , x3 , . . .)
and y = (y1 , y2 , y3 , . . .) are two elements of `2 then
T (x + y) =
=
=
=
T ((x1 + y1 , x2 + y2 , x3 + y3 , . . .))
(0, x1 + y1 , x2 + y2 , x3 + y3 , . . .)
(0, x1 , x2 , x3 , . . .) + (0, y1 , y2 , y3 , . . .)
T (x) + T (y)
If λ ∈ R is a scalar then
T (λx) =
=
=
=
T ((λx1 , λx2 , λx3 , . . .))
(0, λx1 , λx2 , λx3 , . . .)
λ(0, x1 , x2 , x3 , . . .)
λT (x)
We proved in class that a linear operator T : X → Y is continuous
if and only if there exists a real number M with
kT (x)k ≤ M kxk,
∀x ∈ X
Note that for any element x = (x1 , x2 , x3 , . . .) in `2 we have
kT (x)k2 = k(0, x1 , x2 , x3 , . . .)k2
!2
∞
X
2
=
0+
xj
j=1
= kxk2
1
2
MA2223: SOLUTIONS TO PROBLEM SHEET 3
This shows that T is continuous, in fact T is an isometry. We also
have
kT kop =
sup kT (x)k2
kxk2 =1
=
sup kxk2
kxk2 =1
= 1