Lecture 3
Recall that if
A
=
{k
T
(
x
)
k
:
x
∈
X
and
k
x
k
6
1
}
and
B
=
{
M
:
x
∈
X
=
⇒ k
T
(
x
)
k
6
M
k
x
k}
,
we claim that
sup
A
= inf
B.
We have already shown that sup
A
6
inf
B
, so all that remains is to prove
that sup
A
>
inf
B
.
Proof.
Suppose sup
A
6∈
B
. Then there is an
x
∈
X
such that
k
T
(
x
)
k
>
(sup
A
)
k
x
k
;
so
x
6
= 0 and
T
x
k
x
k
>
sup
A.
But
k
T
(
x/
k
x
k
)
k ∈
A
, a contradiction; hence sup
A
∈
B
and thus sup
A
>
inf
B
.
Definition 1.
The common value
sup
A
= inf
B
in the preceding theorem is
denoted
k
T
k
(the ‘
operator norm
’ of
T
).
Notice that
x
∈
X
=
⇒ k
T
(
x
)
k
6
k
T
k k
x
k
(because sup
A
∈
B
).
Theorem 1.
L
(
X, Y
)
is a vector space upon which the operator norm is a
norm.
1
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Proof.
Note that addition and scalar multiplication are pointwise:
(
αA
)(
x
) =
αA
(
x
)
(
B
+
C
)(
x
) =
B
(
x
) +
C
(
x
)
.
So a sample of the standard types of arguments needed is as follows:
if
B, C
∈
L
(
X, Y
) and
x
∈
X
, then
k
(
B
+
C
)(
x
)
k
=
k
Bx
+
Cx
k
6
k
Bx
k
+
k
Cx
k
(
4
inequality in
Y
)
6
k
B
k k
x
k
+
k
C
k k
x
k
= (
k
B
k
+
k
C
k
)
k
x
k
,
therefore
B
+
C
∈
L
(
X, Y
) and
k
B
+
C
k
6
k
B
k
+
k
C
k
.
Exercise
:
View the matrix
T
=
a
b
c
d
as a linear map from
R
2
to
R
2
;
then
T
x
y
=
ax
+
by
cx
+
dy
.
Give
R
2
the
l
2
norm; then
x
y
=
p
x
2
+
y
2
.
Calculate
k
T
k
.
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 Summer '10
 Staff
 Linear Algebra, Algebra, Vector Space, Tn, λm xm, T xm, Tq ∈

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