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245B, Winter 2009, Assignment 4:
Notes and selected model answers
(Model solutions follow question numbers in
bold
.)
Folland Chapter 5
22.
c.
We assume from parts a. and b. that we have already deﬁned the adjoint
operator
T
†
∈
L
(
Y
*
,X
*
)
associated to a bounded linear operator
T
∈
L
(
X,Y
)
.
We will prove that
T
†
is injective if and only if
T
has dense range.
(
⇒
)
We prove the contrapositive. Suppose that
R
(
T
)
$
Y
. Then since
R
(
T
)
is a proper closed subspace of
Y
, by the HahnBanach Theorem there exists some
f
∈
Y
*
with
f
6
= 0
but
f

R
(
T
)
= 0
, and hence in particular
f

R
(
T
)
= 0
. This latter
restrictions tells us that for any
x
∈
X
we have
T
†
f
(
x
) =
def
f
(
Tx
) = 0
, and thus
that
T
†
f
= 0
, so
T
†
is not injective.
(
⇐
)
Once again we prove the contrapositive. If
T
†
f
=
T
†
g
for some
f
6
=
g
in
Y
*
, then
T
†
h
= 0
for
h
:=
f

g
. This tells us that
h
(
Tx
) =
T
†
h
(
x
) = 0
for
all
x
∈
X
, so since
h
6
= 0
we deduce that
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 Spring '10
 tao/analysis

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