1Normed vector spaces

II Linear Analysis



1.3 Adjoint
The idea of the adjoint is given a
T B
(
X, Y
), produce a “dual map”, or an
adjoint T
B(Y
, X
).
There is really only one (non-trivial) natural way of doing this. First we can
think about what
T
should do. It takes in something from
Y
and produces
something in
X
. By the definition of the dual space, this is equivalent to taking
in a function g : Y F and returning a function T
(g) : X F.
To produce this
T
(
g
), the only things we have on our hands to use are
T
:
X Y
and
g
:
Y F
. Thus the only option we have is to define
T
(
g
)
as the composition g T , or T
(g)(x) = g(T (x)) (we also have a silly option of
producing the zero map regardless of input, but this is silly). Indeed, this is the
definition of the adjoint.
Definition
(Adjoint)
.
Let
X, Y
be normal vector spaces. Given
T B
(
X, Y
),
we define the adjoint of T , denoted T
, as a map T
B(Y
, X
) given by
T
(g)(x) = g(T (x))
for x X, y Y
. Alternatively, we can write
T
(g) = g T.
It is easy to show that our T
is indeed linear. We now show it is bounded.
Proposition. T
is bounded.
Proof.
We want to show that
kT
k
B(Y
,X
)
is finite. For simplicity of notation,
the supremum is assumed to be taken over non-zero elements of the space. We
have
kT
k
B(Y
,X
)
= sup
gY
kT
(g)k
X
kgk
Y
= sup
gY
sup
xX
|T
(g)(x)|/kxk
X
kgk
Y
= sup
gY
sup
xX
|g(T x)|
kgk
Y
kxk
X
sup
gY
sup
xX
kgk
Y
kT xk
Y
kgk
Y
kxk
X
sup
xX
kT k
B(X,Y )
kxk
X
kxk
X
= kT k
B(X,Y )
So it is finite.