Definition 13.3.7 Let X be a Banach space. Denote by X^{′} the space of continuous linear functions which map X to the field of scalars. Thus X^{′} = ℒ(X, F). By Theorem 13.1.11 on Page 1013, X^{′} is a Banach space. Remember with the norm defined on ℒ

Definition 13.3.8 Let X and Y be Banach spaces and suppose L ∈ℒ(X,Y ). Then define the adjoint map in ℒ(Y ^{′},X^{′}), denoted by L^{∗}, by

for all y^{∗}∈ Y ^{′}.
The following diagram is a good one to help remember this definition.

This is a generalization of the adjoint of a linear transformation on an inner product space. Recall

What is being done here is to generalize this algebraic concept to arbitrary Banach spaces. There are some issues which need to be discussed relative to the above definition. First of all, it must be shown that L^{∗}y^{∗}∈ X^{′}. Also, it will be useful to have the following lemma which is a useful application of the Hahn Banach theorem.
Lemma 13.3.9 Let X be a normed linear space and let x ∈ X ∖ V where V is a closed subspace of X. Then there exists x^{∗}∈ X^{′} such that x^{∗}(x) = x, x^{∗}

In the case that V =
Proof: Let f : Fx + V → F be defined by f(αx + v) = αx. First it is necessary to show f is well defined and continuous. If α_{1}x + v_{1} = α_{2}x + v_{2} then if α_{1}≠α_{2}, then x ∈ V which is assumed not to happen so f is well defined. It remains to show f is continuous. Suppose then that α_{n}x + v_{n} → 0. It is necessary to show α_{n} → 0. If this does not happen, then there exists a subsequence, still denoted by α_{n} such that

By the Hahn Banach theorem, there exists x^{∗}∈ X^{′} such that x^{∗} = f on Fx + V. Thus x^{∗}

In case V =

and so, in this case,

Thus
Theorem 13.3.10 Let L ∈ℒ(X,Y ) where X and Y are Banach spaces. Then
a.) L^{∗}∈ℒ(Y ^{′},X^{′}) as claimed and L^{∗} = L.
b.) If L maps one to one onto a closed subspace of Y , then L^{∗} is onto.
c.) If L maps onto a dense subset of Y , then L^{∗} is one to one.
Proof: It is routine to verify L^{∗}y^{∗} and L^{∗} are both linear. This follows immediately from the definition. As usual, the interesting thing concerns continuity.

Thus L^{∗} is continuous as claimed and
By Lemma 13.3.9, there exists y_{x}^{∗}∈ Y ^{′} such that
If L is one to one and onto a closed subset of Y , then L

Now let x^{∗}∈ X^{′} be given. Define f ∈ℒ(L(X), ℂ) by f(Lx) = x^{∗}(x). The function, f is well defined because if Lx_{1} = Lx_{2}, then since L is one to one, it follows x_{1} = x_{2} and so f

and so the norm of f on L

so L^{∗}y^{∗} = x^{∗} because this holds for all x. Since x^{∗} was arbitrary, this shows L^{∗} is onto and proves b.).
Consider the last assertion. Suppose L^{∗}y^{∗} = 0. Is y^{∗} = 0? In other words is y^{∗}
Corollary 13.3.11 Suppose X and Y are Banach spaces, L ∈ℒ(X,Y ), and L is one to one and onto. Then L^{∗} is also one to one and onto.
There exists a natural mapping, called the James map from a normed linear space, X, to the dual of the dual space which is described in the following definition.
Theorem 13.3.13 The map, J, has the following properties.
a.) J is one to one and linear.
b.) Jx = x and J = 1.
c.) J(X) is a closed subspace of X^{′′} if X is complete.
Also if x^{∗}∈ X^{′},

Proof:

and so J is linear. If Jx = 0, then by Lemma 13.3.9 there exists x^{∗} such that x^{∗}(x) = x and x^{∗} = 1.Then

This shows a.).
To show b.), let x ∈ X and use Lemma 13.3.9 to obtain x^{∗}∈ X^{′} such that x^{∗}(x) = x with x^{∗} = 1. Then

This shows b.).
To verify c.), use b.). If Jx_{n} → y^{∗∗}∈ X^{′′} then by b.), x_{n} is a Cauchy sequence converging to some x ∈ X because

and
Finally, to show the assertion about the norm of x^{∗}, use what was just shown applied to the James map from X^{′} to X^{′′′} still referred to as J.


