Calculus of Real and Complex Variables

13.3.4 The Dual Space And Adjoint Operators

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

∥f ∥ = sup{|f(x)| : ||x|| \leq 1}

X^′ is called the dual space.

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

L*y*(x ) \equiv y*(Lx )

for all y^∗∈ Y ^′.

The following diagram is a good one to help remember this definition.

L* X ' \leftarrow Y ' \to X Y L

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

* (Ax,y) = (x,A y)

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^∗

, and

||x*|| \leq----1---- dist(x,V )

In the case that V =

, ||x^∗|| = 1.

Proof: Let f : Fx + V → F be defined by f(αx + v) = α||x||. First it is necessary to show f is well defined and continuous. If α₁x + v₁ = α₂x + v₂ then if α₁≠α₂, 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_nx + 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

≥ δ > 0. Then x +

v_n → 0 contradicting the assumption that x

V and V is a closed subspace. Hence f is continuous on Fx + V. Being a little more careful,

1 ||f|| = ||αxs+uvp||\leq1|f (αx + v)| = |α|||x+s(uvp∕α)||\leq1|α|||x|| = dist(x,V-) ||x||

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

and also

* ----1---- ||x || \leq ||f|| = dist(x,V )

In case V =

, the result follows from the above or alternatively,

||f || \equiv sup |f (αx)| = sup |α|||x|| = 1 ||αx||\leq1 |α|\leq1∕||x||

and so, in this case,

≤

= 1. Since x^∗(x) = ||x|| it follows

|| ( ) || ||x*|| \geq ||x* -x-- || = ||x||= 1. ||x|| ||x||

Thus

= 1. ■

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.

||L *y*|| = sup |L*y*(x)| = sup |y*(Lx)| \leq ||y*||||L||. ||x||\leq1 ||x||\leq1

Thus L^∗ is continuous as claimed and

≤

By Lemma 13.3.9, there exists y_x^∗∈ Y ^′ such that

= 1 and y_x^∗

.Therefore,

||L*|| = s*up ||L *y*|| = su*p sup |L *y* (x)| ||y||\leq1 * ||y ||\leq1||x||\leq1 * = s*up sup |y (Lx)| = sup su*p |y (Lx)| ||y||\leq1||x*||\leq1 ||x||\leq1||y||\leq1 \geq ||sux|p|\leq1 |yx(Lx)| = |s|xu|p|\leq1||Lx || = ||L ||

showing that

≥

and this shows part a.).

If L is one to one and onto a closed subset of Y , then L

being a closed subspace of a Banach space, is itself a Banach space and so the open mapping theorem implies L⁻¹ : L(X) → X is continuous. Hence

|| || ||x|| = ||L -1Lx|| \leq ||L- 1||||Lx||

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 Lx₁ = Lx₂, then since L is one to one, it follows x₁ = x₂ and so f

= x^∗

= f

. Also, f is linear because

f (aL(x1)+ bL(x2)) = f (L (ax1 + bx2)) \equiv x*(ax + bx ) * 1 2* = ax (x1)+ bx (x2) = af (L (x1))+ bf (L (x2)).

In addition to this,

|| || |f(Lx)| = |x*(x)| \leq ||x*||||x|| \leq ||x*||||L -1||||Lx ||

and so the norm of f on L

is no larger than ||x^∗||

. By the Hahn Banach theorem, there exists an extension of f to an element y^∗∈ Y ^′ such that ||y^∗||≤||x^∗||

. Then

L *y*(x) = y*(Lx) = f(Lx) = x*(x)

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^∗

= 0 for all y ∈ Y ? Pick y ∈ Y . Since L

is dense in Y, there exists a sequence,

such that Lx_n → y. But then by continuity of y^∗, y^∗

= lim_n→∞y^∗

= lim_n→∞L^∗y^∗

= 0. Since y^∗

= 0 for all y, this implies y^∗ = 0 and so L^∗ is one to one. ■

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.

Definition 13.3.12 Define J : X → X^′′ by J(x)(x^∗) = x^∗(x).

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^′,

||x*|| = sup {|x**(x*)| : ||x**|| \leq 1,x** \in X ''}.

Proof:

J (ax +by)(x*) \equiv x*(ax +by) = ax*(x)+ bx*(y) = (aJ (x)+ bJ (y))(x*).

Since this holds for all x^∗∈ X^′, it follows that

J (ax + by) = aJ (x)+ bJ (y)

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

0 = J(x)(x*) = x*(x) = ||x ||.

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

||x|| \geq sup{|y*(x)| : ||y*|| \leq 1} = sup{|J(x)(y*)| : ||y*|| \leq 1} = ||Jx || * * \geq |J (x)(x )| = |x (x)| = ||x||

Therefore,

as claimed. Therefore,

||J|| = sup{||Jx|| : ||x|| \leq 1} = sup{||x|| : ||x|| \leq 1} = 1.

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

||xn - xm|| = ||Jxn - Jxm||

and

is a Cauchy sequence. Then Jx = lim_n→∞Jx_n = y^∗∗.

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.

* * * ||x || = sup {|x (x)| : ||x|| \leq 1} = sup {|J (x)(x )| : ||Jx|| \leq 1}

** * ** * ** ** \leq sup {|x (x )| : ||x || \leq 1} = sup {|J (x )(x )| : ||x || \leq 1}

* * \equiv ||Jx || = ||x ||.■

Definition 13.3.14 When J maps X onto X^′′, X is called reflexive.