450 CHAPTER 17. BANACH SPACES

f (L(x1)) = x∗ (x1) = x∗ (x2) = f (L(x1)). Also, f is linear because

f (aL(x1)+bL(x2)) = f (L(ax1 +bx2))

≡ x∗ (ax1 +bx2)

= ax∗ (x1)+bx∗ (x2)

= a f (L(x1))+b f (L(x2)) .

In addition to this,

| f (Lx)|= |x∗(x)| ≤ ||x∗|| ||x|| ≤ ||x∗||∣∣∣∣L−1∣∣∣∣ ||Lx||

and so the norm of f on L(X) is no larger than ||x∗||∣∣∣∣L−1

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

∣∣∣∣L−1∣∣∣∣. 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 andproves b.).

Consider the last assertion. Suppose L∗y∗ = 0. Is y∗ = 0? In other words is y∗ (y) = 0for all y∈Y ? Pick y∈Y . Since L(X) is dense in Y, there exists a sequence, {Lxn} such thatLxn → y. But then by continuity of y∗, y∗ (y) = limn→∞ y∗ (Lxn) = limn→∞ L∗y∗ (xn) = 0.Since y∗ (y) = 0 for all y, this implies y∗ = 0 and so L∗ is one to one.

Corollary 17.2.11 Suppose X and Y are Banach spaces, L ∈L (X ,Y ), and L is one to oneand 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 17.2.12 Define J : X → X ′′ by J(x)(x∗) = x∗(x).

Theorem 17.2.13 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∗∗|| ≤ 1, x∗∗ ∈ X ′′

}.

Proof:

J (ax+by)(x∗) ≡ 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)