12.3. HAHN BANACH THEOREM 313
By Lemma 12.3.8, there exists y∗x ∈Y ′ such that ∥y∗x∥= 1 and y∗x (Lx)= ∥Lx∥ .Therefore,
∥L∗∥ = sup∥y∗∥≤1
∥L∗y∗∥= sup∥y∗∥≤1
sup∥x∥≤1
|L∗y∗ (x)|
= sup∥y∗∥≤1
sup∥x∥≤1
|y∗ (Lx)|= sup∥x∥≤1
sup∥y∗∥≤1
|y∗ (Lx)|
≥ sup∥x∥≤1
|y∗x (Lx)|= sup∥x∥≤1
∥Lx∥= ∥L∥
showing that ∥L∗∥ ≥ ∥L∥ and this shows part a.).If L is one to one and onto a closed subset of Y , then L(X) being a closed subspace
of a Banach space, is itself a Banach space and so the open mapping theorem impliesL−1 : L(X)→ X is continuous. Hence ∥x∥ = ∥L−1Lx∥ ≤
∥∥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 becauseif Lx1 = Lx2, then since L is one to one, it follows x1 = x2 and so 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∗∥ and also ∥x∥ ≤ ∥x∗∥∥∥L−1
∥∥∥Lx∥ and so thenorm 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 and proves 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 12.3.11 Suppose X and Y are Banach spaces, L ∈L (X ,Y ), and L is one toone 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 12.3.12 Define J : X → X ′′ by J(x)(x∗) = x∗(x).
Theorem 12.3.13 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∗∗∥ ≤ 1, x∗∗ ∈ X ′′
}.
Proof: First note that from the definition,
J (ax+by)(x∗)≡ x∗ (ax+by) = ax∗ (x)+bx∗ (y) = (aJ (x)+bJ (y))(x∗) .
12.3. HAHN BANACH THEOREM 313By Lemma 12.3.8, there exists y* € Y’ such that ||y*|| = 1 and y* (Lx) = ||Lx||. Therefore,|L*|| = sup |[L*y*||= sup sup |L*y* (x)|lly*||<1 \y* [<1 ||x|[<1= sup sup ly" (Lx)|= sup sup |y" (Lx)|lly* [<1 [lx|[<1 lxlI<1 |ly*II<1> sup ly, (Lx)| = sup ||Lx|| = ||L\]lxI|<1 \x||<1showing that ||L*|| > ||L|| and this shows part a.).If L is one to one and onto a closed subset of Y, then L(X) being a closed subspaceof a Banach space, is itself a Banach space and so the open mapping theorem impliesL~! : L(X) — X is continuous. Hence ||x|| = ||L~!Lx|| < ||L~'|| ||Lx|]. Now let x* € X’ begiven. Define f € Y(L(X),C) by f(Lx) = x*(x). The function, f is well defined becauseif Lx; = Lx, then since L is one to one, it follows xj = x2 and so f (L(x1)) = x* (x1) =x* (x2) = f (L(x1)). Also, f is linear becausef (aL (x1) +bL (x2)) = f (L (ax, + bx2)) = x* (ax, + bx2)= ax" (x1) + bx" (x2) = af (L(m1)) + bf (L022).In addition to this, | f(Lx)| = |x*(x)| < ||x*|| and also |[x|] < ||x*|] ||Z~'|| ||Zx|| and so thenorm of f on L(X) is no larger than ||x*|| uo! || By the Hahn Banach theorem, thereexists an extension of f to an element y* € Y’ such that ||y*|| < ||x*|| ||Z~'||. 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* (y) = 0for all y € Y? Pick y € Y. Since L(X) is dense in Y, there exists a sequence, { Lx, } such thatLx, — y. But then by continuity of y*, y* (y) = limy +00 y* (Lt) = limp. L*y* (x) = 0.Since y* (y) = 0 for all y, this implies y* = 0 and so L* is one to one. HfCorollary 12.3.11 Suppose X and Y are Banach spaces, L€ £(X,Y), and L is one toone 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 12.3.12 Define J: x = xX" by J(x)(x*) =x*(x).Theorem 12.3.13 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',||" |] = sup {lx** @*)] sa] <1 a eX").Proof: First note that from the definition,J (ax-+ by) (x*) =x" (ax + by) = ax* (x) + bx" (y) = (ad (x) +I (y)) (2°).
