312 CHAPTER 12. BANACH SPACES
Lemma 12.3.8 Let X be a normed linear space and let x ∈ X \V where V is a closedsubspace of X. Then there exists x∗ ∈ X ′ such that x∗(x) = ∥x∥, x∗ (V ) = {0}, and
∥x∗∥ ≤ 1dist(x,V )
In the case that V = {0} , ∥x∗∥= 1.
Proof: Let f :Fx+V →F be defined by f (αx+v)=α∥x∥. First it is necessary to showf is well defined and continuous. If α1x+v1 = α2x+v2 then if α1 ̸= α2, then x ∈V whichis assumed not to happen so f is well defined. It remains to show f is continuous. Supposethen that αnx+ vn→ 0. It is necessary to show αn→ 0. If this does not happen, then thereexists a subsequence, still denoted by αn such that |αn| ≥ δ > 0. Then 1
|αn| ≤1δ
. But thenx+(1/αn)vn→ 0 so x is a limit of points of V which is closed, and this means x∈V whichis not so. Hence f is continuous on Fx+V. Now dist(x,V ) ≡ inf{∥x+ v∥ : v ∈V} . Thusif it is required that |α|∥x+(v/α)∥ ≤ 1, to make |α| as large as possible one would make∥x+(v/α)∥ as small as possible. Hence, sup|α|∥x+(v/α)∥≤1 |α|= 1
dist(x,V ) . Therefore,
∥ f∥= sup∥αx+v∥≤1
| f (αx+ v)|= sup|α|∥x+(v/α)∥≤1
|α|∥x∥= 1dist(x,V )
∥x∥
By the Hahn Banach theorem, there exists x∗ ∈ X ′ such that x∗ = f on Fx+V. Thus x∗ (x) =∥x∥ and also ∥x∗∥ ≤ ∥ f∥= 1
dist(x,V ) .In case V = {0} , the result follows from the above or alternatively,
∥ f∥ ≡ sup∥αx∥≤1
| f (αx)|= sup|α|≤1/∥x∥
|α|∥x∥= 1,
and so, in this case, ∥x∗∥ ≤ ∥ f∥= 1. Since x∗(x) = ∥x∥ it follows
∥x∗∥ ≥∣∣∣∣x∗( x
∥x∥
)∣∣∣∣= ∥x∥∥x∥ = 1.
Thus ∥x∗∥= 1. ■Note that this says that if x ̸= y, then there exists x∗ ∈ X ′ with x∗ (x− y) = ∥x− y∥ and
so x∗ (x) ̸= x∗ (y) . This proves
Proposition 12.3.9 If x ̸= y, there exists x∗ ∈ X ′ such that x∗ (x) ̸= x∗ (y).
Theorem 12.3.10 Let L ∈L (X ,Y ) where X and Y are Banach spaces. Thena.) 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 immediatelyfrom the definition. As usual, the interesting thing concerns continuity.
∥L∗y∗∥= sup∥x∥≤1
|L∗y∗ (x)|= sup∥x∥≤1
|y∗ (Lx)| ≤ ∥y∗∥∥L∥ .
Thus L∗ is continuous as claimed and ∥L∗∥ ≤ ∥L∥ .