624 CHAPTER 20. REPRESENTATION THEOREMS

Proof: Suppose first that ||φ || = 1. Then by assumption, there is x such that ∥x∥ = 1and φ (x) = 1 = ∥φ∥ . Then φ (y−φ(y)x) = 0 and so

φ(x+ t(y−φ(y)x)) = φ (x) = 1 = ||φ ||.

Therefore, ||x+ t(y−φ(y)x)|| ≥ 1 since otherwise ||x+ t(y−φ(y)x)||= r < 1 and so

φ

((x+ t(y−φ(y)x))

1r

)=

1r

φ (x) =1r

which would imply that ||φ ||> 1.Also for small t, |φ(y)t|< 1, and so

1≤ ||x+ t (y−φ(y)x)||= ||(1−φ(y)t)x+ ty||

≤ |1−φ (y) t|∥∥∥∥x+

t1−φ (y) t

y∥∥∥∥.

Divide both sides by |1−φ (y) t|. Using the standard formula for the sum of a geometricseries,

1+ tφ (y)+o(t) =1

1− tφ(y)

Therefore,

1|1−φ (y) t|

= |1+φ (y) t +o(t)| ≤∥∥∥∥x+

t1−φ (y) t

y∥∥∥∥= ∥x+ ty+o(t)∥ (20.5.15)

where limt→0 o(t)(t−1) = 0. Thus,

|1+φ (y) t| ≤ ∥x+ ty∥+o(t)

Now |1+ tφ (y)|−1≥ 1+ t Reφ (y)−1 = t Reφ (y) .Thus for t > 0,

Reφ (y) ≤ |1+ tφ (y)|−1t

∥x∥=1≤ ||x+ ty||− ||x||

t+

o(t)t

and for t < 0,

Reφ (y)≥ |1+ tφ (y)|−1t

≥ ||x+ ty||− ||x||t

+o(t)

tBy assumption, letting t→ 0+ and t→ 0−,

Reφ (y) = limt→0

||x+ ty||− ||x||t

= ψ′y (0) .

Nowφ (y) = Reφ(y)+ i Imφ(y)

soφ(−iy) =−i(φ (y)) =−iReφ(y)+ Imφ(y)