19.2. THE HILBERT SPACE L(U) 527

the latter set being a closed convex nonempty set thanks to Lemma 19.2.4. Then by theseparation theorem, Theorem 18.2.5 there exists z ∈ H such that

Re(T1 (u0) ,z)H > 1 > Re(T2 (v) ,z)H

for all ||v||2 ≤ c. Therefore, replacing v with vθ where θ is a suitable complex numberhaving modulus 1, it follows

||T ∗1 z||> 1 >∣∣∣(v,T ∗2 z)U2

∣∣∣ (19.2.15)

for all ||v||2 ≤ c. If c = 0, 19.2.15 gives a contradiction immediately because of 19.2.12.Assume then that c > 0. From 19.2.15, if ||v||2 ≤ 1, then∣∣∣(v,T ∗2 z)U2

∣∣∣< 1c<

1c||T ∗1 z||

Then from 19.2.15,

||T ∗2 z||U2= sup||v||≤1

∣∣∣(v,T ∗2 z)U2

∣∣∣≤ 1c<

1c||T ∗1 z||

which contradicts 19.2.12. Therefore, it is clear that T1 (U1)⊆ T2 (U2).Now consider the second claim. The first part shows T1 (U1) = T2 (U2). Denote by

ui ∈ Ui, the point T−1i x. Without loss of generality, it can be assumed x ̸= 0 because if

x = 0, then the definition of T−1i gives T−1

i (x) = 0. Thus for x ̸= 0 neither ui can equal 0. Ineed to verify that ||u1||1 = ||u2||2. Suppose then that this is not so. Say ||u1||1 > ||u2||2 > 0.

x||u2||2

= T2

(u2

||u2||2

)∈ T2

(B(0,1)

)But from the first part of the theorem this equals T1

(B(0,1)

)and so there exists u′1 ∈

B(0,1) such thatx

||u2||2= T1u′1

Hence

T1

(u′1−

u1

||u2||2

)=

x||u2||2

− x||u2||2

= 0.

From Theorem 19.2.3 this implies

0 =

(u1,u′1−

u1

||u2||2

)≤ ||u1||1

∣∣∣∣u′1∣∣∣∣1−||u1||1||u1||1||u2||2

= ||u1||1(∣∣∣∣u′1∣∣∣∣1− ||u1||1

||u2||2

)≤ ||u1||1

(1− ||u1||1||u2||2

)which is a contradiction because it was assumed ||u1||1

||u2||2> 1. This proves the theorem.