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Unformatted text preview: Analysis 1
Homework 02
1. (i) Let z ∈ c0 be nonzero; ﬁnd φ ∈ c∗ with φ = 1 and φ(z ) = z .
0
(ii) Let z ∈ 1 be nonzero; ﬁnd φ ∈ ∗ with φ = 1 and φ(z ) = z .
1
2. For z = (zn )∞ ∈ c0 write
n=1
∞ φ(z ) =
n=1 zn
.
2n Prove:
(i) φ is a bounded linear functional of norm 1;
(ii) if z ∈ c0 is nonzero then φ(z ) < z .
Solutions
1. (i) As z ∈ c0 there exists N such that zN  = z ∞ > 0. The speciﬁc
‘coordinate’ linear functional φN : c0 →: w → wN clearly does the job in this
case: if w ∈ c0 then φN (w) = wN 
w ∞ while φN (z ) = zN  = z ∞ .
(ii) Not quite so straightforward. Seek: a ∈ ∞ such that a ∞ = 1 and
Ta (z ) = z 1  that is, n∈N an zn = n∈N zn . Found: for each n let an be
any scalar of absolute value 1 such that an zn = zn   that is, an = zn /zn if
zn = 0 and (say) an = 1 if zn = 0.
2. If z ∈ c0 then each zn  z
∞ φ(z )
n=1 ∞ so that
∞ zn 
2n n=1 z∞
=z
2n ∞ This proves that φ
1. Moreover, if z ∈ c0 is nonzero then (as zn → 0)
some zN  < z ∞ whence φ(z ) < z ∞ (inspect the second inequality in
the chain displayed above).
Note: Thus, φ ∈ c0 is not normattaining. 1 ...
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This note was uploaded on 07/08/2011 for the course MHF 3202 taught by Professor Larson during the Spring '09 term at University of Florida.
 Spring '09
 LARSON

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