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Unformatted text preview: Analysis 1
1. (i) Let z ∈ c0 be nonzero; ﬁnd φ ∈ c∗ with φ = 1 and φ(z ) = z .
(ii) Let z ∈ 1 be nonzero; ﬁnd φ ∈ ∗ with φ = 1 and φ(z ) = z .
2. For z = (zn )∞ ∈ c0 write
∞ φ(z ) =
(i) φ is a bounded linear functional of norm 1;
(ii) if z ∈ c0 is nonzero then |φ(z )| < z .
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∞
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 norm-attaining. 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