≤m for all FEF. subset of X. Then |||| ≤a for all & E nd F € F, ||F((rx/a)+a) || ≤m, so all /TI
≤m for all FEF. subset of X. Then |||| ≤a for all & E nd F € F, ||F((rx/a)+a) || ≤m, so all /TI
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter5: Rings, Integral Domains, And Fields
Section5.4: Ordered Integral Domains
Problem 1E: Complete the proof of Theorem 5.30 by providing the following statements, where and are arbitrary...
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![9.1 Theorem (Uniform boundedness principle)
Let X be a Banach space, Y be a normed space and ♬ be a subset
of BL(X, Y) such that for each x € X, the set {F(x) : F € F} is
bounded in Y. Then for each bounded subset E of X, the set {F(x) :
x € E, F € F} is bounded in Y, that is, F is uniformly bounded on
E. In particular, sup{||F|| : F € F} <∞0.
Proof:
For n = 1,2,..., let
D₁ = {x € X : ||F(x)|| > n for some F € F}.
For each FF, the function x→→ ||F(x)|| is continuous on X, so that
the set {x € X : ||F(x)|| > n} is open in X. Since D₁ is the union of
all these sets, it follows that Dn is open in X.
Let x E X. Then, by hypothesis, ||F(x)| ≤ n for all F € F
and some positive integer n, that is, xDn. Thus ₁ D₂ = 0.
Consequently, Dn cannot be dense in X.
Since X is a Banach space, Baire's theorem (3.4) implies that
some Dm must not be dense in X. Then there is some a € X and
some r> 0 such that Ux(a,r)Dm = 0, that is, if y € X and
|ly - all ≤r, then ||F(y)|| ≤ m for all F € F.
Let E be a bounded subset of X. Then |||| ≤a for all z € E and
some a > 0. For x € E and F = F, ||F((rx/a) + a)|| ≤m, so that
||F(x) || = = ||F(77)|||
7|||F (77 + a) − F(a)||
≤ | F (+a) || + || F(a)}||] =
2am
r
Thus sup{||F(x)|| : x ≤ E, F = F} ≤ 2am/r, that is, F is uniformly
bounded on E. If we let E = Ux(0, 1), it follows that
sup{||F|| : F € F}
sup{||F(x)|| : x € X, ||x|| ≤ 1, F=F}
<∞.
2m
T](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2Fbcac2821-07bf-4505-8d44-b2640f007d79%2Fbt5esj_processed.png&w=3840&q=75)
Transcribed Image Text:9.1 Theorem (Uniform boundedness principle)
Let X be a Banach space, Y be a normed space and ♬ be a subset
of BL(X, Y) such that for each x € X, the set {F(x) : F € F} is
bounded in Y. Then for each bounded subset E of X, the set {F(x) :
x € E, F € F} is bounded in Y, that is, F is uniformly bounded on
E. In particular, sup{||F|| : F € F} <∞0.
Proof:
For n = 1,2,..., let
D₁ = {x € X : ||F(x)|| > n for some F € F}.
For each FF, the function x→→ ||F(x)|| is continuous on X, so that
the set {x € X : ||F(x)|| > n} is open in X. Since D₁ is the union of
all these sets, it follows that Dn is open in X.
Let x E X. Then, by hypothesis, ||F(x)| ≤ n for all F € F
and some positive integer n, that is, xDn. Thus ₁ D₂ = 0.
Consequently, Dn cannot be dense in X.
Since X is a Banach space, Baire's theorem (3.4) implies that
some Dm must not be dense in X. Then there is some a € X and
some r> 0 such that Ux(a,r)Dm = 0, that is, if y € X and
|ly - all ≤r, then ||F(y)|| ≤ m for all F € F.
Let E be a bounded subset of X. Then |||| ≤a for all z € E and
some a > 0. For x € E and F = F, ||F((rx/a) + a)|| ≤m, so that
||F(x) || = = ||F(77)|||
7|||F (77 + a) − F(a)||
≤ | F (+a) || + || F(a)}||] =
2am
r
Thus sup{||F(x)|| : x ≤ E, F = F} ≤ 2am/r, that is, F is uniformly
bounded on E. If we let E = Ux(0, 1), it follows that
sup{||F|| : F € F}
sup{||F(x)|| : x € X, ||x|| ≤ 1, F=F}
<∞.
2m
T
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