3 In general case one uses Zorns lemma Let F be the family of linear extensions

3 in general case one uses zorns lemma let f be the

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3. In general case one uses Zorn’s lemma. Let F be the family of linear extensions, F , of f to a subspace M F with F ( x ) p ( x ). It is partially ordered by inclusion: F 1 F 2 if { ( x, F 1 ( x )) : x M F 1 } ⊂ { ( x, F 2 ( x ) : x M F 2 } . Since the union of increasing family of subspaces of X is a subspace, one has that the union of totally ordered subfamilies of F lies in F . Hence by Zorn’s lemma F has a maximal element. It is defined on the whole space since otherwise it would admit an extension and, hence, would not be the maximal element of F . Remark 12.4. 1. Step 2 is presented here with the main reason to demonstrate that, for separable spaces, the theorem can be proved without using the Zorn lemma. 2. If p is a seminorm/norm then f ( x ) p ( x ) ⇔ | f ( x ) | ≤ p ( x ) . In fact, in this case, | f ( x ) | = ± f ( x ) = f ( ± x ) and f ( x ) p ( x ) ⇐ − f ( x ) p ( x ) = p ( x ) giving the statement. 68
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Theorem 12.5. ( The complex Hahn-Banach Theorem ) Let X be a complex vector space, p a seminorm on X , M a subspace of X , and f a complex linear functional on M such that | f ( x ) | ≤ p ( x ) for x M . Then there exists a complex linear functional F on X such that | F ( x ) | ≤ p ( x ) for all x X and F ( x ) = f ( x ) for all x M . Proof. Consider X and M as vector spaces over reals and denote them by X R and M R respectively. Note that X R ( M R ) and X ( M respectively) are different as linear spaces but coincides as sets. Clearly p is a sublinear functional on X R . Let u = Re f . Then u is a real linear functional on M R satisfying the condition | u ( x ) | ≤ p ( x ) and hence u ( x ) p ( x ). By the real Hahn-Banach theorem there exists a real linear functional U on X R such that U ( x ) p ( x ), for all x X R and U ( x ) = u ( x ) for all x M R . Clearly U ( x ) = U ( x ) p ( x ) = p ( x ) and hence | U ( x ) | ≤ p ( x ) , x X R . Define a functional F on X by letting F ( x ) = U ( x ) iU ( ix ) . F is a complex linear functional: F is real linear since so is U and F ( ix ) = U ( ix ) iU ( x ) = U ( ix ) + iU ( x ) = i ( U ( x ) iU ( ix )) = iF ( x ) , x X. F is an extension of f to X , i.e. F ( x ) = f ( x ), x M : as Im f ( x ) = Re( if ( x )) = Re( f ( ix )) = u ( ix ) we have F ( x ) = u ( x ) iu ( ix ) = Re f ( x ) + i Im f ( x ) = f ( x ) . It remains to prove that | F ( x ) | ≤ p ( x ) for all x X . Assume contrary that for some x 0 X , | F ( x 0 ) | > p ( x 0 ). Write F ( x 0 ) = | F ( x 0 ) | e . Let y 0 = e x 0 . Then U ( y 0 ) = Re F ( y 0 ) = Re[ e F ( x 0 )] = | F ( x 0 ) | > p ( x 0 ) = p ( y 0 ) . A contradiction that gives the statement. Corollary 12.6. Let X be a normed space and let M be its subspace. Then for every linear continuous functional f defined on M there exists a functional F X such that F ( x ) = f ( x ) for all x M and F = f . Proof. Let p ( x ) = f ∥∥ x , x X . We have | f ( x ) | ≤ p ( x ) for any x M . By the complex Hahn-Banach theorem there exists a linear functional F on X such that | F ( x ) | ≤ p ( x ) = f ∥∥ x for all x X and F | M = f . Hence, F is bounded and F ∥ ≤ ∥ f . As F = sup {| F ( x ) | : x ∥ ≤ 1 , x X } ≥ sup {| F ( x ) | : x ∥ ≤ 1 , x M } = f , we have F = f . 69
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12.1 Corollaries of the Hahn-Banach Theorem Theorem 12.7. Let X be a normed space.
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