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Unformatted text preview: Chapter II Elements of Functional Analysis Lecture 8 1. HahnBanach Theorems The result we are going to discuss is one of the most fundamental theorems in the whole field of Functional Analysis. Its statement is simple but quite technical. Definitions. Let K be either of the fields R or C . Suppose X is a Kvector space. A. A map q : X R is said to be a quasiseminorm , if (i) q ( x + y ) q ( x ) + q ( y ), for all x,y X ; (ii) q ( tx ) = tq ( x ), for all x X and all t R with t 0. B. A map q : X R is said to be a seminorm if, in addition to the above two properties, it satisfies: (ii) q ( x ) =   q ( x ), for all x X and all K . Remark that if q : X R is a seminorm, then q ( x ) 0, for all x X . (Use 2 q ( x ) = q ( x ) + q ( x ) q (0) = 0.) There are several versions of the HahnBanach Theorem. Theorem 1.1 (HahnBanach, Rversion) . Let X be an Rvector space. Suppose q : X R is a quasiseminorm. Suppose also we are given a linear subspace Y X and a linear map : Y R , such that ( y ) q ( y ) , for all y Y . Then there exists a linear map : X R such that (i) Y = ; (ii) ( x ) q ( x ) for all x X . Proof. We first prove the Theorem in the following: Particular Case : Assume dim X / Y = 1 . This means there exists some vector x X such that X = { y + sx : y Y , s R } . What we need is to prescribe the value ( x ). In other words, we need a number R such that, if we define : X R by ( y + sx ) = ( y )+ s , y Y , s R , then this map satisfies condition (ii). For s > 0, condition (ii) reads: ( y ) + s q ( y + sx ) , y Y , s > , and, upon dividing by s (set z = s 1 y ), is equivalent to: (1) q ( z + x ) ( z ) , z Y . For s < 0, condition (ii) reads (use t = s ): ( y ) t q ( y tx ) , y Y , t > , 53 54 LECTURE 8 and, upon dividing by t (set w = t 1 y ), is equivalent to: (2) ( w ) q ( w x ) , w Y . Consider the sets Z = { q ( z + x ) ( z ) ; z Y} R W = { ( w ) q ( w x ) : w Y} R . The conditions (1) and (2) are equivalent to the inequalities (3) sup W inf Z. This means that, in order to find a real number with the desired property, it suffices to prove that sup W inf Z , which in turn is equivalent to (4) ( w ) q ( w x ) q ( z + x ) ( z ) , z.w Y . But the condition (4) is equivalent to ( z + w ) q ( z + x ) + q ( w x ) , which is obviously satisfied because ( z + w ) q ( z + w ) = q ( ( z + x ) + ( w x ) ) q ( z + x ) + q ( w x ) . Having proved the Theorem in this particular case, let us proceed now with the general case. Let us consider the set of all pairs ( Z , ) with Z is a subspace of X such that Z Y ; : Z R is a linear functional such that (i) Y = ; (ii) ( z ) q ( z ), for all z Z ....
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 Three '10
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