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Math Help - Normed space

  1. #1
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    Normed space

    Let X be a normed space, F be a closed linear subspace of X,
    Let z be in X, z is not in F.
    Let S={x+az:a is in the field Phi}=Span of F and z
    We show S is closed.

    I would define a function f : S--> Phi by f(x+az)=a
    and show that |f|< or equal to 1/d where d= distance(z,F)
    hence f is in S* (dual of S).
    So we let {x_n +a_n z} be a sequence converging to w (x_n is in F,a_n is in Phi). We show w is in S.
    We note {f(x_n +a_n z)} is a cauchy sequence.

    But I am not sure how to proceed.
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  2. #2
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    Quote Originally Posted by math8 View Post
    Let X be a normed space, F be a closed linear subspace of X,
    Let z be in X, z is not in F.
    Let S={x+az:a is in the field Phi}=Span of F and z
    We show S is closed.

    I would define a function f : S--> Phi by f(x+az)=a
    and show that |f|< or equal to 1/d where d= distance(z,F)
    hence f is in S* (dual of S).
    So we let {x_n +a_n z} be a sequence converging to w (x_n is in F,a_n is in Phi). We show w is in S.
    We note {f(x_n +a_n z)} is a cauchy sequence.

    But I am not sure how to proceed.
    That's a very good start. In fact, (f(x_n +a_n z)) is a Cauchy sequence of scalars, and since the scalar field (presumably \mathbb{R} or \mathbb{C}) is complete, that sequence must converge to a scalar a. In other words, a_n\to a.

    Then x_n = w-a_nz\to w-az = y say, as n→∞. But F is closed, and therefore y∈F. Thus w = y + az \in S.
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    Thanks, this makes a lot of sense.

    But how do I show that |f|< or equal to 1/d where d= distance(z,F)

    I know it involves some corollary to the Hahn-Banach theorem:
    " If Y is a linear subspace of the normed space X, x is in X and d=dist(x,Y)>0, then there exists x* in X* such that x*(x)=1 and |x*|=1/d and x*|Y=0 (x* restricted to Y is 0)".

    I think maybe the reason we want this to be true is that in that case, f would be bounded hence continuous, thus f is in the dual S*.

    But how does this help?
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  4. #4
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    Quote Originally Posted by math8 View Post
    Thanks, this makes a lot of sense.

    But how do I show that |f|< or equal to 1/d where d= distance(z,F)

    I know it involves some corollary to the Hahn-Banach theorem:
    " If Y is a linear subspace of the normed space X, x is in X and d=dist(x,Y)>0, then there exists x* in X* such that x*(x)=1 and |x*|=1/d and x*|Y=0 (x* restricted to Y is 0)".

    I think maybe the reason we want this to be true is that in that case, f would be bounded hence continuous, thus f is in the dual S*.

    But how does this help?
    Yes, that is exactly what you need. In fact, if you apply the HB theorem as you have stated it, taking Y to be the subspace F and x to be the element z, then the functional x* satisfies x*|F=0 and x*(z)=1. So for each element v in F, x*(v+az) = a. In other words, x* is the functional f that you want (and you need it to be continuous because you want it to take a Cauchy sequence to a Cauchy sequence).
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  5. #5
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    Thanks a lot that is very helpful
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