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Subspace of Normed Vector Space

  1. Sep 10, 2009 #1
    Let X be a normed vector space. If C is a closed subspace x is a point in X not in C, show that the set C+Fx is closed. (F is the underlying field of the vector space).
     
  2. jcsd
  3. Sep 11, 2009 #2
    Hi there!

    Here's my suggestion, but it needs an overview from someone familiar with functional analysis, to be sure it's correct or incorrect:

    As X is a normed space, C and Fx are also normed as subspaces of a normed space. So the convergence in these subspaces is set with respect to the original norm of X.

    Let [tex]c_n, n\in N;c_n\longwrightarrow\c,c\in C[/tex] be a convergent sequence in C+Fx with limit c. Then, following the structure of the space (because x does not lie in C, C+Fx is a direct sum), we could write [tex]c_n=a_n+\lambda_n x; a_n\in C,\lambda_n\in F[/tex] both convergent sequences [tex]a_n\rightarrow a ;\lambda_n\rightarrow\lambda[/tex] (otherwise c_n could not be convergent).

    Since C is closed then [tex]a\in C[/tex] and (assuming F is closed or complete)*** [tex]\lambda\in F[/tex]. So the limit takes the form [tex]c_n\rightarrow a+\lambda x\in C+Fx[/tex] and therefore lies in the space.


    *** I am not sure, but I think F must also be given as closed. (as a counter example consider the space R^2 over the rationals Q: Set C to be the disc around the origin (closed) and x be some point not in C)

    I hope this proof is true at least to some extend.

    [edit: spelling :)]
     
    Last edited: Sep 12, 2009
  4. Sep 11, 2009 #3
    Marin is right, you need [itex]F[/itex] to be a complete field, for example the reals or complexes. Otherwise, from [itex]\lambda_n x[/itex] convergent we cannot conclude the limit is a scalar multiple of [itex]x[/itex].
     
  5. Sep 12, 2009 #4
    Hi again!

    As I look at the proof once again (assuming the proof is correct), the condition X to be a normed space could actually be reduced. The same proof will also be true for any metric space (X,d) with a metric d , as the metric induces the same metric on the subspaces.
     
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