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Math Help - Banach dual space

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    Banach dual space

    Let X be a normed vector space with Banach dual space X ′,let f be a non-zero element of X′, and let x0 be an element of X \ ker(f ). Show that every element x of X may be written in the form x = αx0 + y, with α in C and y in ker(f ). Deduce that ({f }◦down)◦up coincides with the subspace Cf .
    Prove that, for any finite-dimensional subspace M of X′, (M◦down)◦up = M.

    [M◦up = {f ∈ X′: f (x) = 0∀x ∈ M}, M◦down= {x ∈ X : f (x) = 0∀f ∈ M }].

    Does anyone know how to do this question?
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    Senior Member Shanks's Avatar
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    the problem may be posted again!
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    Okay, I can do the first two bits of this question, but not the last part: for any finite dimensional subspace, (M◦down)◦up = M. (Possibly use induction by choosing a basis of M??)
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    Quote Originally Posted by KSM08 View Post
    Okay, I can do the first two bits of this question, but not the last part: for any finite dimensional subspace, (M◦down)◦up = M. (Possibly use induction by choosing a basis of M??)
    Let \{f_1,\ldots,f_d\} be a basis for M. Define a linear map \theta:X\to F^d (where F = \mathbb{R} or \mathbb{C} is the scalar field) by \theta(x) = (f_1(x),\ldots,f_d(x)). Then \ker(\theta) = M_{\text{down}}. So \theta induces an isomorphism from X/M_{\text{down}} to F^d. Write \pi:X\to X/M_{\text{down}} for the quotient map.

    If g\in (M_{\text{down}})^{\text{up}} then g vanishes on M_{\text{down}} and therefore gives a well-defined linear functional g' on the quotient space, given by g'(\pi(x)) = g(x). Define g'':F^d\to F by g''(z) = g'(\theta^{-1}(z)). Notice that f_i''(z) = z_i (the i'th coordinate of z), for 1\leqslant i\leqslant d.

    Every linear functional on F^d is a linear combination of coordinate functionals, so there exist \alpha_1,\ldots,\alpha_d such that g'' = \textstyle\sum\alpha_if''_i. But this means that (g-\textstyle\sum\alpha_if_i)x = 0 for all x\in X. Thus g = \textstyle\sum\alpha_if_i\in M.

    That proves that (M_{\text{down}})^{\text{up}}\subseteq M. The reverse inclusion is obvious.
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