Banach dual space

**KSM08** · November 30th 2009, 02:43 AM

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?

**Shanks** · November 30th 2009, 04:15 AM

the problem may be posted again!

**KSM08** · November 30th 2009, 05:50 AM

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??)

**Opalg** · November 30th 2009, 01:01 PM

Originally Posted by KSM08

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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