1. ## 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?

2. the problem may be posted again!

3. 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??)

4. 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 $\displaystyle \{f_1,\ldots,f_d\}$ be a basis for M. Define a linear map $\displaystyle \theta:X\to F^d$ (where $\displaystyle F = \mathbb{R}$ or $\displaystyle \mathbb{C}$ is the scalar field) by $\displaystyle \theta(x) = (f_1(x),\ldots,f_d(x))$. Then $\displaystyle \ker(\theta) = M_{\text{down}}$. So $\displaystyle \theta$ induces an isomorphism from $\displaystyle X/M_{\text{down}}$ to $\displaystyle F^d$. Write $\displaystyle \pi:X\to X/M_{\text{down}}$ for the quotient map.

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

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

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