# Functional Analysis: Linearly Independent elements of dual space

Chapter 5 (Folland Real Analysis )Problem 52. (parts (a) and (b) )

I am really clueless about the first 2 parts of problem this problem as to how to approach it. I guess that we do not have to apply the results of section 5.4.

Let $\mathcal{X}$ be a Banach space and $f_{1},\ldots,f_{n}$ be linearly independent elements of $\mathcal{X}^{\star}$. Now

a) Define $T:\mathcal{X}\to \mathbb{C}^{n}$ by $Tx = (f_{1}(x),\ldots,f_{n}(x))$. If $N = \{x: Tx = 0\}$ and $\mathcal{M}$ is the linear span of $f_{1},\ldots,f_{n}$, then $\mathcal{M} = \mathcal{N}^{0}$, where $\mathcal{N}^{0} = \{f\in\mathcal{X}^{\star}: f|\mathcal{N} = 0\}$. Hence $\mathcal{M}^{\star}$ is isomorphic to $(\mathcal{X}/\mathcal{N})^{\star}$.

b) If $F\in \mathcal{X}^{\star\star}$, for any $\epsilon >0$ there exists $x\in \mathcal{X}$ such that $F(f_{j}) = f_{j}(x)$ for $j = 1,\ldots, n$ and $||x ||\leq (1+\epsilon)||F||$

($F|\mathcal{M}$ can be identified with an element of $(\mathcal{X}/\mathcal{N})^{\star\star}$ and hence an element of $\mathcal{X}/\mathcal{N}$, since the latter is finite dimensional.

for part (b), I did not get the last part, i.e. how does definition of ||x+N|| to obtain the inequality ||x+N||≥||x||(1+ϵ) . Note that since F is fixed (i.e. x is fixed, the above inequality should be true for all x and not for some x.)

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Sorry. Folland Real Analysis – user24367 Feb 22 '12 at 3:37

Clearly $\mathcal M\subseteq \mathcal N^0$. Let $f\in \mathcal N^0$ be given. By assumption we have $\ker T \subseteq \ker f$, and it follows from linear algebra that $f = g \circ T$, where $g$ is a functional on $\mathbb C^n$. But then $g$ corresponds to a vector $(a_1,...,a_n)\in \mathbb C^n$ hence $f(x)=g(T(x))=\sum_{i=1}^n a_i\cdot f_i(x)$. Therefore, $f\in \mathcal M$.

For part (b), observe that since $F|\mathcal M\in (X/\mathcal N)^{**}=(X/\mathcal N)$ there is an $x\in X$ such that $x+\mathcal N=F|\mathcal M$, in other words $f_i(x)=x^{**}(f_i) =F(f_i)$ for all $i$. Now, use the fact that $\| x+\mathcal N\|=\|F|\mathcal M\|\leq \| F\|$ and the definition of $\|x+\mathcal N\|$ to obtain the desired conclusion.

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so this solves part (a), and for part (b) I was not able to relate any general element $F$ of $\mathcal{X}^{\star\star}$ to $F|\mathcal{M}$ in the way inequality is presented in the question – user24367 Feb 22 '12 at 3:42
@user17523 I'll add a solution for part (b) – azarel Feb 22 '12 at 4:46
ok...... thanks – user24367 Feb 22 '12 at 4:53
for part (b), I did not get the last part, i.e. how does definition of $||x+ \mathcal{N}||$ to obtain the inequality $$||x+ \mathcal{N}||\geq \frac{||x||}{(1+\epsilon)}$$. Note that since $F$ is fixed (i.e. $x$ is fixed, the above inequality should be true for all $x$ and not for some $x$.) – user24367 Feb 22 '12 at 6:50
@user17523. For all $n\in\mathcal N,\ f_i(x)=f_i(x+n)$ for all $i$ since $n\in \bigcap \ker(f_i)$. So you can choose an appropriate element of $x+\mathcal N$. – azarel Feb 22 '12 at 16:06