x_0 element of closure of span of M

frater_cp · #1 $Old$ June 17th, 2009, 02:52 AM

Let M be any subset of a normed space X.
Show that an x_0 element of X is an element of A=closure(span M)

if and only if

f(x_0)=0 for every f element X' such that

f(x)=0 for all x elements of M (restriction of f to M is 0).

pop · #2 $Old$ June 18th, 2009, 06:23 PM

Proof =>
Let $x_o \in X$ and $x_o \in span(M)$ , we know that $x_o$ can be written as a linear combination of elements in $M$ Say $x_o = \sum a_i b_i$ where $a_i$ is scalar and $b_i \in M$ .

Let $f \in X^1$ be arbitrary and such that $f(x) = 0$ for all $x \in M$ .

We have $f(x_o)=f(\sum a_i b_i ) = a_i \sum f(b_i)$ by linearity. Since $b_i \in M$ , $f(b_i) = 0$ . Therefore $f(x_o) = 0$ .

pop · #3 $Old$ June 18th, 2009, 06:54 PM

Proof <=

Since the statement applies to all linear functionals, choose $f$ such that $f(x) = 0 \ \forall x \in M$ and $f(y) <> 0 \ \forall \ y \notin M$ . This means that $f (z ) = 0$ implies z in M.
Suppose to the contrary that $x_o \in X$ and $x_o \notin span(M)$ with $f(x_o ) = 0$ . Then $x_o = \sum a_i b_i + c$ for some $c \notin M$ . Since $\sum f (b_i) = 0$ , we have $f(x_o) = f(\sum a_i b_i) + f(c)= 0 + f(c) = 0$ . This implies $c \in M$ . Contradiction.

Note: This is my first answer I've posted. YAY!

halbard · #4 $Old$ July 5th, 2009, 08:11 AM

Let $E=\textrm{span}(M)$ and suppose $x_0\in\bar E$ . Then there is a sequence $x_n\in E$ for which $||x_n-x_0||\to 0$ .

If $f\in X^*$ vanishes on $E$ then $f(x_n)=0$ for all $n$ , so that $|f(x_0)|=|f(x_0)-f(x_n)|=|f(x_0-x_n)|\leq ||f||||x_0-x_n||\to 0$ , giving $f(x_0)=0$ , and so $f$ must vanish on $x_0$ .

For the converse, suppose $x_0\notin\bar E$ so that $\inf_{x\in E}d(x_0,x)=d(x_0,E)=k>0$ .

Let $E_1=E+\textrm{span}(x_0)$ . Elements of $E_1$ are of the form $x+\lambda x_0$ for $x\in E$ and scalar $\lambda$ , and this expression is unique since $x_0\notin E$ .

Define a linear functional $f$ on $E_1$ by $f(x+\lambda x_0)=\lambda k$ . Then $f$ vanishes on $E$ and $f(x_0)=k$ . We will show that $||f||=1$ .

Firstly, if $\lambda\neq0$ then $||x+\lambda x_0||=|\lambda|||x_0+\lambda^{-1}x||\geq|\lambda|k$ since $-\lambda^{-1}x\in E$ and $d(x_0,E)=k$ .

Hence $|f(x+\lambda x_0)|=|\lambda|k\leq||x+\lambda x_0||$ (also true if $\lambda=0$ ) and so $||f||\leq 1$ .

For any $\epsilon>0$ with there is an element $x\in E$ for which $||x_0-x||<k+\epsilon$ . Since $f(x_0-x)=f(x_0)=k$ it follows that

$||f||\geq\frac{|f(x_0-x)|}{||x_0-x||}>\frac k{k+\epsilon}$ for all $\epsilon>0$ , so $||f||\geq 1$ . Therefore $||f||=1$ .

By the Hahn-Banach theorem, $f$ can be extended to all of $X$ as a linear functional and the extension has norm $1$ , so $f\in X^*$ . For this extension, $f(x)=0$ for all $x\in E$ but $f(x_0)=k\neq 0$ .

So, if $x_0\notin\bar E$ then there is a linear functional which vanishes on $E$ but not on $x_0$ . So if all linear functionals that vanish on $E$ also vanish on $x_0$ then $x_0\in\bar E$ .