Proof =>
Let and , we know that can be written as a linear combination of elements in Say where is scalar and .
Let be arbitrary and such that for all .
We have by linearity. Since , . Therefore .
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).
Proof <=
Since the statement applies to all linear functionals, choose such that and . This means that implies z in M.
Suppose to the contrary that and with . Then for some . Since , we have . This implies . Contradiction.
Note: This is my first answer I've posted. YAY!
Let and suppose . Then there is a sequence for which .
If vanishes on then for all , so that , giving , and so must vanish on .
For the converse, suppose so that .
Let . Elements of are of the form for and scalar , and this expression is unique since .
Define a linear functional on by . Then vanishes on and . We will show that .
Firstly, if then since and .
Hence (also true if ) and so .
For any with there is an element for which . Since it follows that
for all , so . Therefore .
By the Hahn-Banach theorem, can be extended to all of as a linear functional and the extension has norm , so . For this extension, for all but .
So, if then there is a linear functional which vanishes on but not on . So if all linear functionals that vanish on also vanish on then .