Dual Space Annihilator in C[0,1] Let $V = C[0,1]$ and let U be the subspace of functions of the form 
$y(x) = ax+b$ for some a, b depending on the function.  Give an explicit family of functionals $F\subset U^\perp$ such that for any $y \in V$ satisfying f(y) = 0$\ \  \forall f \in F$, we have$\  y\in U$.  
In other words, in $V^{**}$, we have$$span F^\perp\cap\phi(V) = \phi(U).$$
Help Please
 A: A function is affine if and only if its second derivative vanishes.
We only have continuous functions here, but this idea can still be made work, but we have to resort to a discrete second derivative.
For $a\in[0,1]$, let $\delta_a:V\to\mathbb R$ be the evaluation functional at $a$, $\delta_a(f)=f(a)$.
(This is continuous in the usual topology of $V$ if you are interested in the topological dual. Here it doesn't really matter if it is the algebraic or the topological one.)
We will only use sums of functionals like this.
For $(a,b)\in[0,1]^2$, denote $f_{a,b}=\delta_a-2\delta_{(a+b)/2}+\delta_b$.
Now let
$$
F=\{f_{a,b};(a,b)\in[0,1]^2\}.
$$
Claim:
For $y\in V$ the following are equivalent:


*

*$y\in U$

*$f(y)=0$ for all $f\in F$.


Proof:
If $y\in U$, it is a simple calculation to observe that $f(y)=0$ for all $f\in F$.
Proving the other direction is the harder part.
Suppose $y\in V$ is annihilated by all $f\in F$.
There is an element $z\in U$ so that $y(0)=z(0)$ and $y(1)=z(1)$.
(This $z$ is actually unique.)
Let $w=y-z$.
We know that $w\in V$ and we will show that in fact $w=0$; from this it will follow that $y=z\in U$.
By construction $w(0)=0$ and $w(1)=0$.
We also know that $f_{0,1}(w)=w(0)-2w(1/2)+w(1)=0$, so $w(1/2)=0$.
Now we can use $w(0)=w(1/2)=0$ and $f_{0,1/2}(w)=0$ to get $w(1/4)=0$ and similarly $w(3/4)=0$.
If $w$ vanishes at two points, it has to vanish in their midpoint as well.
Continuing inductively, we find that for any $n>1$ and $1<m<2^n$ our function satisfies $w(2^{-n}m)=0$.
But points like this are dense in $[0,1]$ and $w$ is continuous, so in fact $w(x)=0$ for all $x\in[0,1]$.
$\square$
