Isometry in the dual of $C[0,1]$

Question

Let $\{a_n\}_{n\geq 1}$ be a sequence of different points in $C[0,1]$ and let $y=(y_1,...)\in l_1$.
Define $f_y:C[0,1]\rightarrow \mathbb{C}$ with $f_y(\phi)=\sum_{n=1}^\infty y_n\phi(a_n)$.

How do I show that

1. the map $h:l_1\rightarrow (C[0,1])^{'}$ with $h(y)=f_y$ is an isometric map?
2. $h(l_1)$ is a closed subspace of $(C[0,1])^{'}$
3. $C[0,1]$ is not reflexive?

What I thought:
1. To be an isometry, we must have $||h(y)||=||y||$ for all $||y||\in l_1$.
We know that $||h(y)||=||f_y||=\sup\{||f_y(\phi)||:||\phi||\leq1\}$, where $||f_y(\phi)||\leq||y||\cdot||\phi||$ since $|\sum_{n=1}^\infty y_n\phi(a_n)|\leq\sum_{n=1}^\infty |y_n|\cdot ||\phi||=||y||\cdot||\phi||$.
All we have to do is prove that $||f_y(\phi)||\geq||y||\cdot||\phi||$. How do we do that?

2. Let $(f_{y^1},f_{y^2},...)$ be a converging sequence of elements in $h(l_1)$ s.t. $f_{y^n}\rightarrow F$. I think we need to show that $F\in h(l_1)$, which means that we want to show that there is a $y\in l_1$ such that $F=f_y$. I assume that this $y$ should be the limit of the $y^n$. How would I show this?

3. For $\phi\in C[0,1]$ let $e_\phi:(C[0,1])^{'}\rightarrow\mathbb{C}$ with $e_\phi(f)=f(\phi)$. Further let $J:C[0,1]\rightarrow(C[0,1])^{''}$ with $J(\phi)=e_\phi$.
   To be reflexive, we must have $J(C[0,1])=(C[0,1])^{''}$. What is an element of $(C[0,1])^{''}$ that is not in $J(C[0,1])$?

Edit:
2 Following LeBtz's hint we get:
Let $(f_{y^1},f_{y^2},...)$ be a Cauchy sequence of elements in $h(l_1)$ with pre-images $y^1,y^2n,..$ in $l_1$. Then we know that for all $\epsilon>0$ there exists an $N$ s.t. for all $n,m\geq N$ we have $$||f_{y^n}-f_{y^m}||<\epsilon.$$ Since $h$ is an isometry it follows that $||y^n-y^m||=||f_{y^n}-f_{y^m}||<\epsilon$, so $\{y^n\}$ is Cauchy in $l_1$ and thus converges to a $y\in l_1$. Thus there exists an $N$ s.t. for all $n\geq N$ we get $$||f_{y}-f_{y^n}||=||y-y^n||<\epsilon.$$ So $\{f_{y^n}\}$ converges to $f_y$, which is in $l_1$.

Answer 1

1) Let $\epsilon>0$. Then there is $N\in\mathbb N$ with $\displaystyle\sum_{n=N+1}^\infty |y_n|<\epsilon$. Now the collection $\{a_1,...,a_N\}$ is finite so we can find a continuous function $\phi:[0,1]\to \mathbb C$ with $\phi(a_i) = \overline{\operatorname{sgn}(y_i)}$ if $i\leq N$ and $\|\phi\|_\infty\leq 1$. ¹ Now $$|f_y(\phi)| = \left|\sum_{n=1}^N |y_i| + \sum_{n=N+1}^\infty y_n\phi(a_n)\right| \geq \sum_{n=1}^N|y_i|-\sum_{n=N+1}^\infty |y_n|\underbrace{|\phi(a_n)|}_{<1} >\|y\|-2\epsilon.$$ Since $\epsilon$ was arbitrary this yields the first claim.

2) Images of complete spaces with respect to an isometry are complete and hence closed. Take a cauchy sequence in $h(\ell^1)$, show that its preimages are Cauchy aswell and therefore must converge in $\ell^1$ and conclode that the sequence must converge in $h(\ell^1)$ due to continuity of $h$.

3) Every closed subspace of a relfexive space is itself reflexive. $\ell^1$ is not reflexive.

¹ We can define $\phi$ on the finite set $\{a_1,...,a_N\}$ and then interpolate linearly for example.