Let $X$ be a normed space. If $X$ is reflexive, one can show that
$l\in X' \implies \exists x\in X : ||x|| = 1 \ \ \land \ \ l(x) = ||l||$
Proof of above statement
Given a normed space $X$, we know the following is true:
$$ x\in X \implies \exists l \in X': ||l||=1 \ \land \ l(x)=||x||$$
Applying the theorem on $X'$, we get:
$$ l\in X' \implies \exists L \in X'': ||L||=1 \ \land \ L(l)=||l||$$
$X$ being reflexive means that the canonical map $T : X \to X''$ is a bijective isometry where: $$Tx(l) = l(x)$$
Hence we know that $||Tx|| = ||x||$ and that $T^{-1}$ exists. Now pick $l \in X'$. Hence $\exists L\in X''$ such that $||L||= 1$ and $L(l) = ||l||$. Let $x = T^{-1}L$, i.e., $Tx = L$. Then
$$||x|| = ||Tx|| = ||L|| = 1$$
and $$l(x) = Tx(l) = L(l) = ||l||$$
which proves the assertion.
Original question
I'm trying to find an example of a non reflexive space that does not satisfy the property above. Any ideas?
Answer
From the hint I got in the comments, this is an example.
Let $X = \{x\in C([0,1]\ |\ x(0)=0)\}$ and $l:X\to\mathbb{R},\ l(x) = \int\limits_0^1x(t)dt$.
Then, $|l(x)| \le 1$ if $||x|| = 1$. On the other hand, we can easily construct a sequence $x_n$ such that $l(x_n) \to 1$, hence $||l|| = 1$. But $\not\exists x\in X: ||x||=1 \ \land l(x) = 1$. This is because $||x|| = 1$ and $x(0) = 0$ implies that $l(x) < 0$.