Why is $\phi$ an isometry? I want to prove this statement but honestly I'm not sure if it's true or not and don't see how I can proceed.
Let $X$ a normed vector space and $\phi$ : $X\rightarrow (X')'$ , the linear mapping given by : $(\phi(x))(f)=f(x) \;, f\in X $.

If $X$ is a Hilbert space (with its norm induced by the dot product), we can say that  $\phi$ is an isometry (bijective)  preserving the operator norm on $E'$ and $(E')'$.

Thanks in advance for your help.
 A: Let $X$ be a normed TVS. Fix $x\in X.$ Now, $\|f(x)\|\le \|f\|\cdot \|x\|$ so $\|\phi(x)f\|\le \|f\|\cdot \|x\|$ so
$\tag1 \|\phi(x)\|\le \|x\|.$
On the other hand, the Hahn-Banach theorem gives us an $f\in X'$ such that $\|f\|=1$ and $\phi(x)f=\|f(x)\|=\|x\|$ so
$\tag2\|x\|=\phi(x)f\le \|\phi(x)\|\cdot \|f\|=\|\phi(x)\|$
so $\phi$ is indeed an isometry.
A: We need to show that $\|\phi(x)\| = \|x\|$ for every $x\in X$.
Fix $x \in X$ and notice that for every $f \in X'$ we have
$$|\phi(x)(f)| = |f(x)| \le \|f\|\|x\|$$
which means $\|\phi(x)\| \le \|x\|$.
To show the reverse inclusion, we need to find $f \in X'$ such that $\|f\|=1$ and $f(x)=\|x\|$. In this case we have
$$\|\phi(x)\| \ge \frac{|\phi(x)(f)|}{\|f\|} = |f(x)|=\|x\|.$$
If $X$ is a Hilbert space, we can define such a functional simply as $y \mapsto \left\langle y, \frac{x}{\|x\|}\right\rangle$ via the Riesz representation theorem. In general we have to use a corrolary of the Hahn-Banach theorem which gives us exactly such an $f$.
