$F'$ and $E'$ are isomorphic isometrically. 
Let $F$ be a dense subspace of the normed space $E$.
Prove that $F'$ and $E'$ are isomorphic isometrically.

At first I was trying to define a function to check the isomorphism and then show that it preserves the norm. However, I couldn't think of a function that is an isometric isomorphism and preserves the norm at the same time. So, I found some sources and I saw that this function defined as
$$\varphi:F'\to E' \text{ as }\varphi(f\mid_F)=f.$$
But I don't understand why this function is well defined, is an isometric isomorphism and why it's defined like that. Can someone explain it to me? I don't get why the density property is important as well.
Note: $E'$ and $F'$ are the spaces of continuous linear functionals. The topological dual of $E$ and $F$, respectively.
 A: Since $F$ is a dense subspace of $E$, it is natural to define
$$\varphi: E' \to F': f \mapsto f\vert_F.$$
Clearly this map is linear. To see that it is isometric, you will have to invoke density of $F$ in $E$. To see the surjectivity, use the Hahn-Banach theorem.
A: Consider the map $\psi \colon E' \to F'$ given by $\psi(f) = \left. f \right|_F.$ For surjectivity, let $f \in F'$ be given. Then we can extend $f$ to $\tilde{f} \in E'$ (since $f$ is continuous and densily defined). This extension coincides with $f$ on $F$ hence $\psi(\tilde{f}) = f,$ so $\psi$ is surjective.
Remains to show that $\psi$ is isometric (this will imply that $\psi$ is injective), i.e. $\left\|f\right\| = \left\|\left. f\right|_F\right\|.$ It is clear that $\left\|\left. f\right|_F\right\| \leq \left\|f\right\|.$ For the other inequality. Let $\epsilon>0$ and find $x\in E$ with  $\left\|f\right\| \leq |f(x)| + \epsilon,$ and you can find $y \in F$ such that $\left\|x-y\right\| < \epsilon$ and $|f(x)-f(y)|<\epsilon.$ (Note $f$ is continuous and $F$ is dense in $E$). Then we have
\begin{align*}
\left\|f\right\|& \leq |f(x)| + \epsilon \\
& \leq |f(x)| - |f(y)| + |f(y)| + \epsilon \\
& \leq \left|f(x)-f(y)\right| + |f(y)| + \varepsilon \\
& \leq |f(y)| + 2\epsilon.
\end{align*}
Taking $\epsilon \to 0,$ yields $\left\|f\right\| \leq \left\|\left.f\right|_F\right\|.$
In conclusion $\psi$ is an isometric isomorphism.
