Prove $X \hookrightarrow X^{**}$.

Question

Let $X$ be a vector space, prove $X \hookrightarrow X^{**}$.

$\hookrightarrow$ denotes a isometric embedding

Where $X^* = \{f: X \to \mathbb{R}: f\text{ is linear and bounded}\}$ and $X^{**} = \{\varphi: X^* \to \mathbb{R}: \varphi \text{ is linear and bounded}\}$

As far as I know, I need to find an injectie map $J:X\to X^{**}$ which conserves the structure.

I gues $J: \begin{array}[t]{cl} X\to X^{**};\\ x \mapsto F_x:& X^*\to \mathbb{R}: f \mapsto f(x)& \end{array}$ would be a good choice of embedding.

Proving that it is well defined is straightforward. $F_x$ is linear because $(af+bg)(x) = af(x)+bg(x)$, and $F_x$ is bounded, since $|F_x(f)| = |f(x)| \leqslant \|f\|_{X^*} \|x\|_X $.

But what about the isometric embedding? Why is $J$ injective and why does it preserve structure?

$F_x=F_y$ implies $f(x) = f(y)$ for all $f \in X^*$ but why does that imply $x=y$?

Answer 1

The canonical injection is indeed injective :

If $J$ is not injective, there exists non zero $x\in X$ such that $\forall f\in X^*, f(x) = 0$

But it's straightforward to build a continuous linear form $T\in X^*$ such that $T(x) = 1$ with the help of Hahn-Banach theorem.

Edit : about the morphism part, it suffice to show that $\forall x,y \in X, \forall a,b \in \mathbb{R}, J(ax+by) = aJ(x) + b J(y)$. And that's easy to prove.

Let $f\in X^*$, then $f(ax+by) = af(x)+bf(y)$

So, $F_{ax+by}(f) = a F_x(f) + b F_y(f)$

ie. $F_{ax+by}= a F_x + b F_y$