Show that $j:X \rightarrow X^{**}$ is an injective bounded linear map, where $j$ is the natural embedding.

Question

Let $X$ be a Banach space with bidual $X^{**}$ and let $j:X \rightarrow X^{**}$ be the natural embedding.

Question: Show that $j$ is an injective bounded linear map.

My attempt:

Recall that for all $x \in X, $ we have $j(x)(x^*) = x^*(x)$ for all $x^* \in X^*.$

Suppose that $x , y \in X.$ Note that for all $x^* \in X^*,$ we have $$(j(x) + j(y))(x^*) = x^*(x+y) = x^*(x) + x^*(y) = j(x)(x^*) + j(y)(x^*).$$ Hence, $ j(x+y) = j(x) + j(y)$

For any $\alpha \in \mathbb{K},$ we have $$j(\alpha x )(x^*) = x^*(\alpha x) = \alpha x^*(x) = \alpha j(x).$$ Therefore, $j(\alpha x ) = \alpha j(x).$
Hence, $j$ is linear.

Suppose that $x, y \in X$ such that $j(x) = j(y).$ For any $x^* \in X^*,$ we have $j(x)(x^*) = x^*(x) = x^*(y) = j(y)(x^*).$ Since $x^*$ is linear, we have $x^*(x-y) = 0$, which implies that $x- y =0.$ Hence, $x = y.$ Therefore, $j$ is injective.

For all $x \in X,$ note that $$\| j(x) \| = \sup_{\| x^* \| \leq 1 } |j(x)(x^*)| = \sup_{\| x^* \| \leq 1} |x^*(x)| = \| x \|.$$ Therefore, $$\| j \| = \sup_{\| x \| \leq 1} \| j(x) \| \leq 1.$$ Hence, $j$ is bounded.

Is my proof correct?

Answer 1

A few comments.

1. to prove linearity, you need to show $j(x+\alpha y) = j(x) + \alpha j(y)$. You only proved additivity (case $\alpha=1$ above), and to complete proof of linearity enough to show $j(\alpha x) = \alpha j(x)$;

2. For injectivity, the fact that $x^*(x-y)=0$ for all $x^*\in X^*$ implies $x=y$ follows from Hahn-Banach. You need to state this. Also, enough to show $j(x)\equiv 0 \Rightarrow x=0$.

3. The fact that $\sup_{\|x\|\le 1 } |x^*(x)|=\|x\|$ also follows from Hahn-Banach.

4. You proved that the norm of $j$ is $1$.