I am trying to prove the following theorem and am stuck.
Let $X$ be a Banach space. The the image of the unit ball in $X$ is weak-* dense in the unit ball of $X^{**}$.
My proof idea
Assume $X^{**}$ is equipped with its weak-* topology.
Let $Q$ be the isometric map from $X$ to $X^{**}$ such that $Q(x)(\pi) = \pi(x)$. Denote by $Q(B_x)$ the image of the unit ball $B_x$ in $X$. Now for every $ y \in Q(B_x)$ if $\|y\| < 1$ then $Q(B_x)\subset B_{x^{**}}$ where $B_{x^{**}}$ is the unit ball in $X^{**}$. But this is equivalent to showing that every vector in the complement of the closure of the $Q(B_x)$ has norm greater than $1$. (*)
Let $Q'(B_x)$ denote the closure of $Q(B_x)$ in $X^{**}$. Take $ \pi \in X^{**}\setminus Q'(B_x)$. As $Q'(B_x)$ is closed and $\{\pi \}$ is compact, by the Hanh-Banach seperation theorem, we have that there exists a continuous linear functional $\omega': X^{**} \to \mathbb{C}$ and $\alpha \in \mathbb{R}$ such that $\text{Re}(\omega'(\pi)) \leq \alpha \leq \text{Re}(\omega'(y))$ for all $y \in Q'(B_x)$.
I am stuck here.
I know that for $x \in X$ we have $Q(x)(\pi) = \pi(x)$ and $ \pi(x) = \|x\|$. So if $ x \notin B_x$ then $ \pi(x) > 1$ but I cannot find a way of putting these together. I would appreciate if someone could tell me how to proceed. I thank you for your time.
(*) Cannot use a direct approach as there are possibly infinite number of vectors in $X$.
Edit: I would also appreciate if you would give me suggestions towards writing a formal proof.
Best Answer
You are sort of two inches from the finish line.
We know that $Q'(B_x)$ is a convex, balanced, and (weak*) closed subset of $X^{\ast\ast}$ (that is contained in $B_{x^{\ast\ast}}$).
By (one of) the Hahn-Banach theorem(s), for every $\pi \notin Q'(B_x)$, there is a continuous linear form $\lambda$ on $X^{\ast\ast}$ (endowed with the weak* topology) with $\lvert\lambda(\eta)\rvert \leqslant 1$ for all $\eta \in Q'(B_x)$, and $\lvert\lambda(\pi)\rvert > 1$.
The difference to what you have is (apart from specifying $\alpha = 1$) the strict inequality for $\lvert\lambda(\pi)\rvert$, where you only had a weak inequality. That difference is crucial, however.
Now, by definition of the weak* topology on $X^{\ast\ast}$, the space of continuous linear forms is $X^\ast$.
Then $\lvert \lambda(\eta)\rvert \leqslant 1$ for all $\eta \in Q'(B_x)$ implies $\lvert\lambda(x)\rvert \leqslant 1$ for all $x\in B_x$, which means $\lVert \lambda\rVert \leqslant 1$.
But then
$$1 < \lvert \pi(\lambda)\rvert \leqslant \lVert \pi\rVert\cdot\lVert\lambda\rVert \leqslant \lVert\pi\rVert.$$
So $\pi \notin B_{x^{\ast\ast}}$. That holds for all $\pi \notin Q'(B_x)$, hence
$$Q'(B_x) = B_{x^{\ast\ast}}.$$