A question about functional in normed space and Hahn Banach theory

Question

Let $X\neq {0}$ a normed space, $x_n\in X$ a sequence in $X,x\in X$.

Assume for every $x^*\in X^*$: $x^*(x_n)\to x^*x$.

Show that: $\|x\|\leq \liminf_{n\to \infty} \|x_n\|$.

I tried to use Hahn Banach's corollary that says: If $X$ is a normed space and $0\neq x\in X$, there is a functional $x^*\in X$ such that $\|x^*\|=1$ and $x^*x=\|x\|$.

We can notice that $x$ that satisfies $x^*(x_n)\to x^*x$ is unique.

Because $X\neq 0$ then the $x$ satisfying this is $\neq 0$. Then, by the corollary mentioned above and the given information we get: $$\|x\|=x^*x=\lim_{n\to \infty} x^*x_n.$$ Now, how can I use the fact that $\|x^*\|=1$?

Answer 1

Let us use the corollary you mentioned.

We have $|x^*(x_n)| \le ||x^*|| \cdot |x_n|$ by definition of norm. As $x^*(x_n) \to x^*(x)$ we have

\begin{gather} |x| = |x^*(x)| = \lim_n | x^*(x_n) | = \liminf_n | x^*(x_n) | \le \\ \liminf_n||x^*|| \cdot |x_n| = \liminf_n |x_n|. \end{gather}

Answer 2

As you said, there is some $f\in X^*$ such that $||f||=1$ and $|f(x)|=||x||$. Note that since $f$ has norm $1$ it follows that:

$|f(x_n)|\leq ||x_n|| \ \ \forall n\in\mathbb{N}$

And so $||x||=|f(x)|=\liminf_{n\to\infty} |f(x_n)|\leq \liminf_{n\to\infty} ||x_n||$.

Answer 3

Hints:

$1).\ $ Since $(x_n)\rightharpoonup x,\ J(x_n)\to J(x)\in X^{**}$ where $J(x)f=f(x).$

$2).\ $ The Uniform Boundedness Principle applies to the collection $(J(x_n))$.

$3).\ 2).$ implies $(x_n)$ is bounded.

$4).\ $ Hahn-Banach gives us a $f\in X^*$ such that $f(x)=\|x\|$ and $\|f\|=1$.

$5).\ |f(x_n)|\to |f(x)|$