Let $(X,||\cdot||)$ be a Banach space in $\mathbb{K}$ and $(x_n)$ a sequence in $X$, and let $f$ be any function in $X^*$ (the dual space of $X$) such that the sequence $(f(x_n))$ is bounded.
I am trying to prove that in this case, the sequence $(||x_n||_X)$ is bounded.
I have managed to prove than for a given integer $n$, $||x_n||_X=||\psi_{x_n} ||$, where $\psi$ is a function defined on $X^*$ with values in $\mathbb{K}$ defined by $\psi_{x_n}(f)=f(x_n)$. So if the sequence $(f(x_n))$ is bounded, then so is $(\psi_{x_n}(f))$. Can we conclude that this implies that $||\psi_{x_n} ||$ (and therefore $||x_n||_X$) is bounded?
My intuition tells me it does, as $||\psi_{x_n} ||=\sup\limits_{f\neq0}\{\frac{||\psi_{x_n}(f)||}{||f||}\}$, but I am not sure since $\psi_{x_n}$ is a function while $(\psi_{x_n})$ is a sequence (of functions..).
Let $\hat x_n \in X^{**}$ be defined by
$$ \hat x_n (f) = f(x_n),\ \ \ \forall f\in X^*.$$
Then the condition imply
$$\sup_{n\in \mathbb N} |\hat x_n (f)| < \infty$$
for all $f$. The uniform boundedness principle implies that
$$\sup_{n\in \mathbb N} \|\hat x_n\|_{X^{**}} <\infty,$$
which is just the same as
$$\sup_{n\in \mathbb N} \|x_n\|_{X} <\infty$$
as $\|x_n\| = \|\hat x_n\|$.