Convergent Sequence in Normed Vector Space is Weakly Convergent/Proof 1

Proof
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.

Let $f \in X^\ast$.

If $\norm f_{X^\ast} = 0$, then $f = 0$ and:


 * $\map f {x_n} \to \map f x$

Take $f \ne 0$, so that $\norm f_{X^\ast} \ne 0$.

Let $\epsilon > 0$.

We then have:

Since $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$:


 * there exists $N \in \N$ such that $\ds \norm {x_n - x} < \frac \epsilon {\norm f_{X^\ast} }$ for all $n \ge N$.

Then, for $n \ge N$, we have:

Since $\epsilon > 0$ was arbitrary, we obtain:


 * $\map f {x_n} \to \map f x$

Since $f \in X^\ast$ was arbitrary, we have:


 * $\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$.