Banach space convergence

Question

Having $X$ , a Banach space. Show that $\{x_n\}$ converging to $x$ implies that for all functions $f$ contained in $X^\ast$ (dual), $f(x_n)$ converges to $f(x)$.

Answer 1

Look how you define the norm of $f$ for $f \in V^*$: $$\|f\|_* = \sup_{z \neq 0} \frac{\lvert f(z) \rvert}{\|z\|}.$$ In particular, for any fixed $x \in X$, we see $$\|f\|_* \ge \frac{\lvert f(x) \rvert}{\|x\|} \,\,\,\, \implies \,\,\,\,\, \lvert f(x) \rvert \le \|f\|_* \|x\|.$$ Now suppose that $x_n \to x$ in $X$; that is, suppose that $\|x_n - x\| \to 0$. What can you say about $\lvert f(x_n) - f(x)\rvert?$

Answer 2

Assuming $X^\ast$ denotes the vector space of continuous linear functionals

$f:X \to \Bbb F, \tag 1$

where $\Bbb F = \Bbb R$ or $\Bbb F = \Bbb C$ is the base field, then every $f \in X^\ast$ is bounded, viz,

$\exists 0 < C_f \in \Bbb R, \; \vert f(x) - f(y) \vert \le C_f \vert x - y \vert, \; \forall x, y \in X; \tag 2$

then if

$x_n \to x \; \text{as} \; n \to \infty, \tag 3$

given $0 < \epsilon \in \Bbb R$ we have, for $n$ sufficiently large,

$\vert x_n - x \vert < \epsilon; \tag 4$

thus, for such $n$,

$\vert f(x_n) - f(x) \vert \le C_f \vert x_n - x \vert < C_f \epsilon; \tag 5$

taking $\epsilon$ small enough ensures $C_f \epsilon$ is itself arbitrarily small, whence we see that (5), by definition, implies that

$f(x_n) \to f(x) \; \text{as} \; n \to \infty. \tag 6$