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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 $f(x)$.

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

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

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 $lvert f(x_n) - f(x)rvert?$