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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 $f(x)$.

Robert Lewis
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mimi
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2 Answers2

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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

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

Robert Lewis
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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 $\lvert f(x_n) - f(x)\rvert?$

User8128
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  • I don't understand why the proof has to be so complicated. Isn't this simply continuity of $f$? By definition every element of the dual space is a continuous function. – Kavi Rama Murthy Nov 23 '18 at 07:17
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    @KaviRamaMurthy: what is it you find so complicated about this proof? It seems the author has covered the essentials in a failry minimalist manner. Cheers! – Robert Lewis Nov 23 '18 at 07:22
  • I suppose it's fine to just assert that $f$ is continuous. I'm just giving a quantitative measure of continuity here. I don't see that this is particularly complicated; the whole argument concludes in roughly 6 lines. – User8128 Nov 23 '18 at 07:22
  • I can't see any proof which is less complicated! – Robert Lewis Nov 23 '18 at 07:23
  • @KaviRamaMurthy: I would be happy to read a less complicated proof if you would care to produce one! – Robert Lewis Nov 23 '18 at 07:24
  • @RobertLewis Please read my comment just below the question. – Kavi Rama Murthy Nov 23 '18 at 07:25
  • @KaviRamaMurthy: I did. Now show us how, show us the details, if you would! – Robert Lewis Nov 23 '18 at 07:27
  • @KaviRamaMurthy: it strikes me your comment is a comment on a proof, but not quite a proof in and of itself. – Robert Lewis Nov 23 '18 at 07:29
  • It seems like you've just skirted around the question by invoking a different definition; meaning that this depends on how you define the dual space: $X^$ could be defined as the set of all linear maps $f : X \to \mathbb F$ such that $$|f|_ = \sup_{z \neq 0} \frac{\lvert f(z) \rvert}{|z|} < \infty.$$ In that case, you would have to prove that any element of the dual space is indeed (sequentially) continuous in order to invoke your comment. – User8128 Nov 23 '18 at 07:29
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    @RobertLewis If $f$ is a continuous function on a topological space and $x_n \to x$ then $f(x_n) \to f(x)$. I don't understand what proof you are asking for. Am i really missing something here? – Kavi Rama Murthy Nov 23 '18 at 07:30
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    @RobertLewis The usual definition of the dual of a normed linear space is it is the set of all continuous linear functionals. I haven\t see any book where the definition of the dual space is given in terms of the operator norm. – Kavi Rama Murthy Nov 23 '18 at 07:32