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Let $E$ be a normed space. We have the usual definitions:

1) $f, f_n \in E^*$, $n \in \mathbb{N}$, then $$f_n \xrightarrow{w^*} f :<=> \forall x \in E: f_n(x) \rightarrow f(x)$$ and in this case we say that $(f_n)$ is $weak^*$-$convergent$ to $f$.

2)$x, x_n \in E$, $n \in \mathbb{N}$, then $$x_n \xrightarrow{w} x :<=> \forall f \in E^*: f(x_n) \rightarrow f(x)$$ and in this case we say that $(x_n)$ is $weakly\ convergent$ to $x$.

Now for the two propositions I want to prove or disprove the following statements.

Let $f, f_n \in E^*$, $n \in \mathbb{N}$, such that $f_n \xrightarrow{w^*} f$ and let $x, x_n \in E$. Consider:

[edit: Thanks for pointing out my mistake!]

a) $x_n \rightarrow x$ => $f_n(x_n) \rightarrow f(x)$,

b) $x_n \xrightarrow{w} x$ => $f_n(x_n) \rightarrow f(x)$.

So far, I think that even b) is true which would imply that a) is also true. My reasoning is that, by assumption, we have $f_m(x_n) \rightarrow f_m(x)$ for every fixed $m \in \mathbb{N}$ as well as $f_m(x) \rightarrow f(x)$ for every $x \in E$. Hence we have $$\lim_{m\to \infty} \lim_{n\to \infty} f_m(x_n) = \lim_{m\to \infty} f_m(x) = f(x)$$ which should be the same as $$\lim_{n\to \infty} f_n(x_n) = f(x).$$

However, I'm a little bit suspicious because the setting seems to imply that a) ist true, but b) is not. Is my argument too sloppy?

Amarus
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  • Are you sure that statements a) and b) are the questions you want to ask? I ask this because as they are written, they are true by continuity of $f$ (assuming $E$ is given the norm topology. Every $f$ in $E^*$, the space of bounded linear functionals, is continuous). They also make no mention of $f_n$. – dc2814 Jun 13 '13 at 00:01
  • Oh sorry, that was supposed to be $f_n(x_n) \rightarrow f(x)$. I'll edit it right away. – Amarus Jun 13 '13 at 00:23
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    Part a) is a standard argument using the principle of uniform boundedness. Part b) is not true in general, e.g. consider an orthonormal basis in a separable Hilbert space. – Michael Jun 13 '13 at 01:46

2 Answers2

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a)

$$ \|f_n(x_n) - f(x) \| \leq \|f_n(x_n) - f_n(x)\| + \| f_n(x) - f(x) \| \leq \|f_n\| \|x_n - x\| + \| f_n(x) - f(x) \|. $$

By the principle of uniform boundedness, $\sup_n \|f_n\|$ is finite, so you're done.

b) Not true in general. Let $\{x_n\}$ be an orthonormal basis in some Hilbert space. Then $x_n \rightarrow 0$ weakly and, by reflexivity, $\langle x_n, \cdot \rangle \rightarrow 0$ in the weak-* topology. Evidently, $\langle x_n, x_n \rangle$ does not go to $0$.

Michael
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2

Your argument is wrong. Let $\delta_{ij}$ be Kronecker's $\delta$. Then you have $$\lim_{i\to\infty}\lim_{j\to\infty} \delta_{ij} = \lim_{i\to\infty} 0 = 0,$$ but $$\lim_{i\to\infty} \delta_{ij} = 1.$$ Note that this corresponds to the following situation: Let $E$ be a Hilbert space, $x_n$ be a orthonormal basis and $f_n(x) := (x_n, x)$, i.e., the functional generated by $x_n$. Then, $f_i(x_j) = \delta_{ij}$. This was already pointed out by Michael.

gerw
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