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I have the following question:

Let $(x_n)$ a sequence in $X$ and $x\in X$ such that for all $F\in X'$ (the dual space of the vector space X) we have that $(F(x_n))$ converge to $F(x)$ (that is: the sequence converge weakly on $X$).

Let $F:X\longrightarrow\mathbb{R}$ a continuous function.

Is it true that $\quad\displaystyle\liminf_{n\to\infty}|f(x_n)|\;{\color{red}\geq}\;{\color{red}|}f(x){\color{red}|}$?

Recall that:

$\displaystyle\liminf_{n\to\infty}|f(x_n)|:=\sup_{n\in\mathbb{N}}\inf_{k\geq n}|f(x_k)|$.

Thanks in advance!

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No, if $X=\ell_2(\mathbb{R})$ the set $\{(u_n) \in \mathbb{R}^{\mathbb{N}}| \sum_{n \in \mathbb{N}} u_n^2 < +\infty\}$, and $x_n=e_n$. We have $X'=X$ because $X$ is an Hilbert space. And $(x_n)$ converge weakly to $x=0$. But if $f(y)=\|y\|_2$, $f(x_n)=1$ for all $n \in \mathbb{N}$ and $f(x)=0$.

The new sentence is not true. If $f(y)=1-\|y\|_2$, we have $f(x_n)=0$ for all $n \in \mathbb{N}$, but $f(x)=1$. So, $\liminf |f(x_n)|< |f(x)|$.

If $f$ is continuous (but not necessarily linear) for the weak topology on $X$, it is true.

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  • $\begingroup$ Thanks @marco2013. I changed the "=" by "greater than" in red. Is the new sentence true? $\endgroup$ – yemino Sep 24 '14 at 20:40
  • $\begingroup$ thanks again. And, it is possible to add some hypothesis where the sentence is true? $\endgroup$ – yemino Sep 24 '14 at 20:54
  • $\begingroup$ Thanks! I did a fresh question because I had changed my question too much. The new question is on math.stackexchange.com/questions/944860/… if you want see it. $\endgroup$ – yemino Sep 24 '14 at 21:08

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