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If $(x_n)$ in a Banach Space $X$ is such that $(f(x_n))$ is bounded for all $f\in X^{'}$,show that $(||x_n||)$ is bounded.

Attempt:

Suppose that $||x_n||$ is unbounded. Then for each $n\in \Bbb N$ we have $||x_n||>n$.

Also for each $x_n\in X,\exists f\in X^{'}$ such that $||f||=1,f(x_n)=||x_n||$.(Hahn-Banach)

Hence $f(x_n)=||x_n||>n$ and hence $f(x_n)$ is unbounded which is false.

I am not getting why we need the hypothesis that $X$ is Banach here?Is it redundant? Please help.

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  • $\begingroup$ Slight correction. You don't know that $\|x_n\|>n$, just that it diverges. Also use \| for double bars. It has better spacing than ||. $\endgroup$ – Cameron Williams Jan 6 '17 at 16:04
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The problem in your proof comes from the fact that the hypothesis says that the sequence $f(x_n)$ is bounded. If you take a particular $m$ and $f$ such that $f(x_m)>m$, this does not implies that $lim_n|f(x_n)|=+\infty$. You can for example have $|f(x_n)|\leq m, n>m$ in this case the sequence is still bounded.

This is a consequence of Banach Steinhauss. $X'$ is a Banach space, you can define $x'_n:(X')'\rightarrow R$ such that $x'_n(f)=f(x_n)$. The family of linear functions $(x'_n)$ satisfies the hypothesis of Banach Steinhauss. This implies that $Sup_{\|f\|=1}\|x'_n(f)\|<A<+\infty$. There exists $f_n$ such that $\|f_n\|=1$ and $f_n(x_n)=\|x_n\|$, we deduce that $|x'_n(f_n)|=|f_n(x_n)|=\|x_n\|<A$.

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  • $\begingroup$ To clear up the issue for OP: the choice of $f$ in this version of Hahn-Banach depends on $x$. Perhaps it is better to write it as $f_x$ to make this explicit. This is an extremely common error with this version of Hahn-Banach. $\endgroup$ – Cameron Williams Jan 6 '17 at 16:14

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