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I am to find a squence from $c_{00}$ space (space of sequences with only finite non-zero entries) that is weakly convergent but not bounded.

My idea is to take $c_n=(0,..., 0, n, 0, 0,..., 0)$. My guess it would be weakly convergent to $0$ - but I am not sure.

maq
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1 Answers1

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No, a weakly convergent sequence on a normed (even non-Banach such as $c_{00}$) space is always bounded.

This follows from the uniform boundedness theorem: suppose $x_n \rightharpoonup x_0$ in a normed space $X$. Consider the operator \begin{align*}J:X&\to X^{**}\\ x &\mapsto Jx:f\mapsto f(x),\;f\in X^*\end{align*} It is not difficult (although you need the Hahn-Banach theorem) to verify that $\|Jx\|=\|x\|$ for all $x$, so $\left\{x_n\right\}$ is bounded iff $\left\{Jx_n\right\}$ is. But for all $f\in X^*$ we have $$|Jx_n(f)|=|f(x_n)|\to |f(x_0)| $$ Thus $\left\{Jx_n\right\}$ is a pointwise bounded family of bounded linear functionals on $X^{*}$, which is always a Banach space. By the uniform boundedness theorem, $\left\{Jx_n\right\}$ is bounded in $X^{**}$, and therefore $\left\{x_n\right\}$ is bounded in $X$.

Lorenzo Q
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  • This looks wrong. Or, it depends on definitions of "weakly convergent" and so on. If we define the inner product $\langle \mathbf{x}, \mathbf{y}\rangle = \sum_{i=1}^{\infty}x_iy_i$ then obviously $\lim_{n\rightarrow\infty} \langle \mathbf{c}n, \mathbf{y}\rangle =0$ for all real-valued sequences $\mathbf{y}=(y_1, y_2, ...)$ that have a finite number of nonzeros, where $\mathbf{c}_n$ are the sequences defined by maq. This is "weakly convergent" to zero according to the definition here: https://en.wikipedia.org/wiki/Weak_convergence(Hilbert_space) – Michael Jun 12 '18 at 23:19
  • Likely there is a "completeness" technicality here that says we are not allowed to use this inner product (since it does not yield the completeness property). The original question does not specify requirements on completeness and/or a working definition of "weak convergence." – Michael Jun 12 '18 at 23:38
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    @Michael that characterization of weak convergence only works in Hilbert spaces. But $c_{00}$ is not a Hilbert space. – Lorenzo Q Jun 13 '18 at 07:27