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Let $V$ be subspace of $\ell^2$ which contains all 1 summable sequences. For each natural number $n$, define $T_n: V \to \mathbb R$ by $T_n(x)=\sum_{i=1}^n x_i$. Then $T_n$ is not uniformly bounded on unit ball $\|x\|_2\leq1$.

My intuition says it has something to do with closed and bounded in infinite dimensional banach space need not be compact. But I don't know how to get a firm answer. Could you please tell me the reason? Thank you very much for your time.

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  • $\begingroup$ $T_n$ is the sequence of linear functional. $\endgroup$ – Ziya Oct 11 '18 at 15:38
  • $\begingroup$ Why $T_n$ is not well defined? $\endgroup$ – Ziya Oct 11 '18 at 15:40
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    $\begingroup$ Do you mean $T_nx=\sum_{k=1}^nx(k)$? (writing elements of $\ell^2$ as functions $x:\mathbb N\to\mathbb C$) $\endgroup$ – Aweygan Oct 11 '18 at 15:41
  • $\begingroup$ Yes. Actually I am using this site on mobile and there it is not very easy to type it. $\endgroup$ – Ziya Oct 11 '18 at 15:43
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Basically you need to find $x$ such that $\lVert x \rVert_2 \leq 1$ but $\lvert T_n(x) \rvert \to \infty$. Put explicitly you require $x$ to satisfy $$ \sum_{i=0}^{\infty} x_i^2 \leq 1 \quad \text{and} \quad \sum_{i=1}^{\infty} x_i = \infty$$ Can you think of such $x$?

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  • $\begingroup$ I tried to find such a $x$ but could not find it. $\endgroup$ – Ziya Oct 11 '18 at 17:21
  • $\begingroup$ It's a very famous sequence with a divergent sum $\endgroup$ – bitesizebo Oct 11 '18 at 17:50
  • $\begingroup$ Is it ${\frac{1}{n}}$? $\endgroup$ – Ziya Oct 11 '18 at 23:59
  • $\begingroup$ Yes! Of course we need to rescale by $\left(\sum_n \frac{1}{n^2}\right)^{-1}$ to get a norm $\leq 1$ but that's not a very important point. $\endgroup$ – bitesizebo Oct 12 '18 at 0:11

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