Let $H=\ell^2$ and we denote $x = (x(k))_k$ for $x\in H$. Let $(e_n,f_n)_{n=1}^\infty$ be a bi-orthogonal system in $H$ , that is, $\langle e_m, f_n\rangle=\delta_{mn}$. For $x\in H$ we define, $$x_n = \sum_{r=1}^n\langle x,f_r\rangle e_r.$$It is given that $$\overline{\mathrm{span}}((e_n)_{n=1}^\infty)=H \tag{1}$$ and $$\lim_{n\to\infty}x_n(k)\rightarrow x(k)\text{ for }k\ge 1\tag{2}.$$ Can one prove that $$\lim_{n\to\infty}x_n\rightarrow x$$ in strong topology (i.e. norm convergence)?

As in any (reflexive) separable Banach space, given $(1)$, $x_n\to x$ strongly if and only if $$\sup_n\|x_n\|<\infty.\tag{3}$$ Again by the Uniform boundedness principle one can show that, given$(2)$, $(3)$ is true if and only if $x_n\to x$ weakly, i.e. for any $y\in H$ $$\lim_{n\to\infty}\langle y,x_n\rangle\to\langle y,x\rangle\tag{4}.$$So it is enough to prove the corresponding weak convergence. I am not sure how to proceed after, i feel like I am missing some conditions. Thanks in advance.

  • $\begingroup$ By definition of a biorthogonal system, given that $x$ lies in its closed linear span we have $x=\sum_{r=1}^\infty\langle x,f_r\rangle e_r=\lim_{n\to\infty}\sum_{r=1}^n\langle x,f_r\rangle e_r=\lim_{n\to\infty}x_n$. So, it doesn't look like you need anything after (1). Am I misunderstanding something here? $\endgroup$ – Ben W Jan 1 '16 at 23:04
  • $\begingroup$ No; $x$ is representable by $\{e_r\}$ iff $\overline{span}(\{e_r\})$ is dense in $H$ and each $\sup_n\|x_n\|<\infty$. See theorem 4.1 of Bases in Banach Space by Ivan Singer. $\endgroup$ – Kunnysan Jan 2 '16 at 11:53

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