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Let $H$ be an infinite-dimensional Hilbert space, and $(H_n)$ a sequence of finite-dimensional subspaces. I would like to show that $(H_n)$ converges.

For this, I managed to show that $A \subseteq \liminf_{n\rightarrow \infty} H_n = \bigcup_{n=0}^\infty \bigcap_{k=n}^\infty H_k \subseteq H$, where $A$ is dense in $H$, and similarly that $A\subseteq \limsup_{n\rightarrow \infty} H_n =\bigcap_{n=0}^\infty \bigcup_{k=n}^\infty H_k \subseteq H$.

Technically, to show that the limit of $(H_n)$ exists, I would need to show that $\liminf_{n\rightarrow \infty} H_n=\limsup_{n\rightarrow \infty} H_n$ but I only have $\overline{A}=\overline{\liminf_{n\rightarrow \infty} H_n}=\overline{\limsup_{n\rightarrow \infty} H_n}=H$. Is there a way to still derive my result?

Spida
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  • Are you sure you managed to show that $\limsup_{n \to \infty} H_n = H$ ? It should not be possible, since $\displaystyle \bigcup_{n=0}^{\infty} H_n$ has countable dimension and every infinite dimensional Banach space has dimension $\geqslant 2^{\aleph_0}$. – Adayah Aug 29 '24 at 08:03
  • I was double-checking this right now and I think I might have made a mistake. I guess I should also have $A\subseteq \limsup_{N\rightarrow \infty} H_n \subseteq H$. I would only need to show that the limit exists, it doesn't necessarily have to be equal to $H$. Do you think there's still something to manage here? (I will edit the question accordingly soon). – Spida Aug 29 '24 at 08:17

1 Answers1

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No, it is not enough. Here is a counterexample.

Take $H = \ell^2(\mathbb{N})$ and let

$$\begin{align*} h & = \left( 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots \right) \in H, \\[1ex] e_n & = (0, 0, \ldots, 0, \underset{\substack{\uparrow \\ n}}{1}, 0, \ldots) \in H & \text{for } n \in \mathbb{N}, \\[1ex] A & = \operatorname{lin} \{ e_n : n \in \mathbb{N} \}, \\[1ex] H_n & = \begin{cases} \operatorname{lin} \{ e_1, \ldots, e_n \} & \text{for even } n, \\ \operatorname{lin} \{ e_1, \ldots, e_n, h \} &\text{for odd } n. \end{cases} \end{align*}$$

Then $A \subseteq H$ is dense and

$$\begin{align*} \liminf_{n \to \infty} H_n & = A, \\[1ex] \limsup_{n \to \infty} H_n & = A \oplus \operatorname{lin} \{ h \}, \end{align*}$$

so the limit $\displaystyle \lim_{n \to \infty} H_n$ does not exist.

Adayah
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