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Let $H$ be a complex infinite dimensional separable Hilbert space

Let $\{S_n\}_{n \in \Bbb N}$ be a sequence of subsets of $H$ such that $S_n$ is compact and connected, $S_{n+1} \subsetneq S_n$ and $\bigcap_{n=1}^\infty S_n=\{v_0\} $ and $S_n$ is linearly independent

Let $\{V_n\}_{n \in \Bbb N}$ be a sequence of subspace of $H$ such that $V_n=\overline{span(S_n)}$

Let $v_n\neq 0$ be an any element of $V_n$ such that $\lVert v_{n+1} \rVert \le \lVert v_n \rVert $

I would like to know if $ \lim_{n \to \infty}v_n=a \cdot v_0$ , $a \in \Bbb C$ (in $H$ metric)

Thanks for any suggestion

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    It seems that you could take $H$ to be Banach, not just Hilbert. – 2017-01-09
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    @AlexM. could you give me any suggestion? – 2017-01-09

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In general, the answer is no! If $v_0 \ne 0$, then as each $V_n$ is a subspace, $v_n = 0$ for each $n$ is a legimate choice, but then $v_n = 0 \to 0 \ne v_0$.

If $v_0 = 0$, this holds also not true in general. Let, for example $H = \ell^2$,
$$ S_n :=\left \{x\in \ell^2 : x_0 = \cdots = x_{n-1}= 0, \forall k \ge n: |x_k|\le \frac 1k\right\}$$
Then $S_n$ is a sequence as wished and $\bigcap_n S_n = \{0\}$. Now $$ V_n = \left\{x\in \ell^2 : x_0 = \cdots = x_{n-1} = 0 \right\} $$ Hence, $v_n := (0, \ldots, 0, 1, 0,\ldots) \in V_n$, $\|v_n\| = 1$ for all $n$. But $v_n \not\to 0 = v_0$.

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    thansk @martini i've forgotten that $ 0 \notin S_n$ and $ v_n \neq 0 $, now i edited – 2017-01-09