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Let $H$ be an infinite dimensional Hilbert space over $\mathbb{C}$

Let $\{v_n\}_{n \in \mathbb{N}} \subset H$ be a sequence of linearly independent vectors in $H$ such that $v_n \to u$

Let $\forall m \in \mathbb{N}: V_m = \operatorname{span} \{v_n\}_{n \geq m}$ and $P_m$ be the orthogonal projection on $V_m$

My question is if it is true that: $$ \forall v \in V_1: \lim_{m \to \infty} P_m(v)= a \cdot u $$ in $H$-norm and with $a \in \mathbb{C}$

Thanks.

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  • $\begingroup$ I suppose that $V_m$ is supposed to be the closed span of $\{v_n: \, n \ge m\}$? Since otherwise there is no orthogonal projection onto $V_m$, in general. $\endgroup$ – Jochen Glueck Feb 24 at 11:31
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The answer is no, in general.

As a counterexample, let $H = L^2([0,1])$, let $(q_n)_{n \in \mathbb{N}}$ be your favourite enumeration of $[0,1] \cap \mathbb{Q}$ and define \begin{align*} v_n := 1 + \frac{1}{n} 1_{[0,q_n]} \end{align*} for each $n \in \mathbb{N}$.

Then the span of $\{v_n: \, n \ge m\}$ is dense in $L^2([0,1])$ for each $m$ (since $\{q_n: \, n \ge m\}$ is dense in $[0,1]$) and hence $P_m$ is the identity operator. However, $v_n$ converges to the constant function with value $1$ as $n \to \infty$.

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  • $\begingroup$ ok @JochenGlueck thanks for your answer $\endgroup$ – Matey Math Feb 24 at 11:48

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