# A sequence of orthogonal projection in Hilbert space

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.

• 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. – Jochen Glueck Feb 24 at 11:31

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 $$n \to \infty$$.