Let $(H, \langle \cdot, \cdot))$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $(E_n)$ be a sequence of mutually orthogonal closed vector subspaces of $H$. This implies $\langle x, y \rangle=0$ for all $x \in E_m$ and $y\in E_n$ with $m \neq n$. Let $\pi_n:H \to H$ be the projection map onto $E_n$, i.e., $\pi_n (x)$ is the orthogonal projection of $x$ onto $E_n$. Let $E$ be the closure of $\operatorname{span}_\mathbb R (\bigcup_n E_n)$. Let $\pi:H \to H$ be the projection map onto $E$. Then $\pi, \pi_n$ are bounded linear operators. Let $(\mathcal L(H), \| \cdot\|)$ be the normed space of all bounded linear operators from $H$ to $H$. Let $\varphi_n := \sum_{k=1}^n \pi_k \in \mathcal L(H)$. I have proved that $|\varphi_n (x) - \pi (x)| \xrightarrow{n \to \infty} 0$ for all $x \in H$.
I would like to ask if any of the following statements is true, i.e.,
- $\| \varphi_n - \varphi\| \xrightarrow{n \to \infty} 0$.
- $\varphi_n \xrightarrow{n \to \infty} \varphi$ in the weak topology $\sigma (\mathcal L(H), \mathcal L(H)^*)$.
Thank you so much for your elaboration!
By considering $x\in E_m$ with $m>n$, you can easily see that $\|\varphi_n-\varphi\|\geq1$.
The sequence $\varphi_n$ does not in general converge to $\varphi$ in the weak topology of $\mathcal L(H)$ (which is a very unusual topology to consider, by the way). To see this consider the case where $\dim E_n<\infty$ for all $n$. Let $\Psi$ be a state on the Calkin algebra $\mathcal L(H)/\mathcal K(H)$. If $\rho$ is the quotient map, we can choose $\Psi$ so that $\Psi(\rho(\varphi))\ne0$, and then $\tilde\Psi=\Psi\circ\rho$ is a state on $\mathcal L(H)$ with $\tilde\Psi(\varphi_n)=0$ for all $n$.
What does hold is that $\varphi_n\to\varphi$ in the weak$^*$-topology.