Non-monotone sequence of orthogonal projections

Question

Let $\mathcal{H}$ be a separable Hilbert space and $A_m:\mathcal{H}\to \mathbb{R}^m$ linear with $\sup_m \|A_m\|<\infty$. Assume that for all $x\in\mathcal{H}\setminus \{0\}$ there is a $\varepsilon>0$ such that $\liminf_{m\to\infty}\|A_m x \| \ge \varepsilon$. What can one say then about the (weak) convergence of the sequence $x_m:=P_{\mathcal{N}(A_m)}x$ (orthogonal projection onto the kernel) for fixed $x\in\mathcal{H}$? I first thought, that the weak limit should be $0$, but I wasn't succesfull in proving this, at least without any additional (kind) of monotonicity assumption...

Answer 1

Let $A_m(x_1, x_2, \ldots) = \left(\frac{x_1 - x_{m + 1}}{\sqrt 2}, x_2, \ldots, x_m\right)$ (essentially projection to subspace $\langle e_1 - e_{m+1}, e_2, \ldots, e_m\rangle$).

Take $x \in \mathcal{H}\setminus \{0\}$.

If $x_n \neq 0$ for some $n > 1$, we can take $\epsilon = |x_n|$ and get $\|A_m x\| \geq \epsilon$ for $m > n$.

If $x_n = 0$ for all $n > 1$ (so $x = x_1 e_1$, as $x \neq 0$), we have $A_m x = \frac{1}{2} e_1 - \frac{1}{2}e_{m + 1}$ and can take $\epsilon = \frac{|x_1|}{2}$. So $\liminf\limits_{m\to\infty}\|A_mx\|>0$.

Now, $\{\frac{e_1 + e_{m + 1}}{\sqrt 2}, e_{m + 2}, e_{m + 3}, \ldots\}$ is orthonormal basis of $\mathcal{N}(A_m)$. So $P_{\mathcal{N}(A_m)}(e_1) = \frac{e_1 + e_{m + 1}}{2}$ which doesn't converge weakly to $0$.