Let $H$ be a separable Hilbert space over $\mathbb R$, $(h_n)_{n\in\mathbb N}$ be a complete orthonormal system in $H$ and $P^{(n)}$ denote the orthogonal projection onto $H^{(n)} := {\rm span}(h_1,\dots,h_n)$.
Further, let $A\colon H\to H$ be a bounded linear operator that is symmetric positive semi-definite and trace-class and let $A^{(n)} = P^{(n)} A P^{(n)}$.
Finally, let $(v_n)_{n\in\mathbb N}$ be a (possibly unbounded) sequence of vectors with $v_n\in H^{(n)}$ such that $\|A^{(n)} v_n\|\le M$ is bounded by some constant $M>0$ (in fact, it converges to some element $w\in H$).
I need to show that $A v_n$ is also bounded by some (possibly different) constant $\tilde M>0$: $\|A v_n\|\le \tilde M$.
I have been struggling to prove this (and also to find counterexamples), but so far I cannot find the right analytical tools.
Any help will be greatly appreciated! (This is not a homework exercise.)
My attempts so far:
Let $(e_m)_{m\in\mathbb N}$ be an (orthonormal) eigenbasis of $A$: $A = \sum_{m=1}^{\infty} \lambda_m\, e_m\otimes e_m$, where $\lambda_m\in\mathbb R$. Then $$ A = \sum_{k,l=1}^{\infty} \underbrace{\sum_{m=1}^{\infty} \lambda_m\, \langle h_k, e_m\rangle \, \langle e_m, h_l\rangle}_{=: a_{kl}} \, h_k\otimes h_l \qquad \text{and} \qquad A^{(n)} = \sum_{k,l=1}^{n} a_{kl} \, h_k\otimes h_l $$ and therefore we obtain for $v_n = \sum_{j=1}^{n} \langle v_n, h_j\rangle h_j$: $$ Av_n = \sum_{k=1}^{\infty} \bigg(\sum_{j=1}^n a_{kj} \, \langle v_n, h_j\rangle\bigg) h_k \qquad \text{and} \qquad A^{(n)}v_n = \sum_{k=1}^{n} \bigg(\sum_{j=1}^n a_{kj} \, \langle v_n, h_j\rangle\bigg) h_k $$ Therefore $$ \|Av_n\|^2 = \|A^{(n)}v_n\|^2 + \sum_{k >n} \bigg|\sum_{j=1}^n a_{kj} \, \langle v_n, h_j\rangle\bigg|^2 $$ and I somehow need to bound the right term. The $a_{kj}$ are probably decaying at some rate, but I don't see how to go on.