Limit of projector operator is not compact Let $(H,⟨·,·⟩)$ be a separable Hilbert space of infinite orthogonal dimension. Let $\{u_n\}$ be an orthonormal basis and $H_N$ be the finite dimensional subspace generated by the first $N$ elements.
$P_N : H → H_N$ is the projector operator defined as usual as the finite sum between the Fourier coefficient $v_n$ and $u_n$
If  $\{a_n\}$  is a (countable) sequence of positive numbers such that $a_n →0$ and we consider the orthogonal system $\{a_nu_n\}$ , is it true that the sequence of corresponding orthogonal projectors $Q_N$ converges in $B(H)$ and the limit operator is compact?
In my opinion no, the weak-* limit is the identity operator $I$ that is not compact.
But it converges in $B(H)$?
I think initially that because $Q_N$ is a finite_rank operator (by definition), it's compact. But $K(H)$ is a closed subspace so it cannot converge to the identity operator. But in this way i don't use at all the information about $a_n$ so there is for sure something that i'm missing
 A: As commented, your definition of $Q_N$ is ambiguous (as is your definition of $P_N$, as it uses undefined $v_n$'s...)
But, guessing what you probably mean: first, I'd say something like
$$
P_N(\sum_{n\ge 0} c_n\cdot u_n) \;=\; \sum_{1\le n\le N}c_n\cdot u_n
$$
for Hilbert-space coefficients $c_n=\langle v,u_n\rangle$, for $v\in \ell^2$. Yes, as bounded (=continuous) operators, the $P_N$ converge in the strong topology (so, certainly the weak topology) to the identity operator. EDIT: But definitely not in the operator-norm topology.
Among other things, I'm supposing you don't want $Q_N$ to be the same thing as $P_N$. With constants $\{a_n\}$, we could define
$$
Q_N(\sum_{n\ge 1} c_n\cdot u_n) \;=\; \sum_{1\le n\le N} a_n\cdot c_n\cdot u_n
$$
for example. When $a_n\to 0$, $Q_N\to Q$ in the operator-norm topology, where
$$
Q(\sum_{n\ge 1} c_n\cdot u_n) \;=\; \sum_{n\ge 1} a_n\cdot c_n \cdot u_n
$$
And this $Q$ is compact. EDIT: any proof of compactness is probably more-or-less equivalent to proving that the image $Q(B)$ of the unit ball $B$ in $\ell^2$ is totally bounded, meaning that, given $\varepsilon>0$, $Q(B)$ can be covered by finitely-many open balls of radius $\varepsilon$. This exactly uses $a_n\to 0$, and is a pretty standard exercise.
