I am trying to understand strong and norm-convergence of operators in infinite dimensional Hilbert spaces. I am reading that the definitions (of the norm of an operator $T$ in a vector space $H$) are respectively $$ ||T_n-T||\rightarrow0, $$ where $||T||=\text{sup}_{x\in D}\frac{||Tx||}{||x||}$, and $$||(T_n-T)x||\rightarrow0\;\;\;\;\;\; \forall x\in H.$$
Consider the projector operators $P_n$, such that it projects a vector onto the 1-dimensional subspace spanned by $e_n$ ($\{e_n\}$ being an orthonormal basis of the vector space).
Why $P_n\rightarrow 0$ strongly, but not in norm?
Is it right that $||P_nx||=a_n||e_n||=a_n$ ($a_n$ being the eigenvalue of the projector operator)?
Does Parseval's identity implies $a_n\rightarrow0$?
If $\{e_n\}$ is an orthonormal basis of a Hilbert space $H$, then for every $x\in H$, we have $$\|{x}\|=\sum_{n=1}^{\infty}|\langle x,e_n\rangle|^2$$ and since the series on the r.h.s converges, the sequence $\langle x,e_n\rangle$ must tend to zero as $n\to\infty$. This can be expressed in terms of the projections $P_n$ as $\lim_{n\to\infty}\|P_nx\|=0$ for every $x\in H$, because the projection of $x$ onto the span of $e_n$ is simply $\langle x,e_n\rangle e_n$. This proves that $P_n\to 0$ strongly. However, it is clear that the norm of each projection $P_n$ is equal to $1$, hence the sequence $P_n$ cannot converge to the zero operator in norm.