The statement of theorem:
Let (Tₙ) be a sequence of compact linear operators from a normed space X into a Banach space Y. If (Tₙ) is uniformly operator converge say ∥Tₙ-T∥→0 then the limit operator T is compact.
if we replace uniform operator convergengence by strong opertor convergence ∥Tₙx-Tx∥→0 then the above theorem becomes false
indeed i need a counter example to prove that strong limit opertor is not compact .
For this I take following counter example: Tₙ: l²→l² defined by Tₙx=(ξ₁,...,ξₙ,0,0,....) where x=(ξᵢ)∈ l² . since Tₙ is linear and bounded, also dim(Tₙx)<∞ so Tₙ is compact.
but now i am stuck to prove that limit is not compact. I also know that the identity operator is not compact since diml²=∞ but don't know how to use it Thanks in advance for any help
Linear and bounded operators need not be compact. The identity operator is a counter-example. Your $T_n$'s are compact because they have finite dimensional range.
$\|T_nx-x\|\to 0$ for every $x \in \ell^{2}$ so the limit in strong operator topology is the identity operator which is not compact.
[$\|T_nx-x\|^{2}=\sum\limits_{k=n+1}^{\infty} |x_k|^{2}$ where $x=(x_1,x_2,...)$. Since $\sum |x_k|^{2} <\infty$ it follows that $\sum\limits_{k=n+1}^{\infty} |x_k|^{2} \to 0$].