I'm stuck with this problem:
Let $H$ a complex or real Hilbert space. Let $L\in\mathcal{L}(H)$ and $(T_n)_{n\in\mathbb{N}}$ a sequence of linear bounded operators such that $T_n\to T$.
For each $n\in\mathbb{N}$, $\lambda_n\in \sigma_{disc}(T_n)$ . If $\lambda_n \to \lambda$, $\lambda\in\sigma_{disc}(T)$?
I'm tried by contradiction but with this follow that $$ \ker(T-\lambda Id) = \left\{0\right\}. $$ Does anyone have any suggestion? I already proved that for the spectrum of the operator is true.
Let $\{e_n\}_{n=1}^\infty$ be an orthonormal basis in a Hilbert space. Define $$T_nx =\sum_{k=1}^n2^{-k}\langle x,e_k\rangle e_k$$ Then $T_n\to T$, where $$Tx =\sum_{k=1}^\infty 2^{-k}\langle x,e_k\rangle e_k$$ namely $\|T_n-T\|=2^{-n-1}.$ The operator $T$ is injective unlike the operators $T_n.$ In other words $\lambda_n=0$ is an eigenvalue of $T_n$ but $\lambda=0$ is not an eigenvalue of $T.$