Claim: Let $A$ be a bounded linear operator on a Banachspace $\mathfrak{X}$. Denote $\sigma(A)$ as the spectrum of A. Let $\lambda$ be a point in the boundary of the $\sigma(A)$. Then there exist a sequence $\{x_n\}_{n\geq 0}\subset \mathfrak{X}$, $\|x_n\| = 1$ such that
$$(\lambda\mathbb{1} - A)x_n\rightarrow 0.$$
How can I prove this?
Because $\lambda\in\partial\sigma(A)$, then there exists $\{ \lambda_n \} \subset\rho(A)$ that converges to $\lambda$. Suppose for the moment that the following holds for all $x$: $$ l(x)=\sup_{n}\|(\lambda_n I-A)^{-1}x\| < \infty. $$ Then, by the uniform boundedness principle, $$ M=\sup_{n} \|(\lambda_n I -A)^{-1} \| < \infty. $$ Hence, for $|\lambda-\lambda_n|M < 1$, it must hold that $\lambda I-A$ is invertible because $(\lambda_n I-A)$ is invertible and \begin{align} (\lambda I-A) & = (\lambda-\lambda_n)I+(\lambda_n I-A)\\ & = \{(\lambda-\lambda_n)(\lambda_nI-A)^{-1}+I\}(\lambda_n I-A) \end{align} This contradiction proves that the original assumption was false. Hence, there exists $x$ such that $\lim_{k} \|(\lambda_{n_k} I-A)^{-1}x\|=\infty$. Let $$ y_k = \|(\lambda_{n_k} I-A)^{-1}x\|^{-1}(\lambda_{n_k}I-A)^{-1}x. $$ Then $\{ y_k \}_{k=1}^{\infty}$ is a sequence of unit vectors for which $$ (\lambda I-A)y_k = (\lambda-\lambda_{n_k})y_k+\|(\lambda_{n_k}I-A)^{-1}x\|^{-1}x $$ Hence, $\lim_{k}(\lambda I-A)y_k = 0$.