Spectrum of an Operator on a Banachspace 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?
 A: 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, $\lim_{k}(\lambda I-A)y_k = 0$.
