Spectrum of $l^p$ multiplication operator (Brezis 6.17) I would just like a brief sanity check for a question in Brezis's functional analysis book so I can be sure I understand the basics of spectral theory.
Take $l^p(\mathbb{R})$, for $1\leq p \leq \infty$. For a fixed, bounded, real sequence $\lambda_n$ define the multiplication operator $M: l^p \to l^p$ by:
$$
M(x_1,x_2,x_3, \cdots ) = (\lambda_1 x_1 , \lambda_2x_2, \lambda_3x_3 ,\cdots)
$$
i.e, a pseudo-dot product type thing. The question asks me to determine the Eigenvalues of $M$, and then the spectrum, $\sigma(M)$. 
First, if we have an eigenvalue $\mu$, we have:
$$
\mu x_1 = \lambda_1 x_1
\\
\mu x_2 = \lambda_2 x_2 
\\
\mu x_3 = \lambda_3x_3
\\ 
\vdots  
$$
But $\lambda_n$ need not be a constant sequence, so this operator in general does not have an eigenvalue. For the Spectrum, we have that:
$$
\sigma(M) = \{\mu \in \mathbb{R} : (M - \mu I) \text{ is not invertible}\}
$$
In general, if $M - \mu I$ has an inverse, it is reasonable to assume its action on some $y$ is given by:
$$
\left(\frac{1}{\lambda_1 - \mu}y_1, \frac{1}{\lambda_2- \mu}y_2,\frac{1}{\lambda_3 - \mu}y_3 , \cdots \right)
$$
Thus, the operator is not invertible when $\mu = \lambda_k$ for some $\lambda_k$. Thus, the spectrum is given by $\{\lambda_n : n \in \mathbb{N}\}$. 
 A: Since we are discussing the spectrum of an operator on a normed space, we consider the operator norm, which in the case of a multiplication operator happens to be the infinity norm. This is relevant because when discussing the spectrum, invertibility means the existence of a bounded inverse.
You will have eigenvectors wherever the sequence $\lambda$ is constant on some subset. In particular, on single points. Namely, for each set $E_m=\{n:\ \lambda_n=\lambda_m\}$, the sequence $x=1_{E_m}$, that is the sequence with $x_n=1$ if $n\in E_m$ and zero otherwise, satisfies $Mx=\lambda_m x$.
Thus $\{\lambda_n:\ n\in\mathbb N\}\subset\sigma(M)$. As the spectrum is always closed, $\overline{\{\lambda_n:\ n\in\mathbb N\}}\subset\sigma(M)$.
Conversely, if $\mu\not\in\overline{\{\lambda_n:\ n\in\mathbb N\}}$, there exists $\delta>0$ such that $|\mu-\lambda_n|>\delta>0$ for all $n$. Then $$\left|\frac1{\mu-\lambda_n}\right|<\frac1\delta$$ for all $n$, which gives that the sequence $\{1/(\mu-\lambda_n)\}_n$ is bounded, and so $\mu I-M$ is invertible. This shows that $\overline{\{\lambda_n:\ n\in\mathbb N\}}\supset\sigma(M)$ and so
$$
\sigma(M)=\overline{\{\lambda_n:\ n\in\mathbb N\}}
$$
