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Let $(\lambda_n)_{n\in\mathbb{R}}$ be a bounded sequence. $T:l^p\rightarrow l^p, (Tx)_i:=\lambda_ix_i\forall i\in\mathbb{N}$. Find the spectrum $\sigma(T)$ and point-spectrum $\sigma_p(T)$. The definitions are: $$\sigma(T):=\mathbb{R}/\{\lambda\in \mathbb{R}|ker(\lambda*Id-T)=\{0\}\cap Im(\lambda*Id-T)=l^p\}$$ $$\sigma_p(T):=\{\lambda\in\sigma(T)|ker(\lambda*Id-T)\neq\{0\}\}$$

I guess the spectrum consists of all $\lambda$, so that $(T-\lambda*Id)$ is not invertable. The point-spectrum consists of all $a$ so there is a $i\in\mathbb{N}$ with $a_i=\lambda_i$. Is that correct? How can I show that? Is there a simple way to proof that?

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Yes, that is correct, but you still need to figure out specifically what $\sigma (T)$ is. This problem can be solved by simply invoking the definition, and analyzing requirements for $\lambda$. Consider $\sigma _p (T)$ first. If $\lambda \in \sigma_p(T)$, then $\exists x \in l^p$, $x \ne 0$ , s.t. $( T - \lambda I)x = 0$. Writing out the components, we obtain for all $n$, $$(\lambda_n -\lambda)x_n = 0$$ Because not all $x_n$ are zero, there must be some $n$ s.t $\lambda = \lambda_n$. It is also easily verified that if $\lambda = \lambda_n$ for some $n$, then $\lambda \in \sigma_p(T)$. Therefore $\sigma_p(T) = \{\lambda_n \}_n$.

Let us turn our attention to $\sigma(T)$. By definition, the spectrum $\sigma (T)$ is the complement of the resolvent, denoted by $\rho (T)$. $\lambda \in \rho (T)$ if and only if $T - \lambda I$ is bijective. Let $\lambda \in \rho(T)$. Then $\lambda \ne \lambda_n$ for all $n$. Since $T - \lambda I$ is surjective, given any $y \in l^p$, there exists $x \in l^p$ s.t. $(T-\lambda I)x = y$. It follows that $$(\lambda - \lambda_n)x_n = y_n$$ $\implies$$$x_n = \frac{y_n}{\lambda-\lambda_n}$$ That $x \in l^p$ requires $$\Vert x \Vert ^p = \sum_{n=1}^\infty \frac{\vert y_n \vert^2}{\vert\lambda - \lambda_n \vert^2} \lt \infty$$ for all $y \in l^p$. You can show that this is true if and only if $\lambda $ is not in the closure of $\left( \lambda_n \right)$.

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