Question about Eigenvalues of left shift $S_L $on $l^\infty$.
I showed that $\lambda =1 $ is not eigenvalue for $S_L$ on $l^1$ but I could solve for this for on $l^\infty$.
Let me also write how did I find $\lambda=1$ is not eigenvalue on $l^1$. Actually, one way is directly $Ker(S_L)=\{ \emptyset \}$ or using $S_L (x) = \lambda.x$
In second way, I used $l^1$ property to have convergency. My question is how can I approach for $l^\infty$. Any help is appreciated.
I assume the space you are focusing on is the standard sequence space of all bounded real sequences: $$ l^\infty(\mathbb R) = \{(x_n)_{n \in \mathbb N} \in \mathbb R^{\mathbb N} : \sup_n |x_n| < \infty\} $$ I will show that every $\lambda \in [-1, 1]$ is an eigenvalue of $S_L$ and also that values outside that range cannot be eigenvalues.
Notice the following. If $\lambda \in \mathbb R$ is an eigenvalue of the left shift operator $S_L : l^\infty(\mathbb R) \to l^\infty(\mathbb R)$, which is explicitly given by: $S_L[(x_n)_{n \in \mathbb N}] = (x_{n + 1})_{n \in \mathbb N}$, then that means there is non-zero $(x_n)_{n \in \mathbb N}$ s.t. $$ S_L[(x_n)_{n \in \mathbb N}] = \lambda (x_n)_{n \in \mathbb N} \implies (x_{n + 1})_{n \in \mathbb N} = (\lambda x_{n})_{n \in \mathbb N} $$ In other words for all $n \in \mathbb N$ $$ x_{n + 1} = \lambda x_n \implies x_{n} = \lambda^{n}x_0 $$ Now we can see what conditions $\lambda$ must satisfy:
First, since $(x_n)_{n \in \mathbb N}$ is non-zero, $x_0$ can't be zero otherwise by the above all values would be zero. Hence $|\lambda| > 1$ is impossible. Otherwise as $|x_0| > 0$, $|x_n| = |\lambda|^n |x_0|$ grows arbitrarily large so that $(x_n)_{n \in \mathbb N}$ would fail to be a bounded sequence i.e. it won't be in $l^\infty(\mathbb R)$.
In other words, if $\lambda$ is an eigenvalue, $\lambda \in [-1, 1]$ necessarily. And in fact, each value in $[-1, 1]$ is an eigenvalue. Clearly $1$ is an eigenvalue with prototypical eigenvector $(1)_{n \in \mathbb N}$. And $-1$ is an eigenvalue with prototypical eigenvector $((-1)^n)_{n \in \mathbb N}$. Do you see why each $-1 < \lambda < 1$ is an eigenvalue?