Question: I saw this definition in one of the computer science book but I am unable to recall the theorem name. Can someone please provide the reference?

$$\lim\_{n \to \infty} \frac{f(n)}{g(n)} = c > 0$$ means there is some $n\_{0}$ beyond which the ratio is always between $\frac{1}{2}c$ and $2c$.

Answer: 
Notice that $\|Ax\| \le \|x\|$. So $\sigma(A)$ is contained in the closed unit disk $D^{c}$. To find the resolvent $(A-\lambda I)^{-1}$, try to solve $(A-\lambda I)x = y$ for $x$, and see if it can be done. First, see if there are any eigenfunctions $Ax=\lambda x$. If there were such an $x$, then $|\lambda| \le 1$ and $x=(x\_1,x\_2,x\_3,\cdots)$ would satisfy
$$
 (x\_2-\lambda x\_1,x\_3-\lambda x\_2,x\_4-\lambda x\_3,\cdots)=0,\\
 x\_2=\lambda x\_1,\; x\_3=\lambda x\_2,\;x\_4=\lambda x\_3.
$$
Assuming $x\_1=1$, there is a solution $(1,\lambda,\lambda^{2},\lambda^{3},\cdots)$. This solution is valid only for $|\lambda| < 1$ because, otherwise, $x \notin l^{2}$. In other words, $\sigma\_{P}(A)=\{ \lambda \in \mathbb{C} : |\lambda| < 1\}$ is the open unit disk. And $\sigma(A)=\{ \lambda \in\mathbb{C} : |\lambda|\le 1\}$ because $\sigma(A)$ is closed, contains the open unit disk, and is contained in the closed unit disk.

The adjoint of $A$ is the shift $B(x\_1,x\_2,x\_3,\cdots)=(0,x\_1,x\_2,x\_3,\cdots)$. Notice that $\|Bx\|=\|x\|$ for all $x \in l^{2}$. $B$ has no point spectrum because $Bx=\lambda x$ implies $(x\_1,x\_2-\lambda x\_1,x\_3-\lambda x\_2,\dots)=0$, which forces $x\_1=0, x\_2=\lambda x\_1=0,etc.$. So the range of $(A-\lambda I)$ is dense for all $\lambda$, which puts the unit circle $\{ \lambda \in\mathbb{C} : |\lambda|=1\}$ in the continuous spectrum of $A$. $A$ has no residual spectrum.
First note that $A$ is an isometry, so $\|A\| = 1$. Hence $\sigma(A)$ lies in the closed unit ball of the complex plane $C$.

Define the adjoint of $A : X \rightarrow Y$ to be $A^\* : Y^\* \rightarrow X^\*$ such that $A^\*(\alpha)(x) = \alpha(Ax)$ where $\alpha \in Y^\*$ and $x \in X$ (here $X,Y$ are Banach spaces). Then $\sigma(A) = \sigma(A^\*)$. Then easy computation should give you the adjoint of left shift is the right shift (D'oh!). It is very easy to see that the spectrum of the right shift is all the closed unit disc by considering $x = (1,\lambda, \lambda^2, \lambda^3,...)$.