spectrum and point spectrum of $(Af)(x)=(2+\cos x)f(x)$ Consider $A:L^2[-\frac{\pi}{2},\frac{\pi}{2}]\to L^2[-\frac{\pi}{2},\frac{\pi} {2}]$ such that $(Af)(x)=(2+\cos x)f(x)$. Do we have $||A||\in \sigma(A)$ and $||A||\in\sigma_p(A)$?
My attempt:
I know that $||A||\in \sigma_p$ iff there exists a nonzero $f\in L^2[-\frac{\pi}{2},\frac{\pi}{2}]$ such that $Af=||A||f$.
First, I showed that $A$ is selfadjoint. 
Then
$$||Af||^2=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|(2+\cos x)f(x)|^2 dx \le \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|3f(x)|^2 dx$$
This implies $||A||\le 3.$
Let $S_\epsilon=\{x\in [-\frac{\pi}{2},\frac{\pi}{2}] : 2+\cos x \ge 3-\epsilon\}$ and $f_\epsilon(x)=\mathbb{1}_{S_{\epsilon}}$.
So, we can show that $||Af||\le (3-\epsilon)||f||$. This impiles $||A||=3$.
If $||A||=3 \in \sigma_p$, then $Af=(2+\cos x)f(x)=3f(x)$ for some $f\in L^2[-\frac{\pi}{2},\frac{\pi}{2}]$.
How to prove or disprove such an $f$ exists? Can we just say the above equation implies $2+\cos x=f(x)$, which is a contradiction?
How about $||A||\in \sigma(A)$? How to prove or disprove it? Do I need to check whether $A$ is compact?
Thank you very much.
 A: To write things in a slightly less cumbersome way, you have a multiplication operator, $Af=gf$ for a certain function $g$ (you have $g(x)=x+\cos x$ in your case). This operator if often denoted as $M_g$. 
It is well-known that $\sigma(M_g)$ is the (closure of) the range of $g$  (short proof at the end). In this case $g$ is continuous with compact domain, so its range is closed. Thus
$$
\sigma(A)=[2,3].
$$
It is also known that $\|M_g\|=\|g\|_\infty$, so you have $\|A\|=3\in\sigma(A)$. 
As $g$ is real, $M_g$ is selfadjoint. 
As for the point spectrum, if $M_gf=\lambda f$ for some nonzero $f$, we have $(g-\lambda)f=0$. So $f$ can be nonzero only on points where $g(x)=\lambda$. So, if $g^{-1}(\{\lambda\})$ has measure zero, then $\lambda$ cannot be an eigenvalue. And if $g^{-1}(\{\lambda\})$ has positive measure $c$, then we can take 
$$
f=1_{g^{-1}(\{\lambda\})},
$$
and then 
$$
M_g f=g\,1_{g^{-1}(\{\lambda\})}=\lambda\,1_{g^{-1}(\{\lambda\})}.
$$
For your $g(x)=x+\cos x$, which takes every value only twice, no point in the spectrum is an eigenvalue. Thus the whole spectrum consists of the continuous spectrum. 
Finally, since the nonzero part of the spectrum of a compact operator consists of eigenvalues, $M_g$ is compact if and only if $g=0$. 

Proof that $\sigma(M_g)$ is the closure of the range of $g$
Note first that a multiplication operator $M_g$ is invertible precisely when $|g|\geq\delta>0$ for some $\delta$. Indeed, when $|g|\geq\delta$ we have that $1/g$ is continuous, and 
$$
M_{1/g}M_g=I.
$$
Conversely, if $g$ is not bounded below, it means that there exists $t_0$ with $g(t_0)=0$. Then $0\in\sigma(M_g)$ and $M_g$ is not invertible. 
Now, we have 
$$
M_g-\lambda I=M_{g-\lambda},
$$
so by the above $\lambda\in\sigma(M_g)$ if and only if there exists $t_0$ with $g(t_0)=\lambda$. 
