Calculating spectrum The question is: Let $H = L^2(-\pi,\pi)$ and $[Au](x) = (1+x^3)u(x)$. Determine $\sigma(A)$. I'm reviewing for a test, so don't be concerned with "overhelping."
I can determine the point spectrum easily, $(A- \lambda I) =0 \implies u(x)=0$, so $\sigma_p(A)=0$. Also, it's self-adjoint, so $\sigma_r(A) = \varnothing$.
However, I'm struggling to find the continuous spectrum and the spectrum in its entirety ($\sigma(A)$). Is the best approach to try and construct an inverse to first get the resolvent?
 A: Since $Af=\varphi f$ where $\varphi(x)=1+x^3$, $A$ is a multiplication operator. It is self-adjoint because $\varphi$ is real-valued - indeed, if $f,g\in L^2(-\pi,\pi)$, then $\langle Af, g\rangle = \langle f, A^\star g\rangle$ implies
$$\int_{[-\pi,\pi]}(1+x^3)f(x)\overline{g(x)}\ \mathsf dx = \int_{[-\pi,\pi]}f(x)\overline{A^\star g(x)}\ \mathsf dx $$ and hence $$A^\star g(x) = (\overline{1+x^3})g(x)=(1+x^3)g(x)=\varphi(x)g(x), $$
so $A^\star = A$. Moreover, the norm of $A$ is the essential supremum of $\varphi$:
$$\|A\| = \|\varphi\|_\infty = \operatorname{ess sup}\{\varphi(x):x\in[-\pi,\pi]\}=\varphi(\pi)=1+\pi^3. $$
Recall now that $\|A^\star A\|=\|A\|^2$ for any bounded operator, so self-adjointness yields $\|A^2\|=\|A\|^2$, and by induction, $\|A^{2m}\|=\|A\|^{2m}$ for all positive integers $m$. So from Gelfand's formula for the spectral radius: $$r(A) = \lim_{n\to\infty}\|A^n\|^{\frac1n} $$ it follows that $r(A)=\|A\|=1+\pi^3$.
The point spectrum of $A$ is empty, but to be more precise, $\lambda I-A$ fails to be injective if $\lambda-\varphi(x)=0$ on a set with positive measure. So $$\sigma_p(A)=\{\lambda\in\mathbb C: \mu(\{x\in X:\varphi(x)=\lambda\})>0\}, $$ which in this case is empty since a polynomial can only take a given value a finite number of times.
To find the inverse operator of $\lambda I-A$, we have
$$(\lambda I-A)f = g \iff (\lambda-\varphi(x))f(x)=g(x) \textrm{ a.e.} \iff f(x) = \frac{g(x)}{\lambda-\varphi(x)} \textrm{ a.e.} $$ So $(\lambda I-A)^{-1}$ is defined by multiplication by $\varphi_\lambda(x)=(\lambda-\varphi(x))^{-1}$. Its image is dense in $L^2[-\pi,\pi]$ (this is actually true for all normal operators, i.e. bounded operators that commute with their adjoint), so the residual spectrum is empty.
Finally, if $\lambda\in\varphi([-\pi,\pi])=[1-\pi^3,1+\pi^3]$, then $\lambda=\varphi(x_0)$ with $x_0\in[-\pi,\pi]$, so for $f\in L^2([-\pi,\pi])$ we have $$(\lambda I - A)f(x_0)=\lambda f(x_0)-\varphi(x_0)f(x_0)=0.$$ It follows that $\lambda I-A$ is not surjective, and so $\sigma(A)\supset[1-\pi^3,1+\pi^3$.
The approximate point spectrum $\sigma_{ap}(A)$, being a closed subset of $\sigma(A)$ which contains the boundary of $\sigma(A)$, is in fact equal to $\sigma(A)$.
Edit: If $T$ is a bounded normal operator on a Hilbert space $\mathcal H$, then its residual spectrum $\sigma_r(T)=\varnothing$. Suppose not, that $\lambda I-T$ is injective but the image $(\lambda I-T)(\mathcal H)$ is not dense. Then there exists a non-zero $f\in(\lambda I-T)(\mathcal H)^\perp$, so for all $g\in \mathcal H$, $$0=\langle g,(\lambda I-T)f\rangle = \langle (\lambda I-T)^\star g,f\rangle = \langle (\overline \lambda I-T^\star)g,f\rangle. $$ This implies that $(\overline \lambda I - T^\star)f=0$, so $f\in Z:=\{h\in\mathcal H:(\overline\lambda I-T^\star)h=0\}$ and the $0$-eigenspace $Z$ of $\overline\lambda I-T^\star$ is non-trivial. From the above we see that $Z$ is also invariant under $\lambda I-T$, so if $h,h'\in Z$, $$\langle (\lambda I-T)h,h'\rangle = \langle h,(\overline \lambda I- T^\star)h'\rangle=\langle h,0\rangle=0.  $$ Since $h'$ was arbitrary, it follows that $(\lambda I-T)h=0$ and so $\lambda I-T$ is not injective, a contradiction.
