Spectrum of operator $Ax(t)=x(-t)$ on $L_1(-1,1)$

Question

My teacher gave me the homework with the following exercise:

Find the spectrum of an operator $Ax(t)=x(-t)$ on $X = L_1(-1,1)$.

According to the definition: spectrum is $\lambda$ such that

1. $Im(A-I\lambda)\neq X$
2. $Ker(A-I\lambda)\neq 0$

I can find $\lambda$ for the 2. condition:

$$ x(-t) - \lambda x(t) = 0 \\ x(-t) = \lambda x(t)\\ A^2x(t) = x(t) = \lambda Ax(t) = \lambda^2x(t) $$ So $\lambda = \pm 1$. For the $\lambda=1$, $x(t)$ is an even function, for the $\lambda=-1$, $x(t)$ is an odd function. So $\lambda = \pm 1$ in the spectrum of $A$.

But I can't find the solution of $$ x(-t)-\lambda x(t) = y(t) $$ From which we can find the $\lambda$ for 1. condition.

Answer 1

An operator that satisfies some minimal polynomial has spectrum consisting of its roots.

This follows from the spectral mapping theorem: If $p(T)=0$ then $$\{0\}=\sigma(p(T))=\{p(\lambda):\lambda\in\sigma(T)\}$$

In this case, $A^2x(t)=x(t)$, that is, $A^2-I=0$, so $\sigma(A)=\{\pm1\}$.

---

Concerning $\mathrm{im}(A-\lambda I)$, note that $$(A-\lambda I)(A+\lambda I)=(1-\lambda^2)I$$ So, if $\lambda\ne\pm1$, then both $A\pm\lambda I$ must be surjective and injective.
If $\lambda=\pm1$, then $A-\lambda I$ is not surjective; their images are the odd/even functions.