Spectrum and resolvent set of operator $(Ax)(t) = x(1 - t)$ on $C[0,1]$ Firstly by direct calculations I proved that equation $Af = \lambda f$ can be solved if and only if $\lambda = +1 or -1$.
So $\sigma_p(A) = 1,-1$
Next I wanted to solve the equation $(A-\lambda I)x(t) = g(t)$. For arbitrary functions $g\in C[0,1]$. We assume that $\lambda \neq  1 or -1 $. Then we get $x(1-t)-\lambda x(t) = g(t)$
I stuck there. Can anyone help?
 A: The original posting had $A$ as $Ax(t)=x(t+1)$ for $x\in C([0,1]$ which led to the interpretation of $Ax(t)=v(t)x(t)$ with $v(t)=1+t$, which was reinforces by test in the OP.  After communications with the OP, it seems that the intention was to define $A$ as a reflection around $1/2$, that is $Ax(t)=x(1-t)$ for $x\in C[0,1]$, that is composition of $x$ with $t\mapsto 1-t$. Her his a solution in terms of  this last setting.
Clearly $\|A\|=1$, and  $A^2=I$. Then
$$(\lambda I+A)(\lambda I-A)=(\lambda^2-1)I$$
Hence, for $\lambda\notin\{-1,1\}$, $(\lambda I-A)$ is invertible, and its inverse
$$R_\lambda:=(\lambda I-A)^{-1}=\frac{1}{\lambda^2-1}(\lambda I+A)$$
is bounded: $\|R_\lambda\|\leq \frac{|\lambda|+1}{|\lambda^2-1|}\leq \frac{1}{||\lambda|-1|}$.
For $\lambda^2=1$, notice that for $x_n(t)=e^{2\pi int}$
$Ax_n(t)=e^{-2\pi in t}$, whence one obtains that $-1$ and $1$ are eigenvalues of $A$ and $u_n(t)=\cos2\pi nt$ and $v_n(t)=\sin2\pi nt$ are eigenvectors of $A$ corresponding $1$ and $-1$ respectively.
In summary the spectrum $\sigma(A)=\{-1,1\}=\sigma_p(A)$, and the resolvent set $\rho(A)=\mathbb{C}\setminus\{-1,1\}$.
