I am trying to learn some basic stuff about spectral theory, and I am a little bit lost. Please, could you help me and tell me how to find $\sigma(T)$ and $\sigma_p(T)$ of the operator $T:C([0,1]) \rightarrow C([0,1])$ $$Tf=f+f(0)-f(1).$$ I have some result where is written that the right answer is $\sigma(T)=\sigma_p(T)= \{0\}$ but I do not know how to prove it.
And I also do not understand why $\sigma_p(T)$ does not contain $1$.
Because of $$Tf=f+f(0)-f(1)=\lambda f$$ if $\lambda =1$ we can take in account e.g. constant function, so $Tf=f $.
But from some reason, this is not in my teacher´s results.
Thank you for your help!!!
I think you are right. If $Tf=\lambda f$, we have $ \lambda f = f + f(1) - f(0), $ or $$\tag{1} (\lambda - 1) \, f= f(1)-f(0). $$ If $\lambda=1$, then any $f$ with $f(1)=f(0)$ satisfies the equation, so $1\in\sigma_p(T)$. In particular, as you mention, constant functions are eigenfunctions for the eigenvalue $1$.
When $\lambda\ne1$, the equation $(1)$ has no solution: when $t=1$ and $t=0$, we get respectively $$ (\lambda-1)f(1)=f(1)-f(0),\ \ \ \ \ (\lambda-1)f(0)=f(1)-f(0), $$ which imply that $f(1)=f(0)$; but then we go back to $(1)$ and get $f(t)=0$. So $\lambda\ne1$ cannot be an eigenvalue.
Consider the operator $Sf=f+f(0)-f(1)$. Then \begin{align} STf&=S(f+f(1)-f(0))=(f+f(1)-f(0))+(f(0)+f(1)-f(0))-(f(1)+f(1)-f(0))\\ \ \\ &=f+f(1)-f(0)+f(1)-2f(1)+f(0)\\ \ \\ &=f. \end{align} And \begin{align} TSf&=T(f+f(0)-f(1))=f+f(0)-f(1)+(f(1)+f(0)-f(1))-(f(0)+f(0)-f(1))\\ \ \\ &=f+f(0)-f(1)+f(0)-2f(0)+f(1)\\ \ \\ &=f. \end{align} So $T$ is invertible, and $0\not\in\sigma(T)$.
More generally, for $\lambda\ne1$ one can define $$ Sf=\frac{f+\alpha f(1)+\beta f(0)}{1-\lambda} $$ for appropriate $\alpha,\beta$ to get an inverse for $T-\lambda I$.
In summary, $\sigma(T)=\sigma_p(T)=\{1\}$.