The spectrum of an unbounded differential operator on $C (\mathbb{R}^+)$ I am working on spectrum and got the problem with the spectrum of the following unbounded operator. Actually, I do not know how to find its spectrum when the operator is unbounded.
Let 
$$X = \left \{ f \in C( \mathbb{R}^+) : \lim_{x\rightarrow \infty} f(x) =0\right \}$$ 
with the spremum norm and $T : D(F) \rightarrow X$ be definded by $T(f) = f'$ where
$$D(F) = \left \{ f \in C^1(\mathbb{R}^+) : f, f' \in X \right \}$$.
Find the spectrum $\sigma(T)$.
Actually, I tried as the following:
Let us choose $f(x) = \frac {1}{x}$, then $f'(x)=\frac{-1}{x^2}$.
Then $\left \|T\right \| = \sup_{x \neq 0} \frac{\left \|\frac{-1}{x^2}\right \|}{\left \|\frac {1}{x} \right \|}=\|x\|$. It tends to infinity as x tends to infinity.
Therefore, the given operator is not bounded?
So, how can we find the spectrum of this operator? 
 A: Try to solve the resolvent equation for $f$. For example, let $g\in X$, and suppose $\Re\lambda > 0$. Then
$$
        (T-\lambda I)f = g \\
           f'-\lambda f = g \\
           (e^{-\lambda x}f)'=e^{-\lambda x}g \\
          0-e^{-\lambda x}f(x)=\int_{x}^{\infty}e^{-\lambda y}g(y)dy \\
           f(x) = \int_{x}^{\infty}e^{-\lambda(y-x)}g(y)dy.
$$
Based on the assumptions on $g\in X$, the solution $f$ is in $X$ for $\Re\lambda >0$, and it must be unique among solutions $f\in X$. And, $f$ varies continuously with $g$. So the resolvent is defined for $\Re\lambda >0$, and is given by
$$
            R(\lambda)g= \int_{x}^{\infty}e^{-\lambda(y-x)}g(y)dy,\;\; \Re\lambda > 0, \\
       \|R(\lambda)g\|_{X} \le \int_{x}^{\infty}e^{-\Re\lambda(y-x)}dy\|g\|_{X}
    = \frac{1}{\Re\lambda}\|g\|.
$$
To show that $\lambda\in\sigma(T)$ for $\Re\lambda < 0$, consider the resolvent equation
$$
            f'-\lambda f = g,
$$
which must be solved by $f \in X$ for all $g\in X$ in order to be valid. So, for example, let $g = e^{\Re\lambda x/2}=e^{-|\Re\lambda| x}$. Then
$$
            (e^{\lambda x}f)' = e^{\lambda x}e^{-\Re\lambda x/2}=e^{i\Im\lambda x}e^{|\Re\lambda|x/2} \\
     e^{\lambda x}f(x) = f(0)+\int_{0}^{x}e^{i\Im\lambda y}e^{|\Re\lambda|y/2}dy \\
         f(x)=e^{-\lambda x}f(0)+\frac{1}{i\Im\lambda +|\Re\lambda| /2}(e^{i\Im\lambda x+|\Re\lambda|x/2}-1)e^{-\lambda x}.
$$
There is no possible value of $f(0)$ that will force $f \in X$. So $\lambda\in\sigma(A)$ for such $\lambda$. Because the spectrum is closed, then $\Re\lambda \le 0$ describes the spectrum and $\Re\lambda > 0$ describes the resolvent set.
