Why is the point spectrum of the operator $(Tx)(t)=\int_{0}^{t}x(s)ds$, for $x\in C[0,1]$, empty? Let $X=C[0,1]$ and define $T\in\mathcal{B}(X)$ by
$$(Tx)(t)=\int_{0}^{t}x(s)ds\qquad(0\le t\le 1)$$
If $\alpha\in\sigma_{p}(T)$ then there is an $x\in C[0,1]$ with $x\ne 0$ such that
$$\begin{aligned} &(Tx)(t)=\alpha x(t) \\ \iff &\int_{0}^{t}x(s)ds=\alpha x(t) \\ \iff&\alpha=\frac{\int_{0}^{t}x(s)ds}{x(t)}\end{aligned}$$
Now, clearly since $t\in[0,1]$, we can have $t=0$, so it follows that we can have $\alpha=0$, but this would contradict Walter Rudin as he asks in an exercise from Functional Analysis to prove that $\sigma_{p}(T)=\emptyset$. 
What worries me more is that I think that we can prove that $T$ is a compact operator via the Arzela-Ascoli theorem which would then imply that 0 is in the spectrum and thus the point spectrum, although I have not actually explored this possibility yet.
 A: Suppose
$$
          \int_{0}^{t}x(s)ds=\lambda x(t),\;\;\; 0 \le t \le 1,
$$
for some continuous function $x : [0,1]\rightarrow\mathbb{C}$ and for some $\lambda\in\mathbb{C}$. If $\lambda =0$, then the Fundamental Theorem of Calculus forces
$$
          0 = \frac{d}{dt}\int_{0}^{t}x(s)ds = x(t),\;\; 0 \le t \le 1.
$$
That that rules out $\lambda=0$ being the point spectrum of $T$. If $\lambda \ne 0$ then $x(t)$ must be continuously differentiable and vanish at $t=0$, with
$$
               x(t) = \lambda x'(t),\;\; x(0)=0 \\
            x'(t)-\frac{1}{\lambda}x(t),\;\; x(0)=0\\
            \frac{d}{dt}\left(e^{-t/\lambda}x(t)\right)=0,\;\; x(0)=0.
$$
Therefore $x\equiv 0$. So $T$ has no point spectrum.
The one compact operator that does not have point spectrum is a quasinilpotent operator $T$ with $\sigma(T)=\{0\}$. Only non-zero points of the spectrum have to be eigenvalues. The point $0$ is always in the spectrum of a compact operator on an infinite-dimensional space, but it does not have to be in the point spectrum. Your operator is quasinilpotent, which means that the resolvent has an essential singularity at $\lambda = 0$. To find the resolvent, which exists for $\lambda \ne 0$, let $y\in C[0,1]$ be given and solve the following for $x$:
$$
            (T-\lambda I)x= \int_{0}^{t}x(s)ds - \lambda x(t) = y(t) \\
      \frac{d}{dt}\int_{0}^{t}x(s)ds-\frac{1}{\lambda}\int_{0}^{t}x(s)ds=-\frac{1}{\lambda}y(t) \\
    \frac{d}{dt}\left(e^{-t/\lambda}\int_{0}^{t}x(s)ds\right)=-\frac{1}{\lambda}e^{-t/\lambda}y(t)
$$
Integrating both sides in $t$ over $[0,u]$ gives
$$
      e^{-u/\lambda}\int_{0}^{u}x(s)ds=-\frac{1}{\lambda}\int_{0}^{u}e^{-t/\lambda}y(t)dt \\
      \int_{0}^{u}x(s)ds = -\frac{1}{\lambda}e^{u/\lambda}\int_{0}^{u}e^{-t/\lambda}y(t)dt \\
      x(u) = -\frac{1}{\lambda}\frac{d}{du}\left(e^{u/\lambda}\int_{0}^{u}e^{-t/\lambda}y(t)dt\right) \\
     x(u) = -\frac{1}{\lambda^2}e^{u/\lambda}\int_{0}^{u}e^{-t/\lambda}y(t)dt-\frac{1}{\lambda}y(u).
$$
Therefore, the resolvent $R(\lambda)=(T-\lambda I)^{-1}$ is
$$
    R(\lambda)y = -\frac{1}{\lambda^2}\int_{0}^{u}e^{(u-t)/\lambda}y(t)dt-\frac{1}{\lambda}y(u)
$$
The resolvent has an essential singularity at $\lambda=0$, but no other singularities. The negative of the residue of $R(\lambda)$ at $\lambda=0$ is the identity $I$, which has to happen because there are no other points in the spectrum.
