Exercise 13, Section 6.2 of Hoffman’s Linear Algebra 
Let $V$ be the vector space of all functions from $\Bbb{R}$ into $\Bbb{R}$ which are continuous, i.e. $V=C^0(\Bbb{R},\Bbb{R})$. Let $T$ be the linear operator on $V$ defined by $$\forall f \in V, \forall x\in \Bbb{R}, \quad [T(f)](x)=\int_0^xf(t)dt.$$ Prove that $T$ has no characteristic values.

My attempt: Assume towards contradiction, $\lambda \in F$ such that $\lambda$ is eigenvalue of $T$. Then $\exists g\in V\setminus \{0_V\}$ such that $T(g)=\lambda\cdot g$. So $[T(g)](x)=\int_0^xg(t)dt=(\lambda \cdot g)(x)=\lambda\cdot g(x)$, $\forall x\in \Bbb{R}$. By elementary property of integration, $[T(g)](0)=0=\lambda \cdot g(0)$. Which implies $\lambda =0$ or $g(0)=0$. Since $g\neq 0_V$, $\exists y\in \Bbb{R}$ such that $g(y)\neq 0$. So $[T(g)](y)=\int_0^yg(t)dt=\lambda \cdot g(y)$. How to reach contradiction?
 A: If $\lambda=0$, you have
$$\tag1
\int_0^xg(t)\,dt=0
$$
for all $x$. Differentiating, since $g$ is continuous, gives $g(x)=0$. This can be done for any $x$, so $g=0$.
If $\lambda\ne0$, from
$$\tag2
\int_0^xg(t)\,dt=\lambda\,g(x),
$$
the continuity of $g$ makes the left-hand-side differentiable. We also get $g(0)=0$. Differentiating $(2)$,
$$
g (x)=\lambda \,g'(x).
$$
This gives $g(x)=k\,e^{x/\lambda }$. The condition $g(0)=0$ then gives $g=0$.
A: Remark that $T(g) '(x) =g(x) $, So $g=\lambda g' $, Now $\lambda =0\implies g=0$,So $\lambda \neq 0$,$$g(x)=c\exp(\frac{1}{\lambda} x) $$But $0=Tg(0)=\lambda g(0)=\lambda c$, hence $c=0$,Contradiction!
A: Suppose $\lambda$ is an eigenvalue for $T$ and let $g$ be a corrisponding eigenvector i.e.
$$\label{eq}\tag{1}T(g)=\int_0^x g(t)dt=\lambda g(x)$$
Since $g$ is continuous $T(g)$ is differentiable because is a primitive of $g$, since $T(g)=\lambda g(x)$ you get that $g$ is $C^1$ (actually $C^\infty$). Differentiating both sides of \eqref{eq} you get the equation
$$\label{eq2}\tag{2}g=\lambda g'$$ with starting value $g(0)=0$ as you noted. (Notice that $\lambda=0$ here implies $g\equiv 0$, so $0$ is not an eigenvalue).
If $\lambda \neq 0$ we may now \eqref{eq2}:
$$(\log g)'=1/\lambda \implies \log g(x)= x/\lambda + C\implies g(x)=Ae^{x/\lambda}$$
and you get contradiction by evaluating $g$ at $0$.
