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?