Let $V$ be the vector space of all functions from $\mathbb{R}$ into $\mathbb{R}$ which are continuous, i.e, the space of continuous real-valued functions on the real line. Let $T$ be the linear operator on $V$ defined by $(Tf)(x)$ := $ \int^{x}_{0} f(t) \ dt $.
Prove that, $T$ has no characteristic values.
By the fundamental theorem of calculus, $(Tf)'=f$ for every $f$.
If $Tf=0$, then $f=(Tf)'=0'=0$, so $T$ is injective, hence $0$ is not a characteristic value of $T$.
Suppose $(T-\lambda I)f=0$, that is, $Tf=\lambda f$. Then $f(0)=(1/\lambda)\int_0^0f(t)dt=0$, and $f=(Tf)'=(\lambda f)'=\lambda f'$. This ODE has only one solution, namely $f(x)=f(0)e^{(1/\lambda)x}=0$. Thus $T-\lambda I$ is injective, and $\lambda$ is not a characteristic value of $T$
