analysis - Space of continuous functions linear operator eigenvalues

Let $V$ be the vector space of continuous functions from $\mathbb R$ to $\mathbb R$. Let $T$ be the linear operator on $V$ defined as

$$(Tf)(x)=\int_0^x f(t)\,dt$$

Prove that $T$ doesn't have eigenvalues.

I want to show that it doesn't exist any continuous function $f \neq 0$ with

$$(1) \space \int_0^x f(t)\,dt=\lambda f(x), \space \text{for some} \space\lambda \space\text{and for all }x \in \mathbb R$$

So suppose it does, then $\lambda f(0)=\int_0^0f(t)\,dt=0$, from here it follows $f(0)=0$. From equation (1) we also arrive to the inequality

$$\lambda|f(x)| \leq |x|\max |f(t)|$$

where $\max |f(t)|$ is the maximum the function attains on the $[0,x]$ or $[x,0]$ respectively.

I would like to arrive to an absurd but I couldn't, any hints or suggestions would be greatly appreciated.

Answer

It follows from your equation (1) that $f$ is differentiable and
$f(x) = \lambda f'(x)$ for all $x \in \mathbb R$. The general solution of
this differential equation is $f(x) = C \exp(x/\lambda)$ for some $C \in \mathbb R$.
Since $f(0) = 0$, $f$ is identically zero.