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I need to check if my proof is correct, if I'm wrong please,tell me.

The problem:

Let $V = {C^{0}} (\mathbb{R},\mathbb{R})$ the $\mathbb{R} $-vector space of real variable of continuous functions with real values. Let's define the operator:

$T:V \longrightarrow V, T(f)(x) := \int_{0}^{x} {f(t)dt}$

Prove that $T$ doesn't have eigenvalues.

My proof:

Suppose that $\lambda$ is an eigenvalue of $T$. Then there exists $f \neq 0$ such that:

$T(f)= \lambda f$

By the fundamental theorem of calculus, we have:

$(\lambda f(x))' = f(x) \iff \lambda f'(x)=f(x), \forall x\in \mathbb{R}$

Therefore, $\lambda \neq 0$ and we have the following differential equation:

$\frac{df(x)}{dx} = \frac{1}{\lambda}f(x)$

Which has as a solution:

$f(x)=A e^{\frac{x}{\lambda}}, A \in \mathbb{R}$

Then, $\lambda f(x) = T(f)(x)= \int_{0}^{x} A e^{\frac{x}{\lambda}}dx=A\int_{0}^{x} e^{\frac{x}{\lambda}} dx= \lambda f(x) - \lambda A$

But this implies that $A=0$, so that's impossible, because $f(x) \neq 0$, therefore, $T$ doesn't have eigenvalues.

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  • $\begingroup$ Your argument is sound. Since $T$ maps functions $f$ to one of their antiderivatives, any eigenvector $f$ of $T$ (by the FTOC) is differentiable. This justifies introducing $(\lambda f)'$ in the proof. I would recommend explicitly justifying why $f$ being (a multiple of) an image under $T$ establishes that it is differentiable. Also, you need to use a "dummy" variable such as $t$ further down, inside the integral. $\endgroup$
    – bgins
    May 7 at 3:03

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