# Prove that $T$ doesn't have eigenvalues

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.

• 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. May 7 at 3:03