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Find the eigenvalue and eigenvector of $T: L^2[0,1] \to L^2[0,1]$ given by $$\forall x\in [0,1],\forall f\in L^2[0,1],\ (Tf)(x)=\int_0^xf(t)dt$$ I already check that $ T \in \mathcal{LC}(L^2[0,1])$, $\|T\|=\frac{1}{\sqrt{3}}$ and that $T$ is injective, ie, 0 is not an eigenvalue. So, if $VP(T)$ is the set of all eigenvectors and $\sigma(T)$ is his spectrum, then $VP(T) \subseteq \sigma(T) \subseteq [-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}]$, meaning the eigenvalues need to be on $[-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}]$.

If $\lambda \in VP(T)$, then $(Tf)(x)=\lambda f(x)$ which yields to the differential equation $f(x)=\lambda f'(x)$, then $f(x)=Ke^{x/\lambda}$ and now im stuck, what is next?

Nicolas Rodriguez
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  • Hint: Any eigenvector $f$ must be a continuous function and furthermore $f(0) = 0$. Is any solution of the differential equation satisfying this? – David Gao Jun 19 '24 at 03:28
  • If$f(0)=0$, certainly the exponential is not an eigenvector, then the only continuos function that pops right now is the zero function – Nicolas Rodriguez Jun 19 '24 at 03:34
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    Indeed, there are no eigenvectors as the only possibility is the zero function. – David Gao Jun 19 '24 at 03:39
  • Also, a few other corrections: the operator norm of $T$ is actually $2/\pi$ instead of $1/\sqrt{3}$. (See https://math.stackexchange.com/a/151444/465145) And, since $T$ is not self-adjoint, you can’t conclude from $|T| = 2/\pi$ that $\sigma(T) \subset [-2/\pi, 2/\pi]$. (Though that happened to be true - in fact, because $T$ is compact but has no eigenvalue, $\sigma(T) = {0}$.) – David Gao Jun 19 '24 at 08:25

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