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Let $T:L^2((0,1)) \to L^2((0,1))$ be the Volterra operator defined by $$ Tf(x) = \int_0^x f(t)\,dt.$$ Given any $\lambda\neq 0$ it is well known that there exists some positive integer $n=n(\lambda)$ such that $\textrm{Ker} (T-\lambda)^n=\textrm{Ker} (T-\lambda)^{n+1}$ and also that $\textrm{Ran} (T-\lambda)^n=\textrm{Ran} (T-\lambda)^{n+1}$ and finally that $$L^2((0,1))=\textrm{Ker} (T-\lambda)^n \oplus \textrm{Ran} (T-\lambda)^n.$$

Now my question is whether these subspaces and the integer $n$ can be explicitly calculated in the case of the Volterra operator.

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    $\begingroup$ I am a bit puzzled by this question. $T$ has no point spectrum away from $0$, so Isn't the kernel of $(T-\lambda)^n$ always just the zero function? $\endgroup$
    – Yonah Borns-Weil
    Commented Sep 26, 2022 at 3:51
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    $\begingroup$ I agree with @YonahBorns-Weil's comment, and I'd add that even the entire spectrum of $T$ consists of $0$ only. $\endgroup$
    – Jochen Glueck
    Commented Sep 26, 2022 at 8:20
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    $\begingroup$ In other words, $n=1$ works, since $\lambda\notin\sigma(T)$, so $R(T-\lambda)=L^2$, $N(T-\lambda)=0$. $\endgroup$
    – Christian Remling
    Commented Sep 26, 2022 at 14:21
  • $\begingroup$ In fairness, the Volterra operator can be unsettling at first, because it's very easy to implicitly assume that every compact operator has eigenvalues (in particular, that's true for self-adjoint operators and for finite-rank operators, for different reasons.) $\endgroup$
    – Yonah Borns-Weil
    Commented Sep 26, 2022 at 16:13

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