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  "body": "\n## Example\n\nThe *Volterra operator* is the operator $V$ on the [[Lp space|SASP14MA1Z]] $L^2(0, 1)$ given by\n$$(V f)(t) = \\int_0^t f(s) \\, ds, \\quad t \\in [0, 1].$$\nIt is the integral operator of the [[Volterra equation of the first kind|H6ZGEDM4VV]] with kernel $K(t, s) = 1$ for $s \\le t$ and $K(t, s) = 0$ for $s \u003e t$.\n\nCheck that it is a [[compact operator|M2NEP1K5AE]]: the kernel is square-integrable on $[0, 1]^2$, with $\\int_0^1 \\int_0^1 \\abs{K}^2 = 1/2$.\nSo $V$ is a [[Hilbert–Schmidt integral operator|CB5G212Q03]], and Hilbert–Schmidt operators are compact.\n\nCheck that $V$ has no eigenvalue.\nIf $V f = 0$, then $\\int_0^t f = 0$ for all $t$, so $f = 0$ almost everywhere.\nIf $V f = \\lambda f$ with $\\lambda \\ne 0$, then $f = \\lambda^{-1} V f$ is continuous, hence $V f$ is continuously differentiable, and $f' = f / \\lambda$ with $f(0) = 0$; so $f = 0$.\n\nCheck that the [[spectrum|BMKATJ802A]] is $\\set{0}$.\nThe power $V^n$ has kernel $(t - s)^{n-1}/(n-1)!$ for $s \\le t$, so $\\norm{V^n} \\le 1/(n-1)!$.\nHence the [[spectral radius|9V7YQA85PC]] $\\lim_n \\norm{V^n}^{1/n}$ is $0$, while $0$ is in the spectrum because a compact operator on an infinite-dimensional space is not invertible.\n\n## Remarks\n\nThe operator norm of $V$ is $2/\\pi$.\nIts [[singular values|QSJKPKPM7H]] are $2/((2n - 1)\\pi)$, $n \\ge 1$, which are square-summable but not summable, so $V$ is not a [[trace class operator|Z2CP530WDE]].\nThe adjoint is $(V^* f)(t) = \\int_t^1 f(s) \\, ds$, and $V + V^*$ is the rank-one operator $f \\mapsto \\int_0^1 f$.\nThe Volterra operator is the standard example of a nonzero quasinilpotent compact operator: its spectrum is the same as that of the zero operator.\n\n## Explanations\n\n### L1 human\n\nThe Volterra operator takes a function and returns its running integral from $0$.\nIt is the most basic integral operator, the continuous version of taking partial sums.\nIt is a compact operator with no eigenvalues at all, something that cannot happen for a square matrix over the complex numbers.\nThis is why equations $f - \\lambda V f = g$ can be solved for every $\\lambda$.\n",
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