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  "body": "\n## Definition\n\nLet $X$ and $Y$ be [[Banach spaces|06GZ956657]] over $\\R$ or $\\C$, and let $X'$ be the [[continuous dual|4A82S5FVRT]] of $X$.\nA [[bounded linear map|03TM135G0N]] $T \\colon X \\to Y$ is *nuclear* if there are sequences $(f_n)$ in $X'$ and $(y_n)$ in $Y$ such that\n$$\\sum_{n=1}^{\\infty} \\norm{f_n} \\norm{y_n} \u003c \\infty \\quad \\text{and} \\quad T x = \\sum_{n=1}^{\\infty} f_n(x) \\, y_n \\quad \\text{for all } x \\in X.$$\nThe *nuclear norm* $\\nu(T)$ is the infimum of $\\sum_n \\norm{f_n} \\norm{y_n}$ over all such representations.\n\n## Remarks\n\nEvery [[finite-rank operator|Z8FABSYQTS]] is nuclear, with a finite sum.\nExample: on $\\ell^2$, let $T x = \\sum_n d_n x_n e_n$ with $\\sum_n \\abs{d_n} \u003c \\infty$; take $f_n(x) = d_n x_n$ and $y_n = e_n$, so $\\norm{f_n} \\norm{y_n} = \\abs{d_n}$ is summable and $T$ is nuclear.\nNon-example: the identity of an infinite-dimensional Banach space is not nuclear, because it is not compact.\nEvery nuclear operator is a norm limit of the partial sums, which are of finite rank, so it is [[compact|M2NEP1K5AE]].\nWhen $X$ and $Y$ are Hilbert spaces, the nuclear operators are exactly the [[trace class operators|Z2CP530WDE]], and $\\nu(T)$ is the trace norm.\n",
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