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  "body": "\n## Definition\n\nLet $H$ and $K$ be [[Hilbert spaces|HM2HHE3J51]] over $\\C$, with inner products linear in the first argument.\nLet $T$ be a [[densely defined operator|BCJHNHKQFG]] from $H$ to $K$ with domain $D(T)$.\nThe *adjoint* of $T$ is the [[operator|KRJX5RA0PW]] $T^*$ from $K$ to $H$ given as follows.\n\n1. Its domain $D(T^*)$ is the set of $y \\in K$ for which there is $z \\in H$ with $\\ip{Tx}{y} = \\ip{x}{z}$ for all $x \\in D(T)$.\n2. For $y \\in D(T^*)$, $T^* y$ is this vector $z$.\n\n## Remarks\n\nThe vector $z$ is unique because $D(T)$ is dense: two such vectors differ by a vector orthogonal to $D(T)$, hence to $H$.\nEquivalently, $y \\in D(T^*)$ if and only if the functional $x \\mapsto \\ip{Tx}{y}$ is bounded on $D(T)$.\nIf $T$ is bounded with $D(T) = H$, then $T^*$ is the [[Hilbert space adjoint|6HSJBJ8TTQ]].\n\nThe adjoint $T^*$ is always [[closed|VM7B38PA87]].\nIt is densely defined if and only if $T$ is [[closable|3TWMBFQJJ8]], and then $T^{**}$ is the [[closure|TRC892PJNN]] of $T$.\nIf $S \\subseteq T$, then $T^* \\subseteq S^*$.\n\nExample: on $\\ell^2$, let $D(T) = \\set{x : \\sum n^2 \\abs{x_n}^2 \u003c \\infty}$ and $T(x_n) = (n x_n)$.\nFor $y \\in D(T^*)$ with $T^* y = z$, testing with the unit vector $x = e_n$ gives $n \\overline{y_n} = \\overline{z_n}$, so $z = (n y_n) \\in \\ell^2$ and $y \\in D(T)$.\nConversely each $y \\in D(T)$ lies in $D(T^*)$ with $T^* y = Ty$, so $T^* = T$.\n\nNon-example: if $D(T)$ is not dense, there is a nonzero $w \\perp D(T)$, and then $z$ and $z + w$ both satisfy clause 1, so clause 2 does not define a map.\n",
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