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  "body": "\n## Definition\n\nLet $X$ and $Y$ be [[normed spaces|03J5RAKB30]] over $\\R$ or $\\C$, with [[continuous dual spaces|4A82S5FVRT]] $X'$ and $Y'$.\nLet $T \\colon X \\to Y$ be a [[continuous linear map|03TM135G0N]].\nThe *dual operator* (or *transpose*, or *Banach space adjoint*) of $T$ is the map\n$$T' \\colon Y' \\to X', \\qquad T'(f) = f \\circ T.$$\n\n## Remarks\n\nThe map $T'$ is well defined since $f \\circ T$ is linear and continuous, and $T'$ is linear and bounded.\nBy the [[Hahn–Banach theorem|QB5AE3VYZZ]], its [[operator norm|06EKW1AKXF]] is $\\norm{T'} = \\norm{T}$.\nIt is the restriction of the algebraic [[dual map|0AZ5ARP287]] to continuous functionals; other notations are $T^*$ and $T^t$.\n\nExample: for $1 \u003c p \u003c \\infty$ and $1/p + 1/q = 1$, identify the dual of $\\ell^p$ with $\\ell^q$ via $f(x) = \\sum_n f_n x_n$.\nLet $S$ be the right shift on $\\ell^p$.\nThen $f(Sx) = \\sum_{n \\ge 2} f_n x_{n-1} = \\sum_{m \\ge 1} f_{m+1} x_m$, so $S' f = (f_2, f_3, \\ldots)$ is the left shift on $\\ell^q$.\n\nNon-example: the rule $f \\mapsto f \\circ T$ does not give a map $Y' \\to X'$ when $T$ is not continuous.\nLet $X = Y$ be the space of polynomials on $[0, 1]$ with the supremum norm, $T p = p'$, and $f(p) = p(0)$.\nThe polynomials $p_n(x) = x (1 - x)^n$ have $\\norm{p_n} \\le 1/(n+1)$, but $(f \\circ T)(p_n) = p_n'(0) = 1$, so $f \\circ T$ is not continuous.\n\nThe rule $(ST)' = T' S'$ holds, and $\\ker T'$ is the annihilator of the range of $T$.\n",
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