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  "body": "\n## Statement\n\nLet $\\mathbb{K}$ be $\\R$ or $\\C$, and let $H$ be a [[Hilbert space|HM2HHE3J51]] over $\\mathbb{K}$ whose inner product $\\ip{\\cdot}{\\cdot}$ is linear in the first argument.\nLet $\\varphi \\colon H \\to \\mathbb{K}$ be a continuous [[linear form|07YQ1C78G0]].\nThen there is exactly one vector $y \\in H$ such that $\\varphi(x) = \\ip{x}{y}$ for all $x \\in H$.\nMoreover $\\norm{\\varphi} = \\norm{y}$, where $\\norm{\\varphi}$ is the operator norm of $\\varphi$.\nHence the map that sends $\\varphi$ to $y$ is a bijective, conjugate-linear isometry from the [[continuous dual|4A82S5FVRT]] $H'$ onto $H$.\n\n## Remarks\n\nIf $\\varphi = 0$, take $y = 0$.\nOtherwise the kernel $N = \\ker \\varphi$ is a closed subspace different from $H$.\nBy the [[Hilbert projection theorem|J5YNTHQMFX]], its [[orthogonal complement|0XJXTBCB2T]] $N^\\perp$ contains a vector $z \\neq 0$, and $N^\\perp$ has dimension $1$.\nThen $y = \\overline{\\varphi(z)}\\, z / \\norm{z}^2$ works, and uniqueness follows because $\\ip{x}{y - y'} = 0$ for all $x$ forces $y = y'$.\n\nExample: on $L^2([0,1])$ the form $\\varphi(f) = \\int_0^1 f(t)\\, dt$ is represented by the constant function $y = 1$, and $\\norm{\\varphi} = 1$.\n\nCompleteness is necessary.\nLet $c_{00}$ be the space of finitely supported sequences with the inner product of $\\ell^2$.\nThe form $\\varphi(x) = \\sum_n x_n / n$ is continuous on $c_{00}$ by the Cauchy–Schwarz inequality.\nThe only candidate is $y = (1/n)_n$, which is not finitely supported, so no $y \\in c_{00}$ represents $\\varphi$.\n\nIn finite dimension the result reduces to the [[Riesz representation theorem in finite dimensions|04GYPXDXNH]].\n\n## Explanations\n\n### L1 human\n\nOn a Hilbert space, every reasonable way to turn vectors into numbers linearly is \"take the inner product with a fixed vector\".\nSo the space of such linear measurements looks exactly like the space itself.\nFor example, taking the average of a function on $[0,1]$ is the inner product with the constant function $1$.\n\n### L2 human\n\nThe theorem identifies a Hilbert space with its dual, up to complex conjugation.\nIt gives the adjoint of a bounded operator: $\\ip{Tx}{y} = \\ip{x}{T^*y}$ defines $T^*y$ because $x \\mapsto \\ip{Tx}{y}$ is a continuous linear form.\nIt also shows that every Hilbert space is reflexive.\nThe Lax–Milgram theorem, used for weak solutions of elliptic equations, extends it to bounded coercive bilinear forms.\n",
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