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  "body": "\n## Definition\n\nLet $H$ be a complex [[Hilbert space|HM2HHE3J51]] and let $B(H)$ be the [[algebra of bounded operators|BMDG4JQPEH]] on $H$.\nThe *strong operator topology* on $B(H)$ is the coarsest topology such that each map $B(H) \\to H$, $T \\mapsto T x$, with $x \\in H$, is continuous.\nHere $H$ carries its norm topology.\n\n## Remarks\n\nEquivalently, it is the [[topology generated by the seminorms|ZVFXQ7S6FY]] $T \\mapsto \\|T x\\|$ for $x \\in H$.\nA net $(T_i)$ converges to $T$ in this topology if and only if $\\|T_i x - T x\\| \\to 0$ for every $x \\in H$.\n\nExample: on $H = \\ell^2(\\N)$ with standard basis $(e_k)$, let $P_n$ be the orthogonal projection onto the span of $e_0, \\dots, e_{n-1}$.\nFor every $x$, $\\|x - P_n x\\|^2 = \\sum_{k \\ge n} \\abs{x_k}^2 \\to 0$, so $P_n \\to I$ strongly.\nBut $\\|I - P_n\\| = 1$ for all $n$, so there is no convergence in norm.\n\nNon-example: let $S$ be the right shift on $\\ell^2(\\N)$.\nThe powers $S^n$ tend to $0$ in the [[weak operator topology|4XM1VN1C1N]].\nIf $S^n$ converged strongly to some $T$, then $T$ would also be the weak limit, so $T = 0$; but $\\|S^n x\\| = \\|x\\|$ does not tend to $0$ for $x \\ne 0$.\nSo $(S^n)$ has no strong limit, while the adjoint powers $(S^*)^n$, which delete the first $n$ entries, do tend to $0$ strongly.\nIn particular $T \\mapsto T^*$ is not continuous in the strong operator topology.\n\nThe strong operator topology is finer than the weak operator topology and coarser than the norm topology.\nA convex subset of $B(H)$ has the same closure in both the weak and the strong operator topology.\n",
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