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  "body": "\n## Statement\n\nLet $X$ and $Y$ be [[Banach spaces|06GZ956657]] over $\\R$ or $\\C$, and let $T \\colon X \\to Y$ be a [[linear map|0NRYHJPQG2]].\nGive $X \\times Y$ the norm $\\norm{(x, y)} = \\norm{x} + \\norm{y}$.\nThen $T$ is [[continuous|03TM135G0N]] if and only if its [[graph|0JBJYT69ZV]] $\\Gamma(T) = \\set{(x, Tx) : x \\in X}$ is [[closed|0ZCK8DQYN1]] in $X \\times Y$.\n\n## Remarks\n\nIf $T$ is continuous, its graph is closed because $Y$ is Hausdorff.\nConversely, if $\\Gamma(T)$ is closed, it is a Banach space, and the projection $(x, Tx) \\mapsto x$ is a continuous linear bijection from $\\Gamma(T)$ onto $X$.\nThe [[bounded inverse theorem|MR5FJXR348]] makes its inverse $x \\mapsto (x, Tx)$ continuous, and so $T$ is continuous.\n\nIn practice the closed graph condition is easier to check: it says that if $x_n \\to x$ and $T x_n \\to y$, then $y = Tx$.\nOne does not have to show that $T x_n$ converges.\nCompleteness is necessary.\nLet $X = C^1[0, 1]$ and $Y = C[0, 1]$, both with the sup norm, and let $T f = f'$.\nIf $f_n \\to f$ and $f_n' \\to g$ uniformly, then $f$ is differentiable with $f' = g$, so the graph is closed.\nBut $T$ is not continuous, since $f_n(t) = t^n$ has $\\norm{f_n} = 1$ and $\\norm{f_n'} = n$; here $X$ is not complete.\n\n## Explanations\n\n### L1 human\n\nTo show that a linear map between complete normed spaces is continuous, it is enough to show that its graph is a closed set.\nThis means: whenever inputs converge and outputs converge, the limit output is the image of the limit input.\nThat is often much easier than finding a bound directly.\n\n### L2 human\n\nThe theorem is used to prove automatic continuity: for example, a linear map $T$ defined on a whole Hilbert space with $\\ip{Tx}{y} = \\ip{x}{Ty}$ for all $x, y$ is bounded (Hellinger–Toeplitz theorem).\nIt is also used to show that a projection onto a closed subspace along a closed complement is bounded, see [[complemented subspace|Q0TY362K0V]].\nUnbounded operators in quantum mechanics are typically closed operators defined only on dense subspaces, which is consistent with the theorem.\n",
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