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  "body": "\n## Statement\n\nLet $X$ be a [[Banach space|06GZ956657]] and let $Y$ be a [[normed vector space|03J5RAKB30]], both over $\\R$ or $\\C$.\nLet $\\mathcal{F}$ be a family of [[continuous linear maps|03TM135G0N]] from $X$ to $Y$.\nIf $\\sup_{T \\in \\mathcal{F}} \\norm{Tx} \u003c \\infty$ for every $x \\in X$, then $\\sup_{T \\in \\mathcal{F}} \\norm{T} \u003c \\infty$, where $\\norm{T}$ is the [[operator norm|06EKW1AKXF]].\n\n## Remarks\n\nFor $k \\ge 1$ let $F_k = \\set{x \\in X : \\norm{Tx} \\le k \\text{ for all } T \\in \\mathcal{F}}$.\nEach $F_k$ is closed, and the hypothesis says that the $F_k$ cover $X$.\nSince $X$ is complete, the [[Baire category theorem|0A7KGAYHGV]] gives some $F_k$ that contains a ball $B(x_0, r)$.\nThen for $\\norm{x} \u003c r$ and $T \\in \\mathcal{F}$, $\\norm{Tx} \\le \\norm{T(x_0 + x)} + \\norm{T x_0} \\le 2k$, so $\\norm{T} \\le 2k/r$.\n\nA main use is the Banach–Steinhaus theorem: if continuous linear maps $T_n \\colon X \\to Y$ converge at every point to $Tx$, then $T$ is linear and continuous, since $\\norm{Tx} \\le \\sup_n \\norm{T_n}\\,\\norm{x}$.\nCompleteness of $X$ is necessary.\nLet $X$ be the space of finitely supported real sequences with the norm $\\norm{x}_\\infty = \\sup_k \\abs{x_k}$, and let $T_n(x) = n x_n$.\nFor each $x$ the values $T_n x$ are eventually $0$, so they are bounded, but $\\norm{T_n} = n$.\n\n## Explanations\n\n### L1 human\n\nSuppose a family of linear operators on a complete space is bounded at each single point.\nThe theorem says that the family is then bounded uniformly, by one constant times the norm.\nPointwise control turns into global control, for free.\n\n### L2 human\n\nIt is one of three consequences of the Baire category theorem at the core of a first course on Banach spaces.\nThe other two are the [[bounded inverse theorem|MR5FJXR348]] and the [[closed graph theorem|GXM8BX8YM7]].\nA classical application shows that there is a continuous periodic function whose Fourier series diverges at a point: the partial-sum functionals at that point have unbounded norms.\nIt also shows that weakly convergent sequences in a normed space are bounded, by applying it to the sequence viewed in the bidual.\n",
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