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  "body": "\n## Definition\n\nLet $X$ and $Y$ be normed spaces, and let $T$ be an [[unbounded operator|KRJX5RA0PW]] from $X$ to $Y$.\nThe operator $T$ is *closable* if it has an extension $S \\supseteq T$ that is a [[closed operator|VM7B38PA87]].\n\n## Remarks\n\nThe following are equivalent.\n\n1. The operator $T$ is closable.\n2. The [[closure|0QEAJB91GZ]] of the [[graph|3RCAY97W1E]] $\\Gamma(T)$ in $X \\times Y$ is the graph of an operator.\n3. For every sequence $(x_n)$ in $D(T)$ with $x_n \\to 0$ and $T x_n \\to y$, one has $y = 0$.\n\nExample: on $C[0, 1]$ with the supremum norm, let $D(T)$ be the space of polynomials and $Tp = p'$.\nIf $p_n \\to 0$ and $p_n' \\to g$ uniformly, then $p_n(x) - p_n(0) = \\int_0^x p_n'$ gives $0 = \\int_0^x g$ for all $x$, so $g = 0$; hence $T$ is closable.\n\nNon-example: on $L^2(0, 1)$, let $D(T) = C[0, 1]$ and $Tf = f(0) \\cdot \\mathbf{1}$, where $\\mathbf{1}$ is the constant function $1$.\nThe functions $f_n(x) = \\max(0, 1 - n x)$ satisfy $f_n \\to 0$ in $L^2(0, 1)$ but $T f_n = \\mathbf{1}$ for all $n$, so $T$ is not closable.\n",
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