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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 $Y$ with domain $D(T)$.\nThe operator $T$ is *densely defined* if $D(T)$ is a [[dense subset|0BSJSX7ME2]] of $X$, that is, the closure of $D(T)$ in $X$ is $X$.\n\n## Remarks\n\nExample: on $X = C[0, 1]$ with the supremum norm, let $D(T)$ be the space of polynomials and $Tp = p'$.\nBy the Weierstrass approximation theorem, every continuous function on $[0, 1]$ is a uniform limit of polynomials, so $T$ is densely defined.\n\nNon-example: on $X = L^2(0, 1)$, let $D(T)$ be the closed subspace of functions that vanish almost everywhere on $(0, \\tfrac12)$, and $Tf = f$.\nThe closure of $D(T)$ is $D(T)$ itself, which is not $X$, so $T$ is not densely defined.\n\nA bounded operator that is densely defined extends uniquely to a bounded operator on all of $X$ when $Y$ is a Banach space.\n",
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