Densely defined operator

Definition

Let $X$ and $Y$ be normed spaces, and let $T$ be an unbounded operator from $X$ to $Y$ with domain $D(T)$ . The operator $T$ is densely defined if $D(T)$ is a dense subset of $X$ , that is, the closure of $D(T)$ in $X$ is $X$ .

Remarks

Example: on $X = C[0, 1]$ with the supremum norm, let $D(T)$ be the space of polynomials and $Tp = p'$ . By the Weierstrass approximation theorem, every continuous function on $[0, 1]$ is a uniform limit of polynomials, so $T$ is densely defined.

Non-example: on $X = L^2(0, 1)$ , let $D(T)$ be the closed subspace of functions that vanish almost everywhere on $(0, \tfrac12)$ , and $Tf = f$ . The closure of $D(T)$ is $D(T)$ itself, which is not $X$ , so $T$ is not densely defined.

A bounded operator that is densely defined extends uniquely to a bounded operator on all of $X$ when $Y$ is a Banach space.

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Tag	`VNTA4KYMQT`
URI	`m2://VNTA4KYMQT`
Kind	definition
Writing	W0 stub
MSC	47A05, 47B25
Defines	Densely defined operator
Uses	Unbounded operator, Dense set
Changed	2 November 2026

Densely defined operator

Definition

Remarks

Reading path

Further reading

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