Editing Linear map (section)

==Continuity==
{{Main|Continuous linear operator|Discontinuous linear map}}

A ''linear transformation'' between [[topological vector space]]s, for example [[normed space]]s, may be [[continuous function (topology)|continuous]]. If its domain and codomain are the same, it will then be a [[continuous linear operator]]. A linear operator on a normed linear space is continuous if and only if it is [[bounded operator|bounded]], for example, when the domain is finite-dimensional.<ref>{{harvnb|Rudin|1991|page=15}}

'''1.18 Theorem''' ''Let <math display="inline">\Lambda</math> be a linear functional on a topological vector space {{mvar|X}}. Assume <math display="inline">\Lambda \mathbf x \neq 0</math> for some <math display="inline">\mathbf x \in X</math>. Then each of the following four properties implies the other three:''
{{ordered list
|list-style-type=lower-alpha
|<math display="inline">\Lambda</math> is continuous
|The null space <math display="inline">N(\Lambda)</math> is closed.
|<math display="inline">N(\Lambda)</math> is not dense in {{mvar|X}}.
|<math display="inline">\Lambda</math> is bounded in some neighbourhood {{mvar|V}} of 0.}}</ref> An infinite-dimensional domain may have [[discontinuous linear operator]]s.

An example of an unbounded, hence discontinuous, linear transformation is differentiation on the space of smooth functions equipped with the supremum norm (a function with small values can have a derivative with large values, while the derivative of 0 is 0). For a specific example, {{math|sin(''nx'')/''n''}} converges to 0, but its derivative {{math|cos(''nx'')}} does not, so differentiation is not continuous at 0 (and by a variation of this argument, it is not continuous anywhere).