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== A concrete example ==
Examples of discontinuous linear maps are easy to construct in spaces that are not complete; on any Cauchy sequence of independent vectors which does not have a limit, a linear operator may grow without bound.{{what|reason=This is not a proof nor even clear statement of anything, yet later in the article it is treated as an established principle.|date=May 2015}} In a sense, the linear operators are not continuous because the space has "holes".
For example, consider the space ''X'' of real-valued [[smooth function]]s on the interval [0, 1] with the [[uniform norm]], that is,
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== General existence theorem ==
Discontinuous linear maps can be proven to exist more generally even if the space is complete.{{what|reason=A general "constructive" proof in the incomplete case was not given above, so this contrast seems kind of hand-wavey.|date=May 2015}} Let ''X'' and ''Y'' be [[normed space]]s over the field ''K'' where ''K'' = '''R''' or ''K'' = '''C'''. Assume that ''X'' is infinite-dimensional and ''Y'' is not the zero space. We will find a discontinuous linear map ''f'' from ''X'' to ''K'', which will imply the existence of a discontinuous linear map ''g'' from ''X'' to ''Y'' given by the formula ''g''(''x'') = ''f''(''x'')''y''<sub>0</sub> where ''y''<sub>0</sub> is an arbitrary nonzero vector in ''Y''.
If ''X'' is infinite-dimensional, to show the existence of a linear functional which is not continuous then amounts to constructing ''f'' which is not bounded. For that, consider a [[sequence]] (''e''<sub>''n''</sub>)<sub>''n''</sub> (''n'' ≥ 1) of [[linearly independent]] vectors in ''X''. Define
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