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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''. Without loss of generality we van assume that <math>\|e_n\| = 1</math>. Define
:<math>T(e_n)=n\|e_n\|\,</math>
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