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Defining <math>\ker p := p^{-1}(0),</math> then subadditivity also guarantees that for all <math>x \in X,</math> the value of <math>p</math> on the set <math>x + (\ker p \cap -\ker p) = \{x + k : p(k) = 0 = p(-k)\}</math> is constant and equal to <math>p(x).</math><ref group=proof name=ConstantOnEquivClasses>Let <math>x \in X</math> and <math>k \in p^{-1}(0) \cap (-p^{-1}(0)).</math> It remains to show that <math>p(x + k) = p(x).</math> The triangle inequality implies <math>p(x + k) \leq p(x) + p(k) = p(x) + 0 = p(x).</math> Since <math>p(-k) = 0,</math> <math>p(x) = p(x) - p(-k) \leq p(x - (-k)) = p(x + k),</math> as desired. <math>\blacksquare</math></ref>
In particular, if <math>\ker p = p^{-1}(0)</math> is a vector subspace of <math>X</math> then <math>- \ker p = \ker p</math> and the assignment <math>x + \ker p \mapsto p(x),</math> which will be denoted by <math>\hat{p},</math> is a well-defined real-valued sublinear function on the [[Quotient space (linear algebra)|quotient space]] <math>X \,/\, \ker p</math> that satisfies <math>\hat{p} ^{-1}(0) = \ker p.</math> If <math>p</math> is a seminorm then <math>\hat{p}</math> is just the usual canonical norm on the quotient space <math>X \,/\, \ker p.</math>
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