# The norm of a space is a uniformly continuous map with respect to the topology it induces

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Check over before removing this, ensure it's linked to, maybe add "uniform continuity => continuity" somewhere if that's true (not seen uniform continuity for 5 years at this point!) Alec (talk) 20:59, 9 April 2017 (UTC)

## Statement

Let [ilmath]((X,[/ilmath][ilmath]\mathbb{K} [/ilmath][ilmath]),\Vert\cdot\Vert)[/ilmath] be a normed space, we claim that the norm itself, [ilmath]\Vert\cdot\Vert:X\rightarrow\mathbb{R} [/ilmath], is a uniformly continuous map, with respect to the topology induced by the metric which itself is the metric induced by the norm and the topology induced by the absolute value metric on [ilmath]\mathbb{R} [/ilmath].

Symbolically we claim:

• [ilmath]\forall\epsilon\in\mathbb{R}_{>0}\exists\delta\in\mathbb{R}_{>0}\forall x,y\in X[d_{\Vert\cdot\Vert}(x,y)<\delta\implies d_\mathbb{R}(\Vert x\Vert,\Vert y\Vert)<\epsilon][/ilmath].
• Substituting [ilmath]d_{\Vert\cdot\Vert}(x,y):\eq\Vert x-y\Vert[/ilmath] and [ilmath]d_\mathbb{R}(a,b):\eq\big\vert a-b\big\vert[/ilmath] we obtain [ilmath]\forall\epsilon\in\mathbb{R}_{>0}\exists\delta\in\mathbb{R}_{>0}\forall x,y\in X[\Vert x-y\Vert<\delta\implies\big\vert \Vert x\Vert - \Vert y\Vert\big\vert<\epsilon][/ilmath]

## Proof

• Let [ilmath]\epsilon\in\mathbb{R}_{>0} [/ilmath] be given
• Recall the reverse triangle inequality for metrics (although reverse triangle inequality is exactly what we want - as it stands when this is being written), that [ilmath]\big\vert d(a,c)-d(b,c)\big\vert\le d(a,b)[/ilmath] for any [ilmath]a,b,c[/ilmath] in the space.
• We use this with [ilmath]c\eq 0[/ilmath] to obtain [ilmath]\big\vert d(x,0)-d(y,0)\Vert\le d(x,y)[/ilmath] which is: [ilmath]\big\vert \Vert x\Vert - \Vert y\Vert \big\vert\le \Vert x-y\Vert[/ilmath] exactly
• This means, if [ilmath]\delta,x,y[/ilmath] were given and [ilmath]\Vert x-y\Vert<\delta[/ilmath] that we would have [ilmath]\big\vert \Vert x\Vert - \Vert y\Vert \big\vert\le \Vert x-y\Vert <\delta[/ilmath], giving by transitivity [ilmath]\big\vert \Vert x\Vert - \Vert y\Vert \big\vert<\delta[/ilmath]
• Define [ilmath]\delta:\eq\epsilon[/ilmath], as [ilmath]\epsilon\in\mathbb{R}_{>0} [/ilmath] by definition we see [ilmath]\delta\in\mathbb{R}_{>0} [/ilmath] - as required
• Let [ilmath]x,y\in X[/ilmath] be given
• Suppose [ilmath]\Vert x-y\Vert\ge\delta[/ilmath], in this case by the nature of logical implication we do not care about the truth or falsity of the right hand side of the implication, it holds regardless
• Suppose [ilmath]\Vert x-y\Vert<\delta[/ilmath], by the nature of logical implication we require that in this case [ilmath]\big\vert \Vert x\Vert - \Vert y\Vert\big\vert<\epsilon[/ilmath] holds
• As discussed above: [ilmath]\big\vert \Vert x\Vert - \Vert y\Vert \big\vert \le \Vert x-y\Vert < \delta \eq \epsilon[/ilmath] (we know that to be here [ilmath]\Vert x-y\Vert < \delta[/ilmath] after all)
• So [ilmath]\big\vert \Vert x\Vert - \Vert y\Vert \big\vert < \epsilon[/ilmath] - as required
• We have shown the implication holds
• Since [ilmath]x,y\in X[/ilmath] were arbitrary we have shown [ilmath]\forall x,y\in X[\Vert x-y\Vert<\delta\implies\big\vert \Vert x\Vert - \Vert y\Vert\big\vert<\epsilon][/ilmath]
• We have defined a [ilmath]\delta[/ilmath] based on being given an [ilmath]\epsilon[/ilmath], so [ilmath]\exists\delta\in\mathbb{R}_{>0}\forall x,y\in X[\Vert x-y\Vert<\delta\implies\big\vert \Vert x\Vert - \Vert y\Vert\big\vert<\epsilon][/ilmath] holds
• Lastly since [ilmath]\epsilon\in\mathbb{R}_{>0} [/ilmath] was arbitrary we have shown [ilmath]\exists\delta\in\mathbb{R}_{>0}\forall x,y\in X[\Vert x-y\Vert<\delta\implies\big\vert \Vert x\Vert - \Vert y\Vert\big\vert<\epsilon][/ilmath] holds for all such [ilmath]\epsilon[/ilmath]

i.e. that [ilmath]\forall\epsilon\in\mathbb{R}_{>0}\exists\delta\in\mathbb{R}_{>0}\forall x,y\in X[\Vert x-y\Vert<\delta\implies\big\vert \Vert x\Vert - \Vert y\Vert\big\vert<\epsilon][/ilmath] - as required.