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## Homework Statement

Let [tex] f:A\subset{\mathbb{R}}^{n}\mapsto \mathbb{R} [/tex] be a linear function continuous a [tex] \vec{0} [/tex]. To prove that [tex]f[/tex] is continuous everywhere.

## Homework Equations

If [tex] f[/tex] is continuous at zero, then [tex]\forall \epsilon>0 \exists\delta>0 [/tex] such that if [tex] \|\vec{x}\|<\delta[/tex] then [tex] \|f(\vec{x}) \|< \epsilon [/tex].

[tex] f[/tex] also satisfies [tex] f(\vec{a}+\alpha\vec{b})=f(\vec{a})+\alpha f(\vec{b}) [/tex].

## The Attempt at a Solution

I tried using several forms of the triangle inequality to prove that [tex] \|f(\vec{x})\|<\epsilon [/tex] implies that [tex] \|f(\vec{x})-f(\vec{x_0})\|<\epsilon [/tex] by means of adding a zero [tex] f(x)=f(x+0)=f(x+x_0-x_0)=f(x)-f(x_0)+f(x_0) [/tex] but I havent been able to conclude anything special.

Thanks in advance for all your help.