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Chen's question at Yahoo! Answers (continuity of the norm).

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Chen,

First, let's prove that $\left|N(x)-N(y)\right|\leq N(x-y)$ for all $x,y\in\mathbb{R}^n$. We have:

$N(y)=N(x+(y-x))\leq N(x)+N(x-y)\Rightarrow -N(x-y)\leq N(x)-N(y)\qquad (1)$

On the other hand:

$N(x)=N(y+(x-y))\leq N(y)+N(x-y)\Rightarrow N(x)-N(y)\leq N(x-y)\qquad (2)$

From $(1)$ and $(2)$ we clearly deduce that $\left|N(x)-N(y)\right|\leq N(x-y)$.

We'll consider on $\mathbb{R}^n$ the usual distance given by the norm $N$, that is $d(x,y)=N(x-y)$ and the usual distance on $\mathbb{R}$.

Now, fix $x_0\in \mathbb{R}^n$ an let $\epsilon>0$, choosing $\delta=\epsilon$:

$d(x,x_0)<\delta\Rightarrow N(x-x_0)<\delta=\epsilon\Rightarrow \left|N(x)-N(x_0)\right|<\epsilon$

which implies that $N$ is a continuous function with the specified distances.