Describe why norms are continuous function by mathematical symbols.
Why are norms continuous?
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0please cite the definitions of "norm" and "continuous" you are using – 2017-10-28
3 Answers
A function $f$ from a metric space to a metric space is continuous if for all $x$ in the domain, for all $\varepsilon>0$, the exists $\delta>0$ such that for all points $y$ in the domain, if the distance from $x$ to $y$ is less than $\delta$, then the distance from $f(x)$ to $f(y)$ is less than $\varepsilon$.
If $f$ is a norm, then it maps a vector space into $\mathbb R$, and the distance from $x$ to $y$ is $f(x-y)$.
In this case it suffices to take $\delta=\varepsilon$, for the following reason. Suppose the distance from $x$ to $y$ is less than $\delta=\varepsilon$. Then $f(x-y)=f(y-x)<\varepsilon$ (where the equality follows from the definition of "norm"). Now recall that norms satisfy a triangle inequality: $ f(x) \le f(y) + f(x-y) $ $ f(y) \le f(x) + f(y-x) $ So $ f(y)-f(x) \le f(y-x)<\varepsilon\text{ and }f(x)-f(y) \le f(x-y)<\varepsilon, $ so $ |f(x)-f(y)|<\varepsilon, $ i.e. $ \Big(\text{distance from $f(x)$ to $f(y)$}\Big) <\varepsilon. $
Let $(X,\left\|\cdot\right\|)$ be a normed space. We need to prove that: $\forall (x_n):\mathbb{N}\to X\ x_n\to x\implies \left\|x_n\right\|\to \left\|x\right\|$ Let $\varepsilon>0$ and $(x_n)$ be an arbitrary sequence in $X$ that converges to $x\in X$. Then, $$\exists N\in \mathbb{N}:n\ge N\implies \left\|x_n-x\right\|<\varepsilon$$ But $ \left| \left\|x_n\right\|- \left\|x\right\|\right|\le \left\|x_n-x\right\|$ by the triangle inequality. Thus, $$\exists N\in \mathbb{N}:n\ge N\implies \left| \left\|x_n\right\|- \left\|x\right\|\right|<\varepsilon$$ and we are done!
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0@Nameless You proved that the norm is continuous, but you did not answered the question. – 2012-12-26
To keep it short and straight to the point: the norm of the normed space $(X,\|\cdot\|)$ is a continuous function because the topology you (usually) consider on $X$ is the smallest topology in which $\|\cdot\|$ is continuous. So it is continuous because we want it to be continuous.
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8@Godot The topology induced by a norm is not generally the smallest topology that makes the norm continuous. The latter makes no distinction between vectors that have the same norm, and hence is not even Hausdorff. What holds is that the topology induced by a norm (and more generally by a metric) is the smallest topology that makes the distance function $((x,y)\mapsto \|x-y\|)$ continuous. – 2014-11-08