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 $\epsilon>0$ and $(x_n)$ be an arbitrary sequence in $X$ that converges to $x\in X$. Then, $$\exists N\in \mathbb{N}:n\ge N\implies \left\|x_n-x\right\|<\epsilon$$ But $$ \left| \left\|x_n\right\|- \left\|x\right\|\right|\le \left\|x_n-x\right\|$$ by the triangle inequality. Thus, $$\exists N\in \mathbb{N}:n\ge N\implies \left| \left\|x_n\right\|- \left\|x\right\|\right|<\epsilon$$ and we are done!
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1at the last line you get norm of numbers! you should get absulote value of them. And you had use of question! – 2012-12-26
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0@VahidDashad "you had use of question" What does that mean? – 2012-12-26
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1,In the middle of your proof in trianglular inequality, you should change norm with absolote value again! – 2012-12-26
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0@VahidDashad Sure – 2012-12-26
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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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2The question included the phrase "mathematical symbols". How is your post an answer to the question? – 2012-12-26
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2@Nameless Please read my post again, it is a precise answer. Most languages, even most formalized languages, are using latin letters among other symbols. Therefore latin letters are mathematical symbols. I don't understand your objections. – 2012-12-26
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1I believe the OP asked for a formal proof of why norms are continuous, not an argument like yours. I don't plan to debate on this or to continue a meaningless discussion. Please keep your answer as it may be of help to the OP (or even what he wants) – 2012-12-26
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1@Nameless OK, so we are ending our discussion. Keep in mind though, that there is no such thing as "a formal proof of WHY norms are continuous". There are only formal proofs of the fact, that norms ARE continuous, for example you just demonstrated one. Best regards. – 2012-12-26
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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