Is the norm continuous? Can I switch limits like this?

Question

I am in the middle of my proof and I want to know if the following is true, suppose $f_n$ is a Cauchy sequence, can i do this?

If $$\| f_n(x) - f(x) \| \to 0,$$ then can I also say this limit is true

$$\lim_{m \to \infty} \| f_n(x) - f_m(x) \| \to 0?$$

@copper.hat, trying to show completeness of a particular space. — Lemon, Nov 06 '14 at 06:08
The question above has nothing to do with completeness. What is $f$? — copper.hat, Nov 06 '14 at 06:08
@copper.hat, I know it has nothing to do completeness, it is part of a proof but this is just a tiny part of the technique I want to use. — Lemon, Nov 06 '14 at 06:09
Assuming that $f_k(x) \to f(x)$, then it is true since the norm is continuous. But the $\to 0$ is irrelevant here. If you have a property $P$ such that $P(x)$ is true and $y=x$ then you know that $P(y)$ is also true. — copper.hat, Nov 06 '14 at 06:12
@copper.hat, $f_n$ is Cauchy in $C[0,1]$. That's why I had the $0$ thing, uniform convergence. — Lemon, Nov 06 '14 at 06:19

score 1 · Answer 1 · answered Nov 06 '14 at 06:18

1

Short Answer: Yes the norm is continuous in the usual topology on a normed vector space.

Medium Answer: Let $X$ be a normed vector space. Let $\epsilon > 0$ and put $\delta = \epsilon$ then $\forall x,y \in X$ we have $$ \delta > \lVert x - y \rVert \ge \lvert \lVert x \rVert - \lVert y \rVert \rvert \implies \epsilon > \lvert \lVert x \rVert - \lVert y \rVert \rvert $$ where we used the reverse triangle inequality.

answered Nov 06 '14 at 06:18

DanZimm

5,781

I am not sure what the med answer is for…? – Lemon Nov 06 '14 at 06:20
@jip it's a proof that the norm is continuous – DanZimm Nov 06 '14 at 06:20
Oh right, that's part of the question in my title…lol – Lemon Nov 06 '14 at 06:21
@jip :P all good, anyways yes you can switch the limit there as long as you have $f_m(x) \to f(x)$ (I'm guessing you've defined $f(x) := \lim_{m \to \infty} f_m(x)$ in which case this is true). – DanZimm Nov 06 '14 at 06:23
The limit is point wise convergence, well I don't think it matters in this case. – Lemon Nov 06 '14 at 06:25

score 1 · Accepted Answer · answered Nov 06 '14 at 06:43

1

The triangle inequality gives $\|x\|-\|y\| \le \|x-y\|$. Switching $x,y$ gives $\|y\|-\|x\| \le \|x-y\|$, hence $|\|x\|-\|y\|| \le \|x-y\|$, and so the norm is (Lipschitz) continuous.

answered Nov 06 '14 at 06:43

copper.hat

172,524

Is the norm continuous? Can I switch limits like this?

2 Answers2