Let $f: \mathbb{F}^n\rightarrow \mathbb{R}$ be defined by $f(a_1,\cdots, a_n)=\|\sum a_jv_j\|$. Show $f$ is continuous on $\mathbb{F}^n$.
1. $\|\cdot\|$ is an arbitrary norm on $\mathcal{V}$.


  1. Let $u=\sum a_iv_i$ and $v=\sum b_iv_i$.
  2. Show $\forall \epsilon>0, \forall u\in \mathcal{V}$, there exists $\delta>0$ such that if $\|v-u\|\leq \delta$, then $|f(v)-f(u)|\leq \epsilon$.
  3. \begin{align} |f(v)-f(u)|&=|\|v\|-\|u\|| \\ &\leq \|v-u\| \ \ \ \ \text{this is implied by triangle inequality}\\ &=\|\sum (b_j-a_j)v_j\| \\ &\leq \sum \|(b_j-a_j)v_j\| \\ & = \sum|b_j-a_j|\|v_j\| \end{align}
  4. So if letting $\sum |b_j-a_j|\|v_j\|=\delta$, the above shows that there exists $\delta\geq 0$ such that if $\|v-u\|\leq \delta$, then $|f(v)-f(u)|\leq \epsilon$.

Is my proof of using $\delta-\epsilon$ correct? Where should I say something more?

  • 1
    $\begingroup$ $\|\sum (b_j-a_j)v_j\|=\Vert v-u\Vert$, so you are done midway through step 3. as soon as you set $\delta=\epsilon$. $\endgroup$ Jun 15, 2016 at 1:50

1 Answer 1


This is one of those cases where maybe it's better --and more instructive--to prove the general case first. Then, adapting the proof to your specific function is easy.

So, suppose $v,w\in \mathbb F^{n}$. Then,

$w=v+(w-v)$ so $\Vert w\Vert \leq \Vert v\Vert +\Vert w-v\Vert \Rightarrow\Vert w\Vert-\Vert v\Vert\leq \Vert w-v\Vert$.

Interchanging the roles of $v$ and $w$ gives $\Vert v\Vert-\Vert w\Vert\leq \Vert v-w\Vert=\Vert w-v\Vert$

so in fact

$\left | \Vert w\Vert-\Vert v\Vert\ \right |\leq \Vert w-v\Vert$, which proves that $\Vert \cdot\Vert $ is continuous (with $\delta =\epsilon)$.


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