0
$\begingroup$

What is the meaning of saying "Norm is continuous"?Under what conditions the norm is continuous?& Under what conditions the norm is Discontinuous?

$\endgroup$
  • 1
    $\begingroup$ Any norm is continuous because they are Lipschitz. A norm never is discontinuous. $\endgroup$ – Masacroso Apr 25 '17 at 18:56
  • $\begingroup$ @Masacroso:Thanks!!Actually this question arised from user66372's second comment in math.stackexchange.com/questions/328479/… as he putting light on the "continuity of norm" $\endgroup$ – P.Styles Apr 25 '17 at 19:01
1
$\begingroup$

Let $(V,\|\cdot\|)$ be a normed vector space. The norm is continuous as a mapping $\|\cdot\| : V \to \mathbb{R}$ of normed vector spaces $(V, \|\cdot\|)$ and $(\mathbb{R},|\cdot|)$.

In fact, the triangle inequality implies that the inequality $$ \big|\|x\| - \|y\|\big| \leq \|x - y\| $$ holds for any $x,y\in V$.

Let $U \subset \mathbb{R}$ be non-empty open set and take $x_0 \in V$ such that $\|x_0\| \in U$. Then there is some $\varepsilon > 0$ so that $B_{\|x_0\|}(\varepsilon) \subset U$. If you now take any $y \in B_{x_0}(\varepsilon) \subset V$, then the inequality above implies that $\|y\| \in B_{x_0}(\varepsilon) \subset U$. In other words, with every $x_0$ in the pre-image of $U$ there is whole neighborhood of $x_0$ in that pre-image. This pre-image is an open set. This verifies the continuity of $\|\cdot\|$.

$\endgroup$
2
$\begingroup$

It means that given an $\epsilon>0$ arbitrary, there is some $\delta>0$ with $||x-y||<\delta$, implying $$ |\;||x||-||y||\;|<\epsilon $$ where the regular absolute value on the reals is used in the final inequality.

This is an easy application of the reverse triangle inequality.

$\endgroup$
0
$\begingroup$

We says that a function $f:A\to B$ is Lipschitz when

$$\frac{d_B(f(x),f(y))}{d_A(x,y)}\le K$$

for any $x,y\in A$ for some fixed $K>0$. Now, cause the reversed triangle inequality, for any norm $\|{\cdot}\|:H\to\Bbb R$ we have that

$$\big|\|x\|-\|y\|\big|\le \|x-y\|\implies \frac{\big|\|x\|-\|y\|\big|}{\|x-y\|}\le 1,\quad x,y\in H$$

And because the Lipschitz condition imply continuity then a norm is continuous.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.