# “Norm is continuous” means?

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

• Any norm is continuous because they are Lipschitz. A norm never is discontinuous. – Masacroso Apr 25 '17 at 18:56
• @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" – P.Styles Apr 25 '17 at 19:01

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 $\|\cdot\|$.

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.

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.