Showing that norm is continuous

Question

Consider a normed vector space $(X,\|\cdot\|).$ Show that $\|\cdot\| : X \to \mathbb{R}$ is continuous when $X$ is given the topology generated by the metric $d(x,y) = \|x−y\|.$

To show it is continuous I tried taking a basis element of $\mathbb{R}$ as $(a,b)$ and showing that the set of points such that $\|x-y\|$ lies in $(a,b)$ is an open set. But I'm really confused with how to go about it. It would be really helpful if someone could tell me how to proceed. Another extension of this question is showing the scalar product is also continuous.

Answer 1

It's easier to approach this problem by using the fact if $X, Y$ are metric spaces then $f: X \rightarrow Y$ is continuous if and only if for each convergent sequence $x_n \rightarrow x$ one has $f(x_n) \rightarrow f(x)$. The continuity of the norm follows almost immediately from this.

But if you want to approach with your method, i.e. show $S = || \cdot ||^{-1}((a, b))$ is open in $X$, just recall the topology generated by a metric $d$ consists of open $d$-balls, i.e. $B_x(r) = \{y \in X \mid d(x, y) < r\}$. Let $x \in S$. We want to find $r > 0$ such that $B_x(r) \subseteq S$. (That is, there is an open neighbourhood of $x$ contained in $S$.) Note $r$ depends on how far $||x||$ is to $a$ and $b$; it might help to think of a simple case such as $X = \mathbb R^2$ to get a feel of what $r$ should be.

Once you've chosen $r$, to prove $B_x(r) \subseteq S$ we want to show if $y \in B_x(r)$ then $||y|| \in (a, b)$, and you can use the triangle inequality to establish the upper bound and the reverse triangle inequality to show the lower bound. It might help that $\min(a, b) \leq a$ and $\min(a, b) \leq b$.

Answer 2

First a word on notation, for any set $W$ let $2^W$ denote the powerset. Let $B(x,r)$ refer to the open ball of radius $r$ centered at $x$ . An open map is a map that sends open sets to open sets. Let $\langle \xi \rangle$ denote the topology generated by the basis $\xi$ . Let the function $j(a, l, b)$ denote the smaller of $|a-l|$ and $|l-b|$ (40). We will use the function $j$ to compute the maximum allowable radius we can given an epsilon ball in $X$ .

$$ j(a, l, b) \stackrel{\text{def}}{=\!=} \min(|a-l|, |l-b|) \tag{40} $$

Let $ F:2^R \to 2^X $ be defined as follows (50)

$$ u \in F(L) \iff \|u\| \in L \tag{50} $$

$\|\cdot\|$ is continuous if and only if $F$ is an open map.

Using the premise of the question, we are allowed to assume that $X$ is the topology associated with the distance function $d(x,y)=\|x−y\|$ .

Let $\tau$ be the set of epsilon balls with all possible centers in $X$ and all positive radii (60).

$$ \tau \stackrel{\text{def}}{=\!=} \{ B(x, \epsilon) : x \in X \land \varepsilon \in \mathbb{R}_{> 0} \} \tag{60} $$

$\tau$ is a basis of the metric topology $\langle \tau \rangle$ .

I will now show that $F$ is an open map.

$F$ of the empty set is the empty set, and the empty set is open by definition.

$$ F(\{\})=\{\} \tag{101} $$

Let $L$ be an epsilon ball in the standard topology on $\mathbb{R}$ .

Therefore, $L$ is an open interval $(a, b)$ . $F(L)$ can be determined by considering each possible $x$ such that $\|x\| \in (a,b)$ and constructing a small epsilon ball centered at $x$ such that every element in that epsilon ball has a norm that's still inside $(a, b)$ (102).

$$ F((a,b)) = \cup \{ B(x, j(a, \|x\|, b)) : x \in X \land \|x\| \in (a, b) \} \tag{102} $$

The epsilon balls on the standard toplogy of $\mathbb{R}$ form a subbasis. Meaning that any open set in $\mathbb{R}$ is expressible as a union of epsilon balls.

Suppose $L$ is an arbitary open set in $\mathbb{R}$ but isn't an epsilon ball and is not empty, it is a union of open balls, U (103). $$ L = \cup U \;\; \text{and} \;\; \forall u \in U \mathop. u \; \text{is an epsilon ball in $\mathbb{R}$} \tag{103} $$

Because $F$ is defined as a union of inverse images of $\|\cdot\|$ it satisfies the following law with respect to unions (104).

$$ F(\cup V) = \cup \{ F(v) : v \in V \} \tag{104} $$

By (103) and (104), $F$ applied to an arbitrary open set of $\mathbb{R}$ is open.

Therefore F is an open map, therefore $\|\cdot\|$is continuous as desired.

Answer 3

To follow your own approach, fix $x,a,b$ and consider $U=\{y: \|y-x\|\in (a,b)\}$. Let $y_0 \in U$ so that $\|y_0-x\|\in (a,b)$. Since $(a,b)$ is open there exists $\epsilon >0$ such that $|t-\|y_0-x\|| <\epsilon $ implies $t \in (a,b)$ [You can even write down such an $\epsilon$ explicitly]. Now $\|y-y_0\| <\epsilon$ implies $|\|y-x\|-\|y_0-x\| |<\epsilon$ so taking $t=|\|y-x\|$ we see that $|\|y-x\| \in (a,b)$ so $y\in U$.

EDIT: To show that the norm is continuous it is enough to take $x=0$ in this argument.

Answer 4

We have $\lVert x \rVert = \lVert x - y + y \rVert \le \lVert x - y \rVert + \lVert y \rVert$, thus $\lVert x \rVert - \lVert y \rVert \le \lVert x - y \rVert$ for all $x,y$. Hence also $-(\lVert x \rVert - \lVert y \rVert) = \lVert y \rVert - \lVert x \rVert \le \lVert y - x \rVert = \lVert -(x - y) \rVert = \lVert x - y \rVert$ for all $x,y$. This shows $$\lvert \lVert x \rVert - \lVert y \rVert \rvert = \max( \lVert x \rVert - \lVert y \rVert, -(\lVert x \rVert - \lVert y \rVert)) \le \lVert x - y \rVert = d(x,y) .$$ This trivially implies that $\lVert - \rVert : X \to \mathbb R$ is (uniformly) continuous.