Is a norm a continuous function?

Question

Is a norm on a set a continuous function with respect to the topology induced by the norm?

Is a topology on the set that can make the norm continuous (i.e. the topology that is compatible with the norm) not unique? Is it a superset of the unique topology induced by the norm?

I am asking this question, because I heard (I am also not sure if it is correct) that a topology that can make an inner product continuous is not unique (such a topology is called weak topology on the inner product space?), and is a superset of the topology induced by the inner product.

Thanks and regards! Pointers to some references are appreciated!

Answer 1

Yes. The norm is continuous.

Consider a normed space $X$ with norm $\| \cdot \|:X \to \mathbb{R}_{\geq 0}$. Let $O$ be an open subset of $\mathbb{R}$ and let $P = \{ x \in X \;|\; \|x\| \in O \}$ (the inverse image of $O$). Suppose $x_0 \in P$. We then have that $c_0=\|x_0\| \in O$ (since $x_0$ is in the inverse image of $O$). Next, since $O$ is open, there exists some $\epsilon >0$ such that $(c_0-\epsilon,c_0+\epsilon) \subseteq O$. Thus if $y \in X$ and $c_0-\epsilon < \| y \| < c_0+\epsilon$ (that is $\|y\|\in(c_0-\epsilon,c_0+\epsilon)\subseteq O$), then $y \in P$. So consider $y \in B_\epsilon(x_0) = \{ z \in X \;|\; \|z-x_0\|<\epsilon \}$ (the open ball of radius $\epsilon$ centered at $x_0$). We have $|\|y\|-\|x_0\|| \leq \|y-x_0\| < \epsilon$ thus $|\|y\|-c_0|<\epsilon$. Hence $\|y\| \in (c_0-\epsilon,c_0+\epsilon)$. Thus $\|y\| \in O$ so $y \in P$. Therefore, $B_{\epsilon}(x_0) \subseteq P$ so $P$ is open. Thus the norm is continuous.

By the way, the proofs that a metric is a continuous map when using the metric induced topology and that a inner-product is a continuous map when using its corresponding topology are essentially the same. :)

For your other question: If you choose any topology $\mathcal{T}$ on $X$ such that $B_\epsilon(x_0) \in \mathcal{T}$ for all $x_0\in X$ and $\epsilon>0$, the norm will still be continuous. In other words, any finer topology will still make the norm continuous. So "No" the topology coming from the norm is not necessarily the only one which makes the norm continuous. For example, consider $\mathcal{T}=\mathcal{P}(X)$ (the powerset of $X$). This (discrete) topology makes every map continuous!

Answer 2

I find it cleanest to think in terms of Lipschitzness, rather than the $\varepsilon$-$\delta$ definition of continuity.

Let $(X, d)$ be a metric space.

1. For any $a \in X$, the map $d (a, \cdot) : X \to \mathbb R$ is Lipschitz, and hence is continuous.

2. The map $d : X \times X \to \mathbb R$ is Lipschitz w.r.t. the product metric, and hence is continuous.

I will prove (2.) and leave (1.) as a simpler exercise. (In fact, (2.) also directly implies (1.).) Fix $(x, y), (z, w) \in X \times X$. Then $$ \begin{array}{rll} |d(x,y) - d(z,w)| &\leqslant |d(x,y) - d(z,y)| + |d(z,y) - d(z,w)| & \\ &\leqslant d(x, z) + d(y, w) & \text{(triangle inequality on } d \text{)} \\ &\leqslant 2 \max \{ d(x, z) , d(y, w) \} & \\ &= 2 d_{\infty} ((x, y) , (z, w)), & \end{array} $$ showing that $d$ is Lipschitz. (Here we have assumed the “max” metric on the product; certainly other choices are possible but they turn out to be equivalent.)

Coming to the OP's question, to prove that a norm on a linear space is continuous w.r.t. itself, notice that the linear space is also a metric space with $d(x,y) = \| x - y \|$. In other words, $\| x \|$ is just the distance of $x$ from the origin; hence applying item (1.) above (with $a = 0$) shows the continuity of the norm.

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A point to ponder: Equivalent norms. The above discussion might suggest that continuity of a norm is not an interesting thing to study. However this question can be modified a little to give rise to a quite fruitful concept:

Given two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ on a linear space $X$, are they continuous with respect to each other? I.e., when $\| \cdot \|_1$ is viewed as a function, is it continuous w.r.t. the norm $\| \cdot \|_2$ and vice versa?

It can be shown that this is true if and only if there exist numbers $0 < \alpha \leqslant \beta < \infty$ such that $$ \alpha \| v \|_1 \leqslant \| v \|_2 \leqslant \beta \| v \|_1 $$ for all vectors $v$. In this case, we say that the two norms are equivalent.