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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!

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In general, in a metric space $(X,d)$, the metric $d: X \times X \to \mathbb R$ is (jointly) continuous, i.e., it is continuous w.r.t. the product topology on $X \times X$ (where the topology on $X$ is induced by the metric). Proof uses just the triangle inequality. –  Srivatsan Jan 5 '12 at 21:05
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Yes the norm is a continuous function because the triangle inequality. It's also yes for the second question. Keep in mind that if you add open sets to a given topology (the new topology is said to be finer, and the original is coarser), functions from that space to another topological space that are continuous for the original topology will stay continuous for the new one. For example, if you consider the discrete topology (every subset is open) then the norm is still continuous. However, the topology induced by the norm is the coarsest that make the norm continuous. –  Joel Cohen Jan 5 '12 at 21:15

2 Answers 2

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!

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Though your answer is correct, I think it is a more interesting question whether the norm can be continuous w.r.t. coarser topology, not finer. :=) –  Srivatsan Jan 5 '12 at 21:25
    
@Srivatsan I would agree. It's not hard to see that any topology making the norm continuous must include all balls centered at the origin (since they're the inverse image of $(-\epsilon,\epsilon)$), but I'm not sure what more can be said. Maybe a courser topology would work...hmmm...not sure. –  Bill Cook Jan 5 '12 at 21:45
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+1. You're almost done: By definition, the topology is generated by the basis $\{ U(a,b) \}$ where $U(a,b)$ is the set of all vectors $v$ such that $a \leqslant \| v \| \leqslant b$. This treats all vectors of a given norm the same, and hence seems far from Hausdorff. Certainly doesn't appear to be a very interesting topology. :-) –  Srivatsan Jan 5 '12 at 21:55
    
@Srivatsan Thanks! Uh, yeah, that doesn't seem to be a very nice topology. :( –  Bill Cook Jan 5 '12 at 21:58
    
Thanks! I wonder if those topologies that can make a norm/inner product continuous are exactly those topologies that can make each element in the continuous dual space continuous? –  Tim Jan 5 '12 at 22:44

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.


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.

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+1 Really nice for pointing out more insights. –  Tim Jan 5 '12 at 22:39
    
@Tim Not just the norm, but also the addition map $+ : X \times X \to X$ and scalar multiplication map $\bullet : \mathbb R \times X$ are also continuous in the respective topologies. These statements also require only the elementary properties of norms. –  Srivatsan Jan 5 '12 at 23:13
    
Thanks! Do you mean that any topology that can make a norm continuous will also make addition and scalar product continuous? –  Tim Jan 5 '12 at 23:18
    
@Tim No, that is not true. In fact, as t.b. pointed out in chat, the weird topology described by me and Bill under Bill's answer, that should not have this property (although I haven't checked the details myself). What I mean is that if you take the respective "standard" topologies, then the addition and scalar multiplication maps are continuous. In fact, you should think of this as saying that the norm is defined in a way to be "compatible" with these operations. –  Srivatsan Jan 5 '12 at 23:21

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