# Topologies and completeness

Is it possible to have two norms on the same space, the topology of one being strictly finer than the other, yet the space is complete in both norms?

• Do you definitely want norms, or would you settle for metrics? – Brian M. Scott Jun 11 '20 at 18:47
• You usually define norms for vector spaces, so I'll stick to the case of real/ complex VS. It is not possible for finite-dimensional ones, as every norm there defines the same topology, but I think it should work on infinite dimensional vector spaces; I'm not quite fit enough in functional analysis to give an example though. – Markus Zetto Jun 11 '20 at 18:51
• @BrianM.Scott Norm – SolutionExists Jun 11 '20 at 18:57
• Try functions that are analytic on the closed unit disk. – Charlie Frohman Jun 11 '20 at 19:39
• @CharlieFrohman which norms though – SolutionExists Jun 11 '20 at 20:01

## 1 Answer

This is not possible. Fix a space $$X$$ with two complete norms $$\|\cdot\|_1, \|\cdot\|_2$$ such that the topology induced by $$\|\cdot\|_1$$ is finer than the one induced by $$\|\cdot\|_2$$. Let $$B_i$$ denote the unit ball for $$\|\cdot\|_i$$.

In particular, $$B_2$$ is also open for $$\|\cdot\|_1$$ and so there is $$\varepsilon > 0$$ such that $$\varepsilon B_1 \subseteq B_2$$. Hence $$\frac{\varepsilon}{2} S_1 \subseteq \varepsilon B_1 \subseteq B_2$$ where $$S_1 = \{x: \|x\|_1 = 1\}$$. This means that for every $$x \in X$$, we have that $$\left \| \frac{\varepsilon x}{2 \|x\|_1} \right \|_2 \leq 1$$ which implies that $$\|x\|_2 \leq \frac{2}{\varepsilon} \|x\|_1.$$ That is, the identity map from $$(X, \|\cdot\|_1)$$ to $$(X, \| \cdot \|_2)$$ is continuous. Hence by the open mapping theorem, the identity map from $$(X, \| \cdot \|_2)$$ to $$(X, \|\cdot\|_1)$$ is continuous also which means that $$\|x\|_1 \leq C \|x\|_2$$ for some constant $$C$$. Hence the two norms are equivalent and so induce the same topology.