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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?

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    $\begingroup$ Do you definitely want norms, or would you settle for metrics? $\endgroup$ – Brian M. Scott Jun 11 '20 at 18:47
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    $\begingroup$ 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. $\endgroup$ – Markus Zetto Jun 11 '20 at 18:51
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    $\begingroup$ @BrianM.Scott Norm $\endgroup$ – SolutionExists Jun 11 '20 at 18:57
  • $\begingroup$ Try functions that are analytic on the closed unit disk. $\endgroup$ – Charlie Frohman Jun 11 '20 at 19:39
  • $\begingroup$ @CharlieFrohman which norms though $\endgroup$ – SolutionExists Jun 11 '20 at 20:01
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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.

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