Let $V$ be a finite-dimensional real vector space with its Euclidean topology. Then all norms on $V$ are equivalent and consequently given two norms $\lVert-\rVert$, $\lVert-\rVert'$, the number $$ d = d(\lVert-\rVert, \lVert-\rVert') := \sup_{0 \neq v \in V}\big| \log\lVert v\rVert - \log\lVert v\rVert'\big| $$ is finite. This equips the set of all norms on $V$ with the structure of a complete metric space. You can think of $C := \exp(d)$ as the smallest real number $\ge 1$ such that $1/C \cdot \lVert-\rVert' \le \lVert-\rVert \le C \cdot \lVert-\rVert'$.

Now let $\{\lVert-\rVert_t\}_{t \in \mathbf{R}}$ be a family of norms on $V$, parametrized by the real numbers (I would also be interested in more general parameter spaces). Let us say that this is a continuously varying family of norms on $V$ if the map $$ \mathbf{R} \times V \to \mathbf{R}_{\ge 0}, \qquad (t, v) \mapsto \lVert v\rVert_t $$ is continuous.

Question: Let $\{\lVert-\rVert_t\}_{t \in \mathbf{R}}$, $\{\lVert-\rVert'_t\}_{t \in \mathbf{R}}$ be two continuously varying families of norms on $V$. Will the function $$ \mathbf{R} \to \mathbf{R}_{\ge 0}, \qquad t \mapsto d(\lVert-\rVert_t, \lVert-\rVert'_t) $$ be continuous?

  • $\begingroup$ Of course: by compactness (Ascoli), the topologies given respectively by $d$ and the pointwise convergence coincide, so $t\mapsto \| \cdot \|_t$ is continuous for $d$. This question is better suited for math.stackexchange.com. $\endgroup$ Jun 14, 2021 at 7:02
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    $\begingroup$ @MikaeldelaSalle Since the question got some upvotes, people seem to enjoy it, so I would rather not delete it. If you make your comment into an answer, I will be happy to accept it! $\endgroup$ Jun 14, 2021 at 7:10
  • $\begingroup$ Ok, I will add an answer. This is something that is very standard in functional analysis, but I can imagine that it is not so clear when you are an outsider. Sorry about that. I hope you did not find my comment offensive, this was certainly not my aim. $\endgroup$ Jun 14, 2021 at 7:21

1 Answer 1


I expand my comment where I claim that, on the space (call it $N(V)$) of all norms on $V$, the smallest topology making continuous the evaluations $\|\cdot\| \mapsto \|v\|$ (for $v \in V$) coincides with the topology defined by the distance $d$.

This clearly implies that, under your hypothesis, the maps $t \mapsto \|\cdot\|_t$ and $t \mapsto \|\cdot\|'_t$ are continuous for the distance $d$, and in particular your question has a positive answer. This holds more generally if $T$ is an arbitrary topological space, $\{\|\cdot\|_t\}_{t \in T}$ is a family indexed by $T$ of norms on $V$, and if for every $v \in V$, $t\mapsto \|v\|_t$ is continuous.

Let me justify the initial claim. A first observation is that, by homogeneity, if $B$ denotes the closed unit ball for a fixed norm on $V$, the topology given by $d$ coincides with the topology of uniform convergence on $B$. Moreover, $V$ being finite dimensional, $B$ is compact and therefore the topology given by $d$ coincides with the topology of uniform convergence on compact subsets of $V$.

On the other hand let $v_1,\dots,v_n$ be a basis for $V$. Observe that, for an arbitrary $v=\sum_i t_i v_i$, $v'=\sum_i t'_i v_i$ and a norm $\|\cdot\|$ on $V$, we have $$ \left|\|v\| - \|v'\| \right| \leq \|v-v'\| \leq \sum_i |t_i-t'_i| \|v_i\|.$$ This implies that, for every constant $C$, $\{ \|\cdot\| \in N(V) | \forall i\leq n, \|v_i\| < C\}$ is made of equicontinuous functions on $V$. By the the Arzelà-Ascoli Theorem, in restriction to this set, the uniform convergence on compact subsets of $V$ coincides with pointwise convergence.

This proves the claim because the sets $\{ \|\cdot\| \in N(V) | \forall i\leq n, \|v_i\| < C\}$ form a exhaustion of $N(V)$ by open sets.

  • $\begingroup$ Dear Mikael, thank you very much for the detailed explanation! $\endgroup$ Jun 14, 2021 at 9:35

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