# Continuously varying norms

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?

• 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. Jun 14, 2021 at 7:02
• @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! Jun 14, 2021 at 7:10
• 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. Jun 14, 2021 at 7:21

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.