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Background: In the paper "Lusternik-Schnirelman Theory on Banach Manifolds" by Richard Palais, the author defines a Finsler structure (Definition 2.1) in a general way, where roughly speaking, given a topological space $\mathcal{B}$ and a vector space $V$, one assigns a norm $\|\cdot\|_b$ on the space $V$ at each point $b \in \mathcal{B}$, subject to the following continuity condition. At each $b$, and for any $K>1$ there is a neighborhood $U$ of $b$ such that the equivalence $$\frac{1}{K}\|\cdot\|_{b_0} \leq \|\cdot\|_b \leq K\|\cdot\|_{b_0} $$ holds for all $b_0 \in U$. I am wondering if in the finite dimensional case, this property holds true as soon as we assume that the map $(x,u) \mapsto \|u\|_x$ is continuous for the product topology on $\mathcal{B} \times \mathbb{R}^{n}$.

Formally, Let $\mathcal{B}$ be $\mathbb{R}^k$ and let $\|\cdot\|_{b}$ be an assignment of a norm on $\mathbb{R}^{n}$ to each point of $\mathcal{B}$, in such a way that the map $(b,u) \mapsto \|u\|_{b}$ is a continuous function from $\mathcal{B} \times \mathbb{R}^{n} \to \mathbb{R}$. Then for any $K>1$ is it always possible to find a neighbrohood $U$ of $b$ for which there is a constant $K$ such that $\frac{1}{K}\|u\|_a \leq \|u\|_{b_0} \leq K\|u\|_b$ for all $(b,u) \in U \times \mathbb{R}^{n}$?

Thanks!

ttb
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    The notation suffers from using "B" in many ways: as the norm identifier, as the parameter, and as the topological space $\mathcal{B}$. Also I would make it clear one way or the other that $\mathcal{B}$ is assumed finite dimentional throughout. – hardmath Aug 01 '17 at 18:33
  • Thanks. I edited it to say that $\mathcal{B}$ is finite dimensional. The norm is determined by the parameter B, thats why I put in the subscript. I'm open to a better notation if you can suggest one. – ttb Aug 01 '17 at 18:42
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    I don't have a better idea for notation; sometimes the reuse of a symbol is felicitous. However I'd feel better about it here if there were more of a motivation for identifying (constructing?) a norm depending on "point" $B\in \mathcal{B}$ – hardmath Aug 01 '17 at 18:48
  • Dear @hardmath I have refined the question and added more background. – ttb Aug 02 '17 at 02:29
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    Thanks. I think your approach works, taking the infimum over the compact set to get a lower bound just as the supremum gives an upper bound. – hardmath Aug 02 '17 at 07:31

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