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Let us say that two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a real vector space $V$ are strongly equivalent if there exists a constant $\lambda \geq 1$ such that $$ \frac{1}{\lambda} \left( \|x\|_1 + \|y\|_1 - \|x + y\|_1 \right) \leq \|x\|_2 + \|y\|_2 - \|x + y\|_2 \leq \lambda \left( \|x\|_1 + \|y\|_1 - \|x + y\|_1 \right) $$ for all vectors $x$ and $y$ in $V$.

A remark I owe to Suvrit is that if we take $y = -x/2$, this condition implies that $$ \frac{1}{\lambda} \|x\|_1 \leq \|x\|_2 \leq \lambda \|x\|_1 $$ so that the norms are equivalent in the usual sense.

Geometrically, two norms are strongly equivalent if the defect in the triangular inequality for any one of the norms is controlled by the defect of the other. In particular, both normed spaces must have exactly the same geodesics. Thus, for example, the $\ell_\infty$ and the $\ell_2$ norms on the plane are not strongly equivalent.

Question. Is there a simple criterion to determine whether two norms on the plane (or in a finite-dimensional space) are strongly equivalent. What if we assume that the unit spheres of both norms are polygons or that the unit spheres are smooth and positively curved?

Added 07/04/2014. Thanks in good measure to the exchange with Willie Wong, the answer to the OP in the two-dimensional case is the following simple (albeit surprising) result:

Proposition. Two norms $N_1$ and $N_2$ on the plane are strongly equivalent if and only if their distributional Laplacians, considered as measures on the unit circle, are comparable in the sense that there exists a constant $\lambda \geq 1$ such that $\lambda \Delta N_1 - \Delta N_2$ and $\Delta N_2 - \lambda^{-1} \Delta N_1$ are (positive) measures.

This contains basically all the geometric information one may wish (see my answer below).

Added on 08/04/2014. The equivalence relation defined in this question simply says that the two norms are "linked" or in the same Thompson component in the cone of seminorms on the vector space $V$. The question can be then rephrased as :

Describe those Thompson components that contain norms in the cone of seminorms on $V$.

This sounds classical and there is probably a solution somewhere.

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sounds like modulus of convexity might be a good keyword... –  Suvrit Apr 3 at 17:03
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btw the second condition implies the first. –  Suvrit Apr 3 at 21:20
    
@Suvrit: Thanks. I'll modify the question to reflect this last comment. –  alvarezpaiva Apr 4 at 7:27
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When the unit sphere is a polygon (or a polyhedron), in the convex cone attached to each of the faces the norm is a linear function. In particular, for two points inside the same convex cone attached to the faces, we have that $$ \|x\| + \|y\| - \|x + y\| = 0 $$ Thus if norm 1 has polygonal unit sphere and norm 2 is strongly equivalent, then the restriction of norm 2 to the convex cones of the faces of norm 1 are also linear. This immediately implies that the unit sphere of norm 2 is also a polygon (or a polyhedron) with the same (in the projective sense) vertices. –  Willie Wong Apr 7 at 7:36
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Sorry, I meant that for each vertex $v$ of the unit sphere of norm 1, there exists (exactly one) vertex $w$ of the unit sphere of norm 2, such that $v = k w$ for some $k > 0$. (Projective over $\mathbb{R}_+$...) –  Willie Wong Apr 7 at 8:12

2 Answers 2

The simple example of strongly equivalent metrics in a comment of Willie Wong actually holds the key to what is going on:

Definition. Two generalized functions $\mu_1$ and $\mu_2$ on the unit sphere $S^n \subset \mathbb{R}^{n+1}$ will be called comparable if there exists a constant $\lambda \geq 1$ such that both $$ \lambda \mu_1 - \mu_2 \; \; \hbox{and} \; \; \mu_2 - \frac{1}{\lambda}\mu_1 $$ are positive distributions (and hence measures).

Definition. A norm $\|\cdot\|$ on $\mathbb{R}^{n+1}$ is generated by a generalized function $\mu$ on $S^n$ if $$ \|v\| = \langle \mu \, | \, \xi \mapsto |\xi \cdot v| \rangle $$ for every vector $v \in \mathbb{R}^{n+1}$ (see this MO question for more on this topic).

The following simple result generalizes Willie's example:

Proposition. Comparable generalized functions generate strongly equivalent norms.

Proof. Let $\mu_1$ and $\mu_2$ be two generalized functions generating the norms $\|\cdot\|_1$ and $\|\cdot\|_2$, respectively. Asumme the generalized functions are comparable and let $\lambda \geq 1$ be such that $$ \lambda \mu_1 - \mu_2 \; \; \hbox{and} \; \; \mu_2 - \frac{1}{\lambda}\mu_1 $$ are measures. Note that, as a consequence, $$ q(v) := \langle \lambda \mu_1 - \mu_2 \, | \, \xi \mapsto |\xi \cdot v| \rangle \; \; \hbox{and} \; \; p(v) := \langle \mu_2 - \frac{1}{\lambda}\mu_1 \, | \, \xi \mapsto |\xi \cdot v| \rangle $$ are seminorms. Writing down the condition for strong equivalence we see that this immediately yields the strong equivalence of the norms. Q.E.D.

I'm guessing that if things are set up correctly, there is a converse to this proposition that completely settles the OP. Let me just add that in two dimensions we can restrict to working with measures (more complicated generalized functions are not needed to generate norms). In this case, the complete solution to the OP is as follows:

Proposition. Two norms on the plane are strongly equivalent if and only if their distributional Laplacians, considered as measures on the unit circle, are comparable.

Proof. Every seminorm $p$ on the plane has a unique expression as the cosine transform of an even measure $\mu$ on the unit circle: $$ p(v) = \int_{\xi \in S^1} |\xi \cdot v| \, d\mu(\xi) . $$ Differentiating $p$ as a generalized function, we see that the mesure $\mu$ is just one-fourth of the Laplacian of $p$ (here we use that $\mu(\xi) = \mu(-\xi)$).

What I realized reading Willie's example is that two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are strongly equivalent if and only if $q(v) := \lambda \|v\|_1 - \|v\|_2$ and $p(v) := \|v\|_2 - \lambda^{-1}\|v\|_1$ are seminorms for some $\lambda \geq 1$.

On the other hand, if $\mu_1$ and $\mu_2$ are the distributional Laplacians of the norms $\|\cdot\|_1$ and $\|\cdot\|_2$, respectively, the functions $q$ and $p$ are the cosine transforms of the possibly signed measures $\lambda \mu_1 - \mu_2$ and $\mu_2 - \lambda^{-1}\mu_1$. Therefore, $q$ and $p$ are seminorms if and only if both $\lambda \mu_1 - \mu_2$ and $\mu_2 - \lambda^{-1}\mu_1$ are (positive) measures. Q.E.D.

Corollary. Two Minkowski norms on the plane are strongly equivalent.

Proof. Minkowski norms arre characterized by the fact that their Laplacians are smooth, strictly positive functions on the circle. Any two such functions are comparable.

Corollary. Two polygonal norms are strongly equivalent if every vertex in the dual unit disc of one norm is a positive multiple of a vertex in the dual unit disc of the other norm.

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My comment above has the following generalisation to the smooth case. (I really want this to be another comment, but it is a bit long.)

Supposing that the unit sphere is smooth. Then we can Taylor expand the norm near any given point $x_0$ as

$$ \|x_0 + x\| = \|x_0\| + A_1 x + A_2(x,x) + \ldots $$

By scaling homogeneity we have

$$ \|x_0 + (x_0 + x)\| = 2 \|x_0 + \frac12 x\| = 2 \|x_0\| + A_1 x + \frac12 A_2(x,x) + \ldots $$

And so

$$ \|x_0\| + \|x_0 + x\| - \|x_0 + x_0 + x\| = \frac12 A_2(x,x) + \frac34 A_3(x,x,x) + \frac78 A_4(x,x,x,x) + \ldots $$


Now let $A_*$ and $B_*$ be the Taylor coefficients for the two norms in question at the point $x_0$. We see immediately a necessary condition for strong equivalence is that the null space of $A_2$ and $B_2$ (as quadratic forms) agree (note that $A_2(x_0,x_0) = B_2(x_0,x_0) = 0$ by scaling homogeneity), and that restricted to this null space, the null space of $A_3$ and $B_3$ agree, and restricted to this null space $A_4, B_4$ etc. (I am assuming that $A_2, B_2$ are positive semidefinite, and $A_3,B_3$ positive semidefinite on the respective null spaces of $A_2,B_2$ etc, to guarantee that the unit sphere is convex.)

The case in my comment has all the higher $A_*$ vanishing identically (as long as $x_0$ is not a "vertex"), which requires that the higher $B_*$ also vanishing identically.

Applying this to $\ell_p$ for $p \in (1,\infty)$ at the point $(1,0,\ldots,0)$ we get that different $\ell_p$ are not strongly equivalent.


If I am not mistaken, using that

  1. the unit sphere is compact, and
  2. from convexity of the unit sphere if $x,y$ satisfies $$ \|x\| + \|y\| - \|x + y\| = 0 $$ then the norm is linear on the line segment joining $x$ and $y$,

we can probably upgrade the above necessarily condition to a sufficient condition; but the epsilons and deltas are a bit annoying to check at the moment.

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Thanks again! This looks good. I'll think about it. –  alvarezpaiva Apr 7 at 9:02

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