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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.

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.

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. If 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 $\mu_1$ 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.

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.

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).

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

Proof. Every Minkowski norm $\|\cdot\|$ on the plane can be written uniquely as the cosine transform of a smooth, positive function $f : S^1 \rightarrow \mathbb{R}$ that is invariant under the antipodal map ($f(\theta) = f(-\theta)$): $$ \|(v_1,v_2)\| = \int_0^{2\pi} |\cos(\theta)v_1 + \sin(\theta)v_2|f(\theta) \, d\theta. $$ It is then clear that two Minkowski norms are generated by comparable generalized functions and are hence strongly equivalent.

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. If 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 $\mu_1$ 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.