Denote by $\mathcal{K}^{n}$ the family of all convex bodies (i.e. the convex sets with nonempty interior) in the Euclidean space $\mathbb{R}^n$. Other notation are referred to [15].

Denote by $\mathcal Aff(\mathbb{R}^n)(\mathrm{GL}(\mathbb{R}^n))$ the family of all affine (linear) maps from $\mathbb{R}^n$ to $\mathbb{R}^n$ and by $aff(\mathbb{R}^n)$ the family of all affine functionals on $\mathbb{R}^n$. As a rule, elements of $\mathbb{R}^n$ are denoted by lower-case letters, subsets by capitals and real numbers by small Greek letters. Given $C\in\mathcal{K}^{n}$, then by $\lambda C$ we mean the homothetic copy of $C$ of ratio $\lambda$ with the center at the origin $o$, and we write $\lambda_x C:=\lambda(C-x)+x.$

In the well-known paper [9], John proved that for every centrally symmetric convex body $C\in\mathcal{K}^{n}$ with the origin as its center, there is a unique ellipsoid $E$ (i.e. an affine image of the unit ball in $\mathbb{R}^n$) such that $E\subset C\subset \sqrt nE$, which in some sense describes the similarity between $C$ and $E$. Later on, it was realized that the John's approach provided actually a way describing the differences between convex bodies, therefore, as a consequence, several (similarly, translation or affine invariant) distances between convex bodies, such as the so-called Banach-Mazur distance etc, were introduced and studied (see [1-3, 5, 8, 10-14]). It turns out that these distances defined for convex bodies play some roles in convex geometrical analysis and other related mathematics areas (cf. [4, 6, 7]).

In this article, we discuss some well-known (affine invariant) distances which appear different. Precisely, following distances will be discussed.

Definition 1  For $K, L\in \mathcal{K}^{n}$, four (affine invariant) distances of different forms are defined as follows (see [4-6, 9]).

ⅰ) $d_1(K,L):=\inf\{\alpha\beta\;|\;\alpha>0,\beta>0,(1/\beta)L_{x}\subset uK_{z}\subset\alpha L_{x}\}$ where $ L_x$ denotes $ L-x$ and the infimum is taken over all applicable $z,x\in\mathbb{R}^n, u\in\mathrm{GL}(\mathbb{R}^n);$

ⅱ) $d_2 (K,L):=\inf\{\lambda\geq 1\;|\;L\subset \mathrm{T}K\subset\lambda_x L\};$

ⅲ) $ d_3 (K,L):=\inf\{\lambda\geq 1\;|\;\mathrm{T}L\subset K\subset\lambda_x \mathrm{T}L\};$

ⅳ) $d_4 (K,L):=\inf\{\lambda\geq 1\;|\;\mathrm{T_1}L\subset \mathrm{T_2}K\subset\lambda_x \mathrm{T_1}L\},$

where the infimum is taken over all applicable $x\in\mathbb{R}^n, \mathrm{T}, \mathrm{T_1},\mathrm{T_2}\in \mathcal Aff(\mathbb{R}^n).$

The following are some weaker version of the above distances

Definition 2 For $K, L\in \mathcal{K}^{n}$, we define (see [4-6, 9]).

ⅰ) $ \widetilde{d}_1(K,L):=\inf\{|\alpha\beta|>0\;|\;(1/\beta)L_{x}\subset uK_{z}\subset\alpha L_{x}\}$, where the infimum is taken over all applicable $z,x\in\mathbb{R}^n, u\in\mathrm{GL}(\mathbb{R}^n);$

ⅱ) $ \widetilde{d}_2 (K,L):=\inf\{|\lambda|\geq 1\;|\;L\subset \mathrm{T}K\subset\lambda_x L\};$

ⅲ) $ \widetilde{d}_3 (K,L):=\inf\{|\lambda|\geq 1\;|\;\mathrm{T}L\subset K\subset\lambda_x \mathrm{T}L\}; $

ⅳ) $\widetilde{d}_4 (K,L):=\inf\{|\lambda|\geq 1\;|\;\mathrm{T_1}L\subset \mathrm{T_2}K\subset\lambda_x \mathrm{T_1}L\}, $

where the infimum is taken over all applicable $x\in\mathbb{R}^n, \mathrm{T}, \mathrm{T_1},\mathrm{T_2}\in \mathcal Aff(\mathbb{R}^n).$

Remark  All $d_i\,'$s (resp. $\widetilde d$'s) are called (resp. absolute) Banach-Mazur distance (B-M distance for short) between $K, L$ by different authors respectively, however, as far as we know, there seems no proofs available to show that they are indeed the same.

In next section, we will show that all $d_i'$s (resp. $\widetilde d_i'$s) are indeed the same. Furthermore, we provide a sufficient condition for $K$ and $L$ under which $d_i(K,L)=\widetilde d_i(K,L)$.

The first result in this section concerns the equivalence of all $d_i$'s (resp. $\widetilde d_i$'s).

Theorem 1 For any convex bodies $K,L\in\mathcal{K}^n$, we have

ⅰ) $ d_1(K,L)=d_2 (K,L)=d_3 (K,L)=d_4 (K,L);$

ⅱ) $ \widetilde{d}_1(K,L)=\widetilde{d}_2 (K,L)=\widetilde{d}_3 (K,L)=\widetilde{d}_4 (K,L).$

Proof ⅰ) First we prove $d_1(K,L)=d_2 (K,L)$. For any $\lambda$ and affine map $\mathrm{T}=u+x^*$ and $x^*\in \mathbb{R}^n$ (where $u\in \mathrm{GL}(\mathbb{R}^n)$) with $L\subset \mathrm{T}K=uK+x^*\subset\lambda_x L=\lambda(L-x)+x,$ we have

$L-x\subset uK+x^*-x=uK_z\subset\lambda(L-x), $

where $z=u^{-1}(x-x^*)$. Thus $d_1(K,L)\leq d_2(K,L)$ (taking $\beta=1$ and $\alpha=\lambda$!).

Conversely, if $(1/\beta)L_{x}\subset uK_{z}\subset\alpha L_{x}$, i.e. $L_{x}\subset \beta uK_{z}\subset\alpha\beta L_{x}$ or

$L\subset \beta uK_{z}+x\subset\alpha\beta (L-{x})+x,$

then writing $\mathrm{T}:=\beta u-\beta u(z)+x$ and $\lambda:=\alpha\beta$, we get $L\subset \mathrm{T}K\subset\lambda_x L,$ which clearly leads to $d_2(K,L)\leq d_1(K,L).$ So $d_1(K,L)=d_2 (K,L).$

Next, we prove $d_2 (K,L)=d_4 (K,L).$ It is obvious that $d_2 (K,L)\geq d_4 (K,L).$ On the other hand, set $d_4 (K,L)=d^\ast,$ then by the definition of $d_4$, for $\forall \varepsilon>0$, there exist $\mathrm{T}_1^\ast,\mathrm{T}_2^\ast \in \mathcal Aff(\mathbb{R}^n)$ and $x\in \mathbb{R}^n$ such that

$\mathrm{T_1^\ast }L\subset \mathrm{T_2^\ast }K\subset(d^\ast+\varepsilon)_x \mathrm{T_1^\ast }L$

from which we get

$d_2(\mathrm{T}_1^\ast L,K)=\inf\{\lambda\geq 1 \;|\;\mathrm{T_1^\ast }L\subset \mathrm{T}K\subset\lambda_x \mathrm{T_1^\ast }L\}\leq d^\ast+\varepsilon.$

Thus by the affine invariant of $d_2$, we get $d_2 (K,L)=d_2(\mathrm{T}_1^\ast L,K)\leq d^\ast+\varepsilon$ which, by the arbitrariness of $\varepsilon$, leads to $d_2(K,L)\leq d^*=d_4(K,L).$ So $d_2(K,L)=d_4(K,L).$

The same argument works as well in showing $d_3 (K,L)=d_4 (K,L).$

ⅱ) The proof is similar to that for ⅰ).

Remark  ⅰ) Since all $d_i\,$'s (resp. $\widetilde d_i$'s) are equal, we denote them uniformly by $d_{BM}$ (resp. $\widetilde d_{BM}$). It may happen that $d_{BM}(K,L)>\widetilde{d}_{BM}(K,L)$ as shown by the example: in $\mathbb{R}^2$, suppose that $K$ is a regular pentagon and $L$ is a triangle, then it was shown by Lassak in [10] that $d_2(K,L)=1+\sqrt 5/2\approx 2.118 $ while it was confirmed in [4] that $\widetilde d_2(K,L)\leq 2$ for all $K,L\in \mathcal K^2.$

ⅱ) $\sup\limits_{K,L\in \mathcal K^n}\widetilde d_{BM}(K,L)=n$ was confirmed in [4], however it is still a great challenge to find $\sup\limits_{K,L\in \mathcal K^n} d_{BM}(K,L)$. A lot of efforts has been put on such an estimate, among which it is an applicable approach to find the relation between $d_{BM}$ and $\widetilde d_{BM}$. Next theorem provides a sufficient condition for $K$ and $L$ under which $d_{BM}(K,L)=\widetilde d_{BM}(K,L)$ holds.

Theorem 2 Let $K, L\in \mathcal K^n$. Then $d_{BM}(K,L)=\widetilde{d}_{BM}(K,L)$ if one of $K, L$ is centrally symmetric.

In order to prove Theorem 2, we need the following lemma.

Lemma 1 Let $K,L\in \mathcal K^n$. Then there are $\alpha, \beta\in \mathbb{R}\setminus\{0\}$, $x,z\in \mathbb{R}^n$ and $u\in \mathrm{GL}(\mathbb{R}^n)$ such that $(1/\beta)L_{x}\subset uK_{z}\subset\alpha L_x$ iff there exist $x_1,z_1\in \mathbb{R}^n$ such that

$(1/\beta)L_{x_1}\subset uK_{z_1}\subset\alpha L.$

Proof If $(1/\beta)L_{x}\subset uK_{z}\subset\alpha L_x,$ then $(1/\beta)L_{x}+\alpha x\subset uK_{z}+\alpha x\subset\alpha L,$ i.e.,

$(1/\beta)L_{(1-\alpha\beta)x}\subset uK_{z-u^{-1}(\alpha x)}\subset\alpha L.$

Now the proof is done by taking $x_1=(1-\alpha\beta)x$ and $z_1=z-u^{-1}(\alpha x).$

Conversely, suppose $(1/\beta)L_{x_1}\subset uK_{z_1}\subset\alpha L.$ If $\alpha\beta=1$, then $L_{x_1}\subset \beta uK_{z_1}\subset\alpha\beta L=L$ which implies obviously $x_1=o$. Thus, we take $x=o$ and $z=z_1$. If $\alpha\beta\not=1$, it is easy to check that $(1/\beta)L_{x}\subset uK_{z}\subset\alpha L_x, $ where $x=(1/(1-\alpha\beta))x_1$ and $z=z_1+u^{-1}((\alpha/(1-\alpha\beta)) x_1).$

Remark  By similar arguments to that for Lemma 1, we can show that

$\begin{eqnarray*}&& d_{BM}(K,L)=\inf\{\alpha\beta\;|\;\alpha>0,\beta>0,(1/\beta)L_{x}\subset uK_{z}\subset\alpha L_{y}\}, \\ && \widetilde d_{BM}(K,L)=\inf\{|\alpha\beta|>0\;|\;(1/\beta)L_{x}\subset uK_{z}\subset\alpha L_{y}\},\end{eqnarray*}$

which are actually the original definitions.

Proof of Theorem 2 Clearly we need only to show the equality for $d_1$. By definition, it is obvious that $d_1 (K,L)\geq \widetilde{d}_1(K,L)$.

Now, without loss of generality, suppose that $L$ is centrally symmetric with the origin as its center. It is a routine by a compactness argument to show that there are $\alpha^*, \beta^*\in \mathbb{R}\setminus\{0\}$, $x,z\in \mathbb{R}^n$ and $u\in \mathrm{GL}(\mathbb{R}^n)$ such that

$\begin{equation}\tag{*}(1/\beta^*)L_{x}\subset uK_{z}\subset\alpha^* L_x\text{ and } \widetilde d_1(K,L)=|\alpha^*\beta^*|.\end{equation}$

If $\alpha^*>0, \beta^*>0$, then by definition we have clearly $d_1 (K,L)\leq \alpha^*\beta^*=\widetilde{d}_1(K,L)$. If $\alpha^*<0, \beta^*<0$, then by (*) we have also

$(1/(-\beta^*))L_{x}\subset (-u)K_{z}\subset(-\alpha^*) L_x,$

which leads to $d_1 (K,L)\leq (-\alpha^*)(-\beta^*)=\widetilde{d}_1(K,L)$ as well.

If $\alpha^*<0$ and $\beta^*>0$, by Lemma 1, there exist $x_1,z_1\in \mathbb{R}^n$ such that

$(1/\beta^*)L_{x_1}\subset uK_{z_1}\subset\alpha^* L=\alpha^*(- L)=(-\alpha^*) L.$

Thus, by Lemma 1 again, we get $d_1 (K,L)\leq (-\alpha^*)\beta^*=\widetilde{d}_1(K,L)$. Hence $d_1 (K,L)=\widetilde{d}_1(K,L).$

Remark  A question related to Theorem 2 is : if $d_{BM} (K,L)=\widetilde{d}_{BM}(K,L)$ holds for all $L\in \mathcal K^n$, must $K$ be centrally symmetric?

As mentioned above, it is a long-standing open problem to get $\sup\limits_{K,L\in \mathcal K^n}d_{BM}(K,L)$. There are many different approaches to tackling with such a problem, among which, besides relating $d_{BM}$ to $\widetilde d_{BM}$, another method is to relate the B-M distance between convex bodies to that between their polar bodies (cf. [14]). In this section, we discuss the B-M distances between polar bodies.

For $K\in \mathcal K^n$ and $x\in \text{int}K$, the interior of $K$, we write

$K^x:=\{z\in \mathbb{R}^n\,|\,\langle z, y-x\rangle\leq 1 \text{ for all } y\in K\}$

called the polar set of $K$ based on $x$. In particular, if $x=o\in \text{int}K$, we use $K^*$ in stead of $K^o$. It is obvious that if $x\in \text{int}K\subset L$, then $K^x\supset L^x$. It is also easy to check that for any $x\in \text{int} K$, $o\in \text{int} K^x$; and $K^{**}=K$. Furthermore, $K$ is symmetric (with center at $o$) iff so is $K^*$.

Proposition 1 Let $K\in \mathcal{K}^n$ and $o\in {\rm int }K$. Then $(\mathrm{T}K)^*=\mathrm{T}^{-\top}K^*$ for all invertible $\mathrm{T}\in \mathrm{GL}(\mathbb{R}^n)$, where $\mathrm{T}^{-\top}=(\mathrm{T}^\top)^{-1}$ and $\mathrm{T}^\top$ denotes the transpose of $\mathrm{T}$. In particular, for $\lambda\not=0$, $(\lambda K)^*=\frac 1\lambda K^*.$

Proof By the definition of polar body,

$\begin{eqnarray*}&& (\mathrm{T}K)^*=\{x\in \mathbb{R}^n\;|\; \langle x,\mathrm{T}y\rangle\leq 1 \text{ for all } y\in K\}\\ &=&\{x\in \mathbb{R}^n\;|\; \langle \mathrm{T}^\top x,y\rangle\leq 1 \text{ for all } y\in K\}\\ &=& \{\mathrm{T}^{-\top}z\in \mathbb{R}^n\;|\; \langle z,y\rangle\leq 1 \text{ for all } y\in K\}=\mathrm{T}^{-\top}K^*.\end{eqnarray*}$

The following theorem is natural.

Theorem 3 Let $K, L\in \mathcal{K}^n$ be symmetric with the origin $o$ as their centers. Then $d_{BM}(K,L)=d_{BM}(K^*,L^*)$.

Proof Suppose $d_{BM}(K,L)=\lambda _0$. Then by the definition of $d_{BM}(\cdot,\cdot)$, for any $\varepsilon>0$, there is $\mathrm{T}_0\in \mathrm{GL}(\mathbb{R}^n)$ such that $K\subseteq \mathrm{T}_0L \subseteq (\lambda_0+\varepsilon)K.$ Thus, by the property of polar bodys and Proposition 1, we have

$\begin{eqnarray*} &&[(\lambda_0+\varepsilon)K]^* \subseteq (\mathrm{T}_0L)^* \subseteq K^*\\ &\Leftrightarrow &\frac{1}{\lambda_0+\varepsilon}K^* \subseteq \mathrm{T}_0^{-\top}L^* \subseteq K^*\\ &\Leftrightarrow & K^*\subseteq (\lambda_0+\varepsilon)\mathrm{T}_0^{-\top}L^* \subseteq (\lambda_0+\varepsilon)K^*, \end{eqnarray*}$

i.e., $K^*\subseteq \mathrm{T}_1L^* \subseteq (\lambda_0+\varepsilon)K^*$, where $\mathrm{T}_1:=(\lambda_0+\varepsilon)\mathrm{T}_0^{-\top}\in \mathrm{GL}(\mathbb{R}^n)$, which implies $d_{BM}(K^*,L^*) \leq \lambda_0+\varepsilon $. So $d_{BM}(K,L) \geq d_{BM}(K^*,L^*)$ by the arbitrariness of $\varepsilon$.

Conversely, simply by substituting $K$ with $K^*$ in the above argument, we get

$d_{BM}(K^*,L^*)\geq d_{BM}(K,L)$

(using the fact that $K^{**}=K$). Thus $d_{BM}(K,L) = d_{BM}(K^*,L^*)$.

For non-symmetric cases, the situation becomes more complicated and of course more interesting. In general, given convex bodies $K,L\in\mathcal K^n$, we don't know if there exist $x\in \text{int}K, y\in \text{int}L$ such that $d_{BM}(K,L)=d_{BM}(K^x,L^y).$ In the following, we discuss in $\mathcal K^2$ a special case only where one of $K$ and $L$ is a triangle and the other is a quadrangle (observe that the polar sets of a triangle are still triangle).

Theorem 4 Let $Q$ be a quadrangle. Then $d_{BM}(\triangle, Q)=d_{BM}(\triangle, Q^{x_0}) \text{ for some } x_0\in \mathrm{int}Q$.

To prove Theorem 4, we need the following lemmas.

Lemma 2 Let $Q$ be a quadrangle, then there exists $\bar x\in \mathrm{int}Q$ such that $Q^{\bar x}$ is a parallelogram.

Proof Let $e_i$ ($i=1,\cdots,4$) be the vertices of $Q$ (indexed anti-o'clockwise) and $\bar x$ be the intersect point of diagonals of $Q$. Then we have first

$Q^{\bar x}=\{y\;|\; \langle y, e_i-\bar x\rangle\leq 1, i=1,\cdots,4\}=:Q_1.$

In fact, $Q^{\bar x}\subseteq Q_1$ obviously. Conversely, observing that for any $z\in Q$, $z=\sum^4\limits_{i=1}\lambda_ie_i$ for some $\lambda_i\geq 0$ with $\sum^4\limits_{i=1}\lambda_i=1$, we have then, for any

$y\in Q_1, \langle y, z-\bar x\rangle=\langle y, \sum^4_{i=1}\lambda_ie_i-\bar x\rangle=\sum^4_{i=1}\lambda_i\langle y, e_i -\bar x\rangle \leq 1,$

that is, $y\in Q^{\bar x}$. Now, since $(e_1-\bar x)\parallel (e_3-\bar x)$ and $(e_2-\bar x)\parallel (e_4-\bar x)$, It is easy to see $Q_1$ is a parallelogram. The proof is completed.

Lemma 3 Let $P$ be an $n$-polygon and $P'$ an $m$-polygon, and $e_i, e'_j$ the vertices of $P$ and $P'$ respectively, $1\leq i\leq n, 1\leq j\leq m$. Then

$\delta(P, P')\leq \max\{\max_{1\leq i\leq n}\min_{1\leq j\leq m}|e_i-e'_j|, \max_{1\leq j\leq m}\min_{1\leq i\leq n}|e_i-e'_j|\}, $

where $\delta(\cdot,\cdot)$ denotes the Hausdorff metric.

The proof is straightforward.

Lemma 4 Let $x_n, x \in {\rm int}Q$ and $x_n\rightarrow x$, then $Q^{x_n}\rightarrow Q^x$ with respect to the Hausdorff metric.

Proof Write $F'_i:=\{y\;|\; \langle y, e_i-x_n\rangle=1\}$, $F''_i:=\{y\;|\; \langle y, e_i-x\rangle=1\}$ and $e'_i=F'_i\cap F'_{i+1}, e''_i=F''_i\cap F''_{i+1}$, $i=1,\cdots,4$ (if $i=4$, then $i+1:=1$), then $e'_i, e''_i$ are the vertices of $Q^{x_n}$ and $Q^x$ respectively. It is easy to show that $|e_i'-e_i''|\rightarrow 0$ as $x_n\rightarrow x$. So, $\delta(Q^{x_n}, Q^x)\leq \max\limits_{1\leq i \leq 4}|e'_i-e''_i|\rightarrow 0$ as $x_n\rightarrow x$.

Remark  With exactly the same argument, one can show that Lemma 4 holds for all polygons, then further, by the density of polygons in $\mathcal K^2$, holds for all convex bodies in $\mathcal K^2$.

By Lemma 4 (and Remark after Lemma 4) and the continuity of the B-M distance with respect to the Hausdorff metric, we have the following corollary.

Corollary 1 Given $K,L\in \mathcal K^2$, the function $p(x):=d_{BM}(K, L^x)$ is continuous in $\mathrm{int}L$.

Proof of Theorem 4 Notice that, for any quadrangle $Q$, $1<d_{BM}(\triangle, Q)\leq 2$ (assume that $[e_1,e_3]$ is a diagonal of $Q$, then one of $e_2,e_4$, say $e_2$, is further from $[e_1,e_3]$ than $e_4$. Thus, it is easy to see that $\Delta_1\subset Q\subset 2_{e_2}\Delta_1$, where $\Delta_1={\rm cov}\{e_1,e_2,e_3\}$). Hence $p(\text{int}Q)=(1,2]$ by the fact (see [5]) that $d_{BM}(\triangle, Q^{\bar x})=2$ (where $Q^{\bar x}$ is as in Lemma 2) and that $d_{BM}(\triangle, Q^x)\rightarrow 1$ as $x\rightarrow \partial Q$, the boundary of $Q$.

Now by the continuity of $d_{BM}(\triangle, Q^x)$ as shown in Corollary 1, there is some $x_0\in \text{int}Q$ such that $d_{BM}(\triangle, Q)=d_{BM}(\triangle, Q^{x_0})$.