**2. Preliminaries**

In this section, we provide some preliminaries and notations required for functions. More details can be found in [3,4].

A function *u* : R*n* → R ∪ {<sup>+</sup>∞} is convex if

$$
\mu((1-\lambda)x+\lambda y) \le (1-\lambda)\mu(x) + \lambda\mu(y)
$$

for any *x*, *y* ∈ R*n* and *λ* ∈ [0, 1]. Let

$$\text{dom}(u) = \{ \mathbf{x} \in \mathbb{R}^n : u(\mathbf{x}) \in \mathbb{R} \}.$$

Since the convexity of *u*, dom(*u*) is a convex set. If dom(*u*) = ∅, then *u* is said proper. The function *u* is called of class C<sup>2</sup> if it is twice differentiable on int(dom(*u*)), with a positive definite Hessian matrix. The Fenchel conjugate of *u* is the convex function defined by

$$u^\*(y) = \sup\_{\boldsymbol{\chi} \in \mathbb{R}^n} \left\{ \langle \boldsymbol{\chi}, y \rangle - u(\boldsymbol{\chi}) \right\}, \quad \forall y \in \mathbb{R}^n.$$

Clearly, *u*(*x*) + *u*<sup>∗</sup>(*y*) ≥ *<sup>x</sup>*, *y* for all *x*, *y* ∈ R*<sup>n</sup>*. The equality holds if and only if *x* ∈ dom(*u*) and *y* is in the subdifferential of *u* at *x*. Hence, one can checked that

$$
\mu^\*(\nabla \mu(\mathbf{x})) + \mathfrak{u}(\mathbf{x}) = \langle \mathfrak{x}, \nabla \mathfrak{u}(\mathbf{x}) \rangle.
$$

From the definition of log-concave functions (1), we known that a log-concave function *f* : R*n* → R has the form *f*(*x*) = *e*<sup>−</sup>*<sup>u</sup>*(*x*) where *u* : R*n* → R ∪ {<sup>+</sup>∞} is convex. Writing

$$\begin{array}{rcl} \mathcal{L} &=& \left\{ u: \mathbb{R}^{n} \to \mathbb{R} \cup \{ +\infty \} \Big| u \text{ is proper and convex}, \lim\_{\|x\| \to +\infty} u(x) = +\infty \right\}, \\\mathcal{A}\_{-} &=& \left\{ f: \mathbb{R}^{n} \to \mathbb{R} \Big| \quad f = e^{-u}, \ u \in \mathcal{L} \right\}. \end{array}$$

For *u*, *v* ∈ L, the inf-convolution of *u*, *v* is defined by

$$(u \square v)(\mathbf{x}) = \inf\_{y \in \mathbb{R}^n} [u(y) + v(y - \mathbf{x})], \ \forall \mathbf{x} \in \mathbb{R}^n,\tag{5}$$

and the right scalar multiplication *ut* is defined by

$$(\mathfrak{u}t)(\mathfrak{x}) := \begin{cases} t\mathfrak{u}\left(\frac{\mathfrak{x}}{\mathbb{F}}\right), & \text{if } \mathfrak{t} > 0, \\ I\_{\{0\} \prime} & \text{if } \mathfrak{t} = 0. \end{cases}$$

Note that these operations are convexity preserving, and *<sup>I</sup>*{0} acts as the identity element in (5). It is proved that (*usvt*)(*x*) ∈ L for *u*, *v* ∈ L and *s*, *t* ≥ 0 (see [21]). Let *f* = *<sup>e</sup>*<sup>−</sup>*u*, *g* = *e*<sup>−</sup>*<sup>v</sup>* ∈ A and *t* > 0. Form (5), the Asplund sum (defined in (3)) can be rewritten as

$$f \oplus \emptyset = \mathfrak{e}^{-\left[\mathfrak{u} \square \mathfrak{v}\right]},\tag{6}$$

and *t* · *g* = *<sup>e</sup>*<sup>−</sup>*vt*. Let *f* = *e*<sup>−</sup>*<sup>u</sup>* ∈ A. The support function, *hf* , of *f* is defined as (see, e.g., [28])

$$h\_f(\mathbf{x}) = u^\*(\mathbf{x}).\tag{7}$$

We recall that a probability measure *μ* is called isotropic if it satisfies .<sup>R</sup>*n xdμ*(*x*) = 0 and

$$\int\_{\mathbb{R}^n} \langle \mathbf{x}, \theta \rangle^2 d\mu(\mathbf{x}) = \frac{1}{n'} \quad \forall \theta \in S^{n-1}. \tag{8}$$

For a measure *μ* with .<sup>R</sup>*n xdμ*(*x*) = 0, the following claims are equivalent (see, e.g., [3]):


$$\int\_{\mathbb{R}^n} \langle \mathbf{x}, T\mathbf{x} \rangle d\mu(\mathbf{x}) = \frac{\text{tr}(T)}{n};$$

(c)

$$\int\_{\mathbb{R}^n} \mathbf{x}\_i \mathbf{x}\_j d\mu(\mathbf{x}) = \frac{1}{n} \delta\_{ij} \quad \text{for all} \quad i, j = 1, \cdots, n.$$

## **3. Minimal Perimeter of Log-Concave Functions**

In this section, we consider the properties of the minimal perimeter of log-concave functions. Let *f* = *e*<sup>−</sup>*<sup>u</sup>* ∈ A, the perimeter *P*(*f*) has an integral expression (see [21,29]):

$$P(f) = -\frac{1}{2} \int\_{\mathbb{R}^n} \frac{\|\nabla f\|^2}{f} d\mathbf{x}.\tag{9}$$

We define that *f* has minimal perimeter if *P*(*f*) ≤ *P*(*f* ◦ *T*) for every *T* ∈ SL(*n*). This is, if *f* has minimal perimeter, then

$$\int\_{\mathbb{R}^n} \frac{||\nabla f||^2}{f} d\mathbf{x} \le \int\_{\mathbb{R}^n} \frac{||\nabla (f \circ T)||^2}{f \circ T} d\mathbf{x}\_{\boldsymbol{\omega}}$$

for any *T* ∈ SL(n).

> The Borel measure *μf*on R*n* of a log-concave function *f* = *e*<sup>−</sup>*<sup>u</sup>* is defined by (see [21])

$$
\mu\_f = (\nabla \mu)\_\sharp (f \mathcal{H}^n).
$$

Here H*n* denotes the *n*-dimensional Hausdorff measure. For any continuous function *g* : R*n* → R, one has

$$\int\_{\mathbb{R}^n} \mathbf{g}(y) d\mu\_f(y) = \int\_{\mathbb{R}^n} \mathbf{g}(\nabla u(\mathbf{x})) f(\mathbf{x}) d\mathbf{x}.\tag{10}$$

The Borel measure *μf* plays the same role for *f* as the surface area measure for the convex body.

**Proposition 1.** *If f* ∈ A*, then*

$$P(f) = \int\_{\mathbb{R}^n} h\_{\gamma\_n}(x) d\mu\_f(x).$$

**Proof.** From Eqautions (9), (10) and (7), we have

$$\begin{aligned} P(f) &= \frac{1}{2} \int\_{\mathbb{R}^n} \frac{||\nabla f||^2}{f} dx \\ &= \int\_{\mathbb{R}^n} f(\mathbf{x}) \frac{||\nabla u(\mathbf{x})||^2}{2} dx \\ &= \int\_{\mathbb{R}^n} f(\mathbf{x}) h\_{\gamma\_n}(\nabla u(\mathbf{x})) dx \\ &= \int\_{\mathbb{R}^n} h\_{\gamma\_n}(\mathbf{x}) d\mu\_f(\mathbf{x}). \end{aligned}$$

We recall that the gauge function of a convex body *K* is defined by

$$\|\|\mathbf{x}\|\|\mathbf{x} = \min\{a \ge 0 : \mathbf{x} \in aK\}.\tag{11}$$

It is clear that

$$\|\mathbf{x}\|\_{K} = 1 \quad \text{whenever} \quad \mathbf{x} \in \partial K\_{\prime} \tag{12}$$

where *∂K* is the boundary of *K*.

We note that the minimal perimeter of a log-concave function *f* is equivalent to considering the minimization problem:

$$\min\_{T \in SL(n)} P(f \circ T). \tag{13}$$

For *T* ∈ *SL*(*n*), we write *γT* for *γn* ◦ *T*. From (9) and the fact that ∇*x*(*f* ◦ *T*) = *<sup>T</sup><sup>t</sup>*∇*Tx f* for *T* ∈ *SL*(*n*) and *x* ∈ R*<sup>n</sup>*, we have

$$\begin{aligned} P(f \diamond T) = \delta f(f \diamond T, \gamma\_{\mathbb{H}}) &= \frac{1}{2} \int\_{\mathbb{R}^n} \frac{\|\nabla\_{\boldsymbol{x}} f \circ T\|^2}{f(\boldsymbol{Tx})} d\boldsymbol{x} \\ &= \frac{1}{2} \int\_{\mathbb{R}^n} \frac{\|\boldsymbol{T}^t \nabla\_{\boldsymbol{x}} f(\boldsymbol{Tx})\|^2}{f(\boldsymbol{Tx})} d\boldsymbol{x} \\ &= \frac{1}{2} \int\_{\mathbb{R}^n} \frac{\|\boldsymbol{T}^t \nabla f(\boldsymbol{x})\|^2}{f(\boldsymbol{x})} d\boldsymbol{x} \\ &= \delta f(f, \gamma\_T) .\end{aligned}$$

Therefore, we can reformulate problem (13) as follows:

$$\min \{ \delta \} (f, \gamma\_T) : T \in SL(n) \}. \tag{14}$$

**Proposition 2.** *There exists a unique (un to an orthogonal matrix) T*0 ∈ *SL*(*n*) *such that it solves the minimization problem (14).*

**Proof.** We can limit our attention to *T* ∈ *SL*(*n*) when *T* is a positive definite symmetric matrix, since any *T* ∈ *SL*(*n*) can be represented in the form *T* = *PQ* where *P* ∈ *SL*(*n*) is a positive definite symmetric matrix and *Q* is an orthogonal matrix. In this case, we can write the function *<sup>γ</sup>T*(*x*) = *<sup>γ</sup>n*(*Tx*) = *e*<sup>−</sup> -*Tx*-2 2 as *<sup>γ</sup>T*(*x*) = *e*<sup>−</sup> *x*-2*E*◦ 2 , where *E* = *TtB* is an origin-centered ellipsoid and *E*◦ is the polar body of *E* defined as *E*◦ = {*x* ∈ R*n* : *<sup>x</sup>*, *y* ≤ 1 for all *y* ∈ *<sup>E</sup>*}. There exists a *zE* ∈ *Sn*−<sup>1</sup> such that the diameter of *E* satisfies diam (*E*)|*zE*,*<sup>x</sup>*| 2 ≤ *x*-*E*◦ . Let {*Tk*}*k* ∈ *SL*(*n*) be a minimizing sequence for the problem (14), namely,

lim *k*→∞ *δJ*(*f* , *γTk* ) = min{*δJ*(*f* , *<sup>γ</sup>T*) : *T* ∈ *SL*(*n*), *T* is a positive definite symmetric matrix}. (15)

From (15) and the fact that min{*δJ*(*f* , *<sup>γ</sup>T*) : *T* ∈ *SL*(*n*)} ≤ *δJ*(*f* , *<sup>γ</sup>n*), we have

$$\begin{aligned} \frac{\text{diam}\,(E\_k^\circ)^2}{8} \min\_{\boldsymbol{z}\in S^{n-1}} \int\_{\mathbb{R}^n} |\langle \boldsymbol{z}, \boldsymbol{x} \rangle|^2 d\mu\_f(\boldsymbol{x}) & \quad \le \int\_{\mathbb{R}^n} \frac{||\boldsymbol{x}||\_{E\_k}^2}{2} d\mu\_f(\boldsymbol{x}) \\ & \quad = \quad \delta f(f, \gamma\_{T\_k}) \\ & \quad \le \quad \delta f(f, \gamma\_{\mathbb{R}^n}). \end{aligned}$$

Since *x*-*Ek* < diam (*E*◦*k* )*x*-, therefore the upper bound of the convex function -*Tkx*-2 2 ∗ is depended only on *f* . According to Theorem 10.9 in [30], there exist a function *γT*0 such that the Legendre conjugate of a minimizing sequence of functions for problem (14) converge to *γT*0 . Due to Theorem 11.34 in [31], we known that a minimizing sequence of functions for problem (14) converge to *<sup>γ</sup>*<sup>∗</sup>*T*0. According to the dominated convergence theorem, there exists a solution to problem (14).

Next, we prove the uniqueness of *T*0. Assume there are two different solutions *T*1, *T*2 ∈ *SL*(*n*) to the considered problem which satisfy *T*1 = *aT*2 for all *a* > 0. If there exists a *a*0 > 0 such that *T*1 = *<sup>a</sup>*0*T*2, then

$$
\delta J(f, \gamma\_{T\_1}) = \delta J(f, \gamma\_{a\_0 T\_2}) = \delta J(f, \gamma\_{T\_2}) \implies a\_0 = 1.
$$

This contradicts to *T*1 = *T*2. The Minkowski inequality for symmetric positive definite matrices shows that

$$\det\left(\frac{T\_1 + T\_2}{2}\right)^{\frac{1}{n}} > \frac{1}{2} \det\left(T\_1\right)^{\frac{1}{n}} + \frac{1}{2} \det\left(T\_2\right)^{\frac{1}{n}} = 1.$$

Let

$$T\_3 = \det(\frac{T\_1^{-1} + T\_2^{-1}}{2})^{\frac{1}{n}} \left(\frac{T\_1^{-1} + T\_2^{-1}}{2}\right)^{-1}.$$

Then *T*3 ∈ *SL*(*n*) and *γT*3 < *γ*( *T*−<sup>1</sup> 1 +*T*<sup>−</sup><sup>1</sup> 2 2 )−<sup>1</sup> , i.e., *<sup>h</sup>γT*3 < *hγ*( *T*−<sup>1</sup> 1 +*T*<sup>−</sup><sup>1</sup> 2 2 )−<sup>1</sup> . This deduces that

$$\begin{aligned} \mathcal{S}f(f, \gamma\_{\mathbb{T}\_3}) &= \int\_{\mathbb{R}^n} h\_{\gamma\_{\mathbb{T}\_3}}(\mathbf{x}) d\mu\_f(\mathbf{x}) \\ &< \int\_{\mathbb{R}^n} h\_{\gamma\_{\frac{\mathbb{T}\_1^{-1} + \mathbb{T}\_2^{-1}}{2}})^{-1}}(\mathbf{x}) d\mu\_f(\mathbf{x}) \\ &= \int\_{\mathbb{R}^n} \frac{||\frac{\mathbb{T}\_1^{-1} + \mathbb{T}\_2^{-1}}{2}\mathbf{x}||^2}{2} d\mu\_f(\mathbf{x}). \end{aligned}$$

By the convexity of the square of the Euclidean normal, we have

$$\begin{aligned} \delta J(f, \gamma\_{T\_3}) &< \int\_{\mathbb{R}^n} \frac{\frac{1}{2} ||T\_1^{-1} \mathbf{x}||^2 + \frac{1}{2} ||T\_2^{-1} \mathbf{x}||^2}{2} d\mu\_f(\mathbf{x}) \\ &= \frac{1}{2} \int\_{\mathbb{R}^n} h\_{\gamma T\_1}(\mathbf{x}) d\mu\_f(\mathbf{x}) + \frac{1}{2} \int\_{\mathbb{R}^n} h\_{\gamma T\_2}(\mathbf{x}) d\mu\_f(\mathbf{x}) \\ &= \frac{1}{2} \delta f(f, \gamma\_{T\_1}) + \frac{1}{2} \delta f(f, \gamma\_{T\_2}) \\ &= \delta f(f, \gamma\_{T\_1}) \\ &= \delta f(f, \gamma\_{T\_2}). \end{aligned}$$

However, from the assumption on *T*1 and *T*2, we have

$$
\delta J(f, \gamma\_{T\_3}) \ge \delta J(f, \gamma\_{T\_1}) = \delta J(f, \gamma\_{T\_2}),
$$

which is a contradiction.

> Proposition 2 implies that the minimal perimeter of log-concave functions is well-defined. Namely,

**Corollary 1.** *For a log-concave function f* : R*n* → R*, there exists a unique (up to an orthogonal matrix) T*0 ∈ *SL*(*n*) *such that f* ◦ *T*0 *has minimal perimeter.*

Next we are in the position to consider the proof of Theorem 1.

**Proof of Theorem 1.** Let *T* ∈ GL(n), and *ε* > 0 be a suitably small real number. Then

$$T\_{\varepsilon} = \left[ \det(I + \varepsilon T) \right]^{-\frac{1}{n}} (I + \varepsilon T)^{t} \in \operatorname{SL}(\mathbf{n})\_{\prime}$$

and this implies that *P*(*f*) ≤ *P*(*f* ◦ *<sup>T</sup>ε*), i.e.,

$$\int\_{\mathbb{R}^n} \frac{\|\nabla f\|^2}{f} d\mathbf{x} \quad \leq \quad \int\_{\mathbb{R}^n} \frac{\|\nabla (f \circ T\_\varepsilon)\|^2}{f \circ T\_\varepsilon} d\mathbf{x}.$$

By the fact that ∇*x*(*<sup>u</sup>* ◦ *T*) = *<sup>T</sup><sup>t</sup>*∇*Txu*, then

$$\begin{aligned} \int\_{\mathbb{R}^n} \frac{||\nabla f||^2}{f} dx &\leq \int\_{\mathbb{R}^n} \frac{||\nabla(f \circ T\_\varepsilon)||^2}{f \circ T\_\varepsilon} dx \\ &= \int\_{\mathbb{R}^n} \frac{||T\_\varepsilon^t \nabla\_{T\_\varepsilon \mathbf{x}} (f(T\_\varepsilon \mathbf{x}))||^2}{f(T\_\varepsilon \mathbf{x})} dx \\ &= \int\_{\mathbb{R}^n} \frac{||T\_\varepsilon^t \nabla f||^2}{f} dx, \end{aligned}$$

i.e.,

$$\left[\det(I+\varepsilon T)\right]^{\frac{2}{n}} \int\_{\mathbb{R}^n} \frac{\|\nabla f\|^2}{f} d\mathbf{x} \quad \leq \int\_{\mathbb{R}^n} \frac{\|(I+\varepsilon T)\nabla f\|^2}{f} d\mathbf{x}.$$

Because

$$\|(I + \varepsilon T)\nabla f\|^2 = \|\nabla f\|^2 + 2\varepsilon \langle \nabla f, T\nabla f \rangle + o(\varepsilon^2)$$

and

$$\left[\det(I+\varepsilon T)\right]^{\frac{2}{n}} = 1 + 2\varepsilon \frac{\text{tr}(T)}{n} + o(\varepsilon^2),$$

when letting *ε* → 0+, we obtain

$$\frac{\text{tr}(T)}{n} \int\_{\mathbb{R}^n} \frac{\|\nabla f\|^2}{f} d\mathbf{x} \quad \le \int\_{\mathbb{R}^n} \frac{\langle \nabla f, T\nabla f \rangle}{f} d\mathbf{x}.\tag{16}$$

Replacing *T* by −*T* in (16), we conclude that there must be equality in (4) for every *T* ∈ GL(n).

On the other hand, assume that (4) is satisfied and let *T* ∈ SL(n). Since tr*Tn* ≥ 1 for symmetric positive-definite metric, (9) and ∇*x*(*f* ◦ *T*) = *<sup>T</sup><sup>t</sup>*∇*Tx f* , we have

$$\begin{aligned} P(f \circ T) &= \quad \frac{1}{2} \int\_{\mathbb{R}^n} \frac{\|\nabla(f \circ T)\|^2}{f \circ T} d\mathbf{x} \\ &= \quad \frac{1}{2} \int\_{\mathbb{R}^n} \frac{\|T^t \nabla\_{T\mathbf{x}} f\|^2}{f \circ T} d\mathbf{x} \\ &= \quad \frac{1}{2} \int\_{\mathbb{R}^n} \frac{\langle \nabla f, T \nabla f \rangle}{f} d\mathbf{x} \\ &= \quad \frac{1}{2} \int\_{\mathbb{R}^n} \frac{\langle \nabla f, T^t T \nabla f \rangle}{f} d\mathbf{x} \\ &= \quad \frac{1}{2} \frac{\text{tr}(T^t T)}{n} \int\_{\mathbb{R}^n} \frac{\|\nabla f\|^2}{f} d\mathbf{x} \\ &\geq \quad \frac{1}{2} \int\_{\mathbb{R}^n} \frac{\|\nabla f\|^2}{f} d\mathbf{x} \\ &= \quad P(f). \end{aligned} \tag{17}$$

This shows that *f* has minimal perimeter. Moreover, the equality in (17) holds only if *T* is the identity matrix. This prove that the uniqueness of the minimal perimeter position (up to *U* ∈ <sup>O</sup>(n)).

**Corollary 2.** *From Theorem 1 and the definition of isotropic measure, the log-concave function f* ∈ A *has minimal perimeter if and only if* 1*J*(*f*)*μ<sup>f</sup> is an isotropic measure.*

Next, we prove that Theorem 1 recovers the *L*2 surface area measure of *K*, *dS*2(*<sup>K</sup>*, ·) = *hK*(·)−1*d<sup>σ</sup>K*(·), is an isotropic measure on *Sn*−1.

**Corollary 3.** *Let K be a convex body in* R*n containing the origin in its interior. If f*(*x*) = *e*−*x*-*K for x* ∈ R*n, then*

$$P(f) = \frac{1}{2}\Gamma(n)S\_2(K),$$

*and Theorem 1 includes the fact that L*2 *surface area measure of K, <sup>S</sup>*2(*<sup>K</sup>*, ·)*, is an isotropic measure.*

**Proof.** For a convex body *K* in R, let *VK* denote the normalized cone volume measure of *K*, which is given by

$$d\nabla\_K(z) = \frac{\langle z, \nu\_K(z) \rangle}{nV(K)} d\mathcal{H}^{n-1}(z) \quad \text{for} \quad z \in \partial K.$$

Here *<sup>ν</sup>K*(*z*) is the outer unit normal of *K* at the boundary point *z*, and H*<sup>n</sup>*−<sup>1</sup> is the (*n* − 1) dimensional Hausdorff measure. For any *x* ∈ R*<sup>n</sup>*, we write *x* = *rz*, with *z* ∈ *∂K* and *dx* = *nV*(*K*)*rn*−<sup>1</sup>*drdVK*(*z*). Since the map *x* → ∇*x*-*K* is 0-homogeneous, and (12), we have

$$\begin{aligned} P(f) &= \frac{1}{2} \int\_{\mathbb{R}^n} ||\nabla||\mathbf{x}||\_{K} ||^2\_{\mathcal{E}} e^{-||\mathbf{x}||\_K} d\mathbf{x} \\ &= \frac{1}{2} n V(K) \int\_0^\infty r^{n-1} \int\_{\partial K} ||\nabla||z||\_K ||^2 e^{-r} d\overline{\nabla}\_K(z) dr \\ &= \frac{1}{2} \Gamma(n) n V(K) \int\_{\partial K} ||\nabla||z||\_K ||^2 d\overline{\nabla}\_K(z), \end{aligned}$$

where <sup>Γ</sup>(·) is the Gamma function. We need the fact that ∇*z*-*K* = *<sup>ν</sup>K*(*z*) *<sup>z</sup>*,*νK*(*z*) when *z* ∈ *∂K* (see, e.g., [4]). Therefore,

$$\begin{aligned} P(f) &=& \frac{1}{2} \Gamma(n) n V(K) \int\_{\partial K} ||\nabla||z||\_K||^2 d\overline{V}\_K(z) \\ &=& \frac{1}{2} \Gamma(n) \int\_{\partial K} h\_K(\nu\_K(z))^{-1} d\mathcal{H}^{n-1}(z) \\ &=& \frac{1}{2} \Gamma(n) S\_2(K). \end{aligned}$$

From the fact that the map *x* → ∇*x*-*K* is 0-homogeneous, (12) and (10), we have

$$\begin{split} \frac{1}{2} \int\_{\mathbb{R}^n} \langle x, Tx \rangle d\mu\_f(x) &= \frac{1}{2} \int\_{\mathbb{R}^n} \langle \nabla \|x\|\_{K} \, T \nabla \|x\|\_{K} \rangle e^{-\|x\|\_K} dx \\ &= \frac{1}{2} nV(K) \int\_0^\infty r^{n-1} \int\_{\partial K} \langle \nabla \|z\|\_{K} \, T \nabla \|z\|\_{K} \rangle e^{-r} d\overline{V}\_K(z) dr \\ &= \frac{1}{2} \Gamma(n) \int\_{\partial K} \langle \nu\_K(z), T\nu\_K(z) \rangle h\_K(\nu\_K(z))^{-1} d\mathcal{H}^{n-1}(z) \\ &= \frac{1}{2} \Gamma(n) \int\_{\mathbb{S}^{n-1}} \langle \nu\_\tau \text{T} u \rangle dS\_2(K, u) .\end{split}$$

Hence, (4) implies that

$$\frac{\text{tr}(T)}{n}S\_2(K) = \int\_{S^{n-1}} \langle \mathfrak{u}, T\mathfrak{u} \rangle dS\_2(K, \mathfrak{u})$$

for every *T* ∈ GL(n). This means that the *L*2 surface area measure of *K*, *<sup>S</sup>*2(*<sup>K</sup>*, ·), is an isotropic measure on *Sn*−1.
