**Appendix B**

Here, we prove the *Generalized isoperimetric inequality* from Section 2. The starting point is the *entropy-power inequality* (EPI) [64]: Let X<sup>1</sup> and X<sup>2</sup> be two independent continuous random vectors in <sup>R</sup>*<sup>D</sup>* with probability densities <sup>F</sup>(1) <sup>∈</sup> *q*(R*D*) and <sup>F</sup>(2) <sup>∈</sup> *p*(R*D*), respectively. Suppose further that *λ* ∈ (0, 1) and *r* > 1, and let

$$q = \frac{r}{(1 - \lambda) + \lambda r}, \quad p = \frac{r}{\lambda + (1 - \lambda)r},\tag{A5}$$

then the following inequality holds:

$$N\_{\mathbf{f}}(\mathcal{X}\_1 + \mathcal{X}\_2) \ge \left(\frac{N\_{\mathbf{f}}(\mathcal{X}\_1)}{1 - \lambda}\right)^{1 - \lambda} \left(\frac{N\_{\mathbf{f}}(\mathcal{X}\_2)}{\lambda}\right)^{\lambda}.\tag{A6}$$

Let us now consider a Gaussian noise vector Z*<sup>G</sup>* (independent of X ) with zero mean and covariance matrix **Σ**. Within this setting, we can write the following EPIs:

$$N\_r(\mathcal{X} + \mathcal{Z}^G) \ge \epsilon^{\lambda} \left(\frac{1}{1-\lambda}\right)^{1-\lambda} \left(\frac{1}{\lambda}\right)^{\lambda} [N\_q(\mathcal{X})]^{1-\lambda},\tag{A7}$$

$$N\_r(\mathcal{X} + \mathcal{Z}^G) \ge \epsilon^{1-\lambda} \left(\frac{1}{1-\lambda}\right)^{1-\lambda} \left(\frac{1}{\lambda}\right)^{\lambda} [N\_p(\mathcal{X})]^\lambda,\tag{A8}$$

with ≡ det(**Σ**)1/*D*. Here, we have used the simple fact that *Nr*(Z*<sup>G</sup>* ) = det(**Σ**)1/*D*, irrespective of the value of *r*.

Let us now fix *r* and maximize the RHS of inequality (A7) with respect to *λ* and *q* provided we keep the constraint condition (A5). This yields the condition extremum

$$\lambda = \frac{\epsilon}{N\_q(\mathcal{X})} \exp\left[q(1-q)\frac{d\log N\_q(\mathcal{X})}{dq}\right] + \mathcal{O}(\epsilon^2). \tag{A9}$$

With this, *q* turns out to be

$$q = r + \frac{\varepsilon (1 - r) r}{N\_l(\mathcal{X})} \exp\left[ (1 - r) r \frac{d \log N\_r(\mathcal{X})}{dr} \right] + \mathcal{O}(\epsilon^2),\tag{A10}$$

which in the limit → 0 reduces to *q* = *r* ≥ 1. The latter implies that *p* = 1. The result (A10) implies that the RHS of (A7) reads

$$N\_q(\mathcal{X}) + \varepsilon \exp\left[ (1-r)r \frac{d \log N\_r(\mathcal{X})}{dr} \right] \left[ 1 - (1-r)r \frac{d \log N\_r(\mathcal{X})}{dr} \right] + \mathcal{O}(\epsilon^2). \tag{A11}$$

Should we have started with the *p* index, we would arrive at an analogous conclusion. To proceed, we stick, without loss of generality, to the inequality (A7). This implies that

$$\begin{split} N\_{\mathbb{T}}(\mathcal{X} + \mathcal{Z}^{c}) &\geq \quad N\_{\mathbb{T}}(\mathcal{X}) \\ &+ \quad \epsilon \exp\left[ (1-r)r \frac{d \log N\_{\mathbb{T}}(\mathcal{X})}{dr} \right] \left[ 1 - (1-r)r \frac{d \log N\_{\mathbb{T}}(\mathcal{X})}{dr} \right] \\ &+ \quad \mathcal{O}(\epsilon^{2}) \\ &= \quad N\_{\mathbb{T}}(\mathcal{X}) + \, [N\_{\mathbb{T}}(\mathcal{X}) - N\_{\mathbb{T}}(\mathcal{X})] \\ &+ \quad \epsilon \exp\left[ (1-r)r \frac{d \log N\_{\mathbb{T}}(\mathcal{X})}{dr} \right] \left[ 1 - (1-r)r \frac{d \log N\_{\mathbb{T}}(\mathcal{X})}{dr} \right] \\ &+ \quad \mathcal{O}(\epsilon^{2}) \\ &\geq \quad N\_{\mathbb{T}}(\mathcal{X}) + \, \epsilon \exp\left[ (1-r)r \frac{d \log N\_{\mathbb{T}}(\mathcal{X})}{dr} \right] + \, \mathcal{O}(\epsilon^{2}) .\end{split} \tag{A12}$$

To proceed, we employ the identity log *Nr*(X ) = 2/*D*[I*r*(X ) − I*r*(Z*<sup>G</sup>* 1I )] with Z*<sup>G</sup>* 1I representing a Gaussian random vector with zero mean and *unit* covariance matrix, and the fact that I*<sup>r</sup>* is monotonously decreasing function of *r*, i.e., *d*I*r*/*dr* ≤ 0 (see, e.g., Ref. [78]). With this, we have

$$\exp\left[(1-r)r\frac{d\log N\_r(\lambda')}{dr}\right] \quad \geq \quad \exp\left[\frac{2(r-1)r}{D}\frac{d\mathcal{L}\_r(\mathcal{Z}\_1^C)}{dr}\right]$$

$$= \quad \exp\left[(r-1)r\frac{d}{dr}\left(\frac{1}{r-1}\log r\right)\right]$$

$$= \quad cr^{r/(r-1)} \geq \frac{c^2}{r}.\tag{A13}$$

Consequently, Equation (A12) can be rewritten as

$$\frac{N\_r(\mathcal{X} + \mathcal{Z}^G) - N\_\emptyset(\mathcal{X})}{\Sigma\_{ij}} \ge \frac{\epsilon}{\Sigma\_{ij}} \frac{\epsilon^2}{r} + \mathcal{O}(\epsilon^2 / \Sigma\_{ij}) . \tag{A14}$$

At this stage, we are interested in the Σ*ij* → 0 limit. In order to find the ensuing leading order behavior of /Σ*ij*, we can use L'Hospital's rule, namely

$$\frac{d\epsilon}{d\Sigma\_{ij}} = \frac{d\epsilon}{d\Sigma\_{ij}} = \frac{d}{d\Sigma\_{ij}} \exp\left[\frac{1}{D} \text{Tr}(\log \mathbf{\mathcal{D}})\right] \\ = \frac{\epsilon}{D} (\mathbf{\mathcal{D}}^{-1})\_{ij}. \tag{A15}$$

Now, we neglect the sub-leading term of order O(2/Σ*ij*) in (A14) and take det(...)1/*<sup>D</sup>* on both sides. This gives

$$\left. \det \left( \frac{d\mathcal{N}\_{\mathbf{r}}(\mathcal{X} + \mathcal{Z}^{\mathcal{G}})}{d\Sigma\_{ij}} \right)^{1/D} \right|\_{\Sigma=0} = \frac{1}{rD} \mathcal{N}\_{\mathbf{r}}(\mathcal{X}) [\det(\mathbb{J}\_{\mathbf{r}}(\mathcal{X}))]^{1/D} \geq \frac{e^2}{rD} \geq \frac{1}{rD}, \tag{A16}$$

or equivalently

$$N\_r(\mathcal{X})[\det(\mathbb{J}\_r(\mathcal{X}))]^{1/D} \ge 1. \tag{A17}$$

At this stage, we can use the inequality of arithmetic and geometric means to write (note that J*<sup>r</sup>* = cov*r*(*Vr*) is a positive semi-definite matrix)

$$\frac{1}{D} \text{Tr}(\mathbb{J}\_I(\mathcal{X})) \ge \left[ \det(\mathbb{J}\_I(\mathcal{X})) \right]^{1/D}.\tag{A18}$$

Consequently, we have

$$\frac{1}{D}N\_r(\mathcal{X})\text{Tr}(\mathbb{J}\_r(\mathcal{X})) = \frac{1}{D}N\_r(\mathcal{X})|\_r(\mathcal{X}) \ge N\_r(\mathcal{X})[\text{det}(\mathbb{J}\_r(\mathcal{X}))]^{1/D} \ge 1,\tag{A19}$$

as stated in Equation (14).

## **Appendix C**

In this appendix, we prove the *Generalized Stam inequality* from Section 2. We start with the defining relation (13), i.e.,

$$N\_{\boldsymbol{\theta}}(\mathcal{Y}) \;= \; \frac{1}{2\pi} q^{1/(1-q)} \|\mathcal{G}\|\_{q}^{2q/[(1-q)D]} \; , \tag{A20}$$

and consider *q* ∈ [1/2, 1] so that *q*/(1 − *q*) > 0. For the *q*-norm, we can write

$$\|\mathcal{G}\|\_{\mathfrak{q}} = \left(\int\_{\mathbb{R}^D} dy \, |\psi\_{\mathcal{G}}(y)|^{2q}\right)^{1/q} = \|\psi\_{\mathcal{G}}\|\_{2q}^2 \ge \|\hat{\psi}\_{\mathcal{G}}\|\_{2r}^2 = \|\psi\_{\mathcal{F}}\|\_{2r}^2 = \|\mathcal{F}\|\_{r} . \tag{A21}$$

Here, 2*r* and 2*q* are Hölder conjugates so that *r* ∈ [1, ∞]. The inequality employed is due to the Hausdorff–Young inequality (which in turn is a simple consequence of the Hölder inequality [64]). We further have


where *<sup>a</sup>* <sup>∈</sup> <sup>R</sup>*<sup>D</sup>* is an arbitrary *<sup>x</sup>*-independent vector, <sup>∇</sup>*<sup>i</sup>* <sup>≡</sup> *<sup>∂</sup>*/*∂xi* and *VD* denotes a regularized volume of R*D*—*D*-dimensional ball of a very large (but finite) radius *R*. In the first line of (A22), we have employed the triangle inequality <sup>|</sup>E*<sup>r</sup> eia*·*x* | ≤ 1 (with equality if and only if *a* = **0**), namely

$$\left| \int\_{\mathbb{R}^D} d\mathbf{x} \, |\psi\_{\mathcal{F}}(\mathbf{x})|^{2r} e^{i\mathbf{a} \cdot \mathbf{x}} \right| = \left| \int\_{\mathbb{R}^D} d\mathbf{x} \, \rho\_{\mathcal{F}}(\mathbf{x}) e^{i\mathbf{a} \cdot \mathbf{x}} \right| \int\_{\mathbb{R}^D} d\mathbf{x} \, |\psi\_{\mathcal{F}}(\mathbf{x})|^{2r} \leq \int\_{\mathbb{R}^D} d\mathbf{x} \, |\psi\_{\mathcal{F}}(\mathbf{x})|^{2r} \text{ (A23)}$$

The inequality in the last line holds for *ai* = *π*/(2*R*) (for all *i*), since, in this case, cos(*a* · *x*) ≥ 0 for all *x* from the *D*-dimensional ball. In this case, one may further estimate the integral from below by neglecting the positive integrand (*<sup>r</sup>* − <sup>1</sup>)*ρr*(*x*)[*Vi*(*x*)]2.

Note that (A22) implies

$$\frac{r\left|\mathbb{E}\_r\left[\mathcal{F}^{-1}\nabla\_i\nabla\_i\mathcal{F}\cos(\mathfrak{a}\cdot\mathbf{x})\right]\right|}{a\_i^2} \le 1\,\tag{A24}$$

with equality if and only if *a* → **0** (to see this, one should apply L'Hospital's rule). Equation (A24) allows for writing

$$\|\mathcal{F}\|\_{r} \geq \frac{r^{\gamma} \Big| \mathbb{E}\_{\mathbf{r}} \Big[ \mathcal{F}^{-1} \nabla\_{i} \nabla\_{i} \mathcal{F} \cos(\mathbf{a} \cdot \mathbf{x}) \big] \Big|^{\gamma}}{a\_{i}^{2\gamma}} \Big( \int\_{\mathbb{R}^{D}} d\mathbf{x} \, |\psi\_{\mathcal{F}}(\mathbf{x})|^{2\tau} \right)^{1/r}$$

$$\geq \frac{r^{\gamma} \Big| \mathbb{E}\_{\mathbf{r}} \Big[ \mathcal{F}^{-1} \nabla\_{i} \nabla\_{i} \mathcal{F} \cos(\mathbf{a} \cdot \mathbf{x}) \big] \Big|^{\gamma}}{a\_{i}^{2\gamma}} \frac{1}{V\_{D}^{1 - 1/r}}$$

$$= \frac{r^{\gamma} \Big| \mathbb{E}\_{\mathbf{r}} \Big[ \mathcal{F}^{-1} \nabla\_{i} \nabla\_{i} \mathcal{F} \cos(\mathbf{a} \cdot \mathbf{x}) \big] \Big|^{\gamma}}{a\_{i}^{2\gamma}} \frac{1}{C\_{D}^{1 - 1/r} R^{D - D/r}},\tag{A25}$$

where *γ* > 0 is some as yet unspecified constant and *CD* = *π<sup>D</sup>*/2/Γ(*D*/2 + 1). In deriving (A25), we have used the Hölder inequality

$$1 \quad = \quad \left(\int\_{\mathbb{R}^D} d\mathbf{x} \mathbf{1} \cdot |\psi\_{\mathcal{F}}(\mathbf{x})|^2 \right) \le \left(\int\_{\mathbb{R}^D} d\mathbf{x} \mathbf{1}^{r'} \right)^{1/r'} \left(\int\_{\mathbb{R}^D} d\mathbf{x} |\psi\_{\mathcal{F}}(\mathbf{x})|^2 r \right)^{1/r} $$
 
$$= \quad V\_D^{1 - 1/r} \left(\int\_{\mathbb{R}^D} d\mathbf{x} |\psi\_{\mathcal{F}}(\mathbf{x})|^{2r} \right)^{1/r} . \tag{A26}$$

Here, and also in (A22) and (A25), *VD* = *CDR<sup>D</sup>* denotes the regularizated volume of R*D*.

As already mentioned, the best estimate of the inequality (A25) is obtained for *a* → **0**. As we have seen, *ai* goes to zero as *π*/(2*R*) which allows for choosing the constant *γ* so that the denominator in (A25) stays finite in the limit *R* → ∞. This implies that *γ* = *D*/2 − *D*/(2*r*). Consequently, (A25) acquires in the large *R* limit the form

$$\|\mathcal{F}\|\_{r} \ge \frac{[4(r-1)/r]^{D/2-D/2r} [\Gamma(D/2+1)]^{1-1/r}}{\pi^{3D/2-3D/2r}} [(\mathbb{J}\_r)\_{ii}(\mathcal{X})]^{D/2-D/2r},\tag{A27}$$

With this, we can write [see Equations (A20)–(A21)]

$$N\_q(\mathcal{Y}) \ge \frac{1}{(2\pi)^2} q^{1/(1-q)} \left[ (\mathbb{J}\_r)\_{ii}(\mathcal{X}) \right] \ge \frac{1}{16\pi^2} \left[ (\mathbb{J}\_r)\_{ii}(\mathcal{X}) \right],\tag{A28}$$

where, in the last inequality, we have used the fact that *<sup>q</sup>*1/(1−*q*) <sup>≥</sup> 1/4 for *<sup>q</sup>* <sup>∈</sup> [1/2, 1] and that [Γ(*D*/2 + <sup>1</sup>)]2/*<sup>D</sup>* ≥ *<sup>π</sup>*/4. As a final step, we employ Equations (A18) and (A28) to write

$$N\_{\mathbb{q}}(\mathcal{Y}) \ge \frac{1}{16\pi^2 D} \text{Tr}(\mathbb{J}\_r(\mathcal{X})) \ge \frac{1}{16\pi^2} [\det(\mathbb{J}\_r(\mathcal{X}))]^{1/D},\tag{A29}$$

which completes the proof of the generalized Stam's inequality.

#### **References**

