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Symmetry of minimizers of a Gaussian isoperimetric problem

Abstract

We study an isoperimetric problem described by a functional that consists of the standard Gaussian perimeter and the norm of the barycenter. The second term is in competition with the perimeter, balancing the mass with respect to the origin, and because of that the solution is not always the half-space. We characterize all the minimizers of this functional, when the volume is close to one, by proving that the minimizer is either the half-space or the symmetric strip, depending on the strength of the barycenter term. As a corollary, we obtain that the symmetric strip is the solution of the Gaussian isoperimetric problem among symmetric sets when the volume is close to one. As another corollary we obtain the optimal constant in the quantitative Gaussian isoperimetric inequality.

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Acknowledgements

The first author was supported by INdAM and by the project VATEXMATE. The second author was supported by the Academy of Finland Grant 314227.

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Appendix

Appendix

We first prove the inequalities (4) and (5). In fact, the proof gives us a slightly stronger estimate than (4). We recal that we are assuming \(s \ge 10^3\).

Lemma 5

The following estimates hold:

$$\begin{aligned} s + \frac{1}{s} \ln \left( 2-2/s^2\right)< a(s) < s + \frac{\ln 2}{s} \end{aligned}$$
(72)

and

$$\begin{aligned} \left( 1 + \frac{\ln 2}{s^2} - \frac{8}{s^4} \right) e^{-\frac{s^2}{2}}< P_\gamma (D_{\omega ,s}) < \left( 1 + \frac{\ln 2}{s^2}\right) e^{-\frac{s^2}{2}}. \end{aligned}$$
(73)

Proof

The right-hand inequality in (72) follows from the isoperimetric inequality \(P_\gamma (D_{\omega ,s}) > P_\gamma (H_{\omega ,s})\) which we may write as \(2 e^{-\frac{a(s)^2}{2}} > e^{-\frac{s^2}{2}}\). This implies

$$\begin{aligned} a(s)< \sqrt{s^2 + 2 \ln 2} < s + \frac{\ln 2}{s}. \end{aligned}$$

In order to show the right-hand inequality in (73) we first note that the function \(\psi : [0, \infty ) \rightarrow {\mathbb {R}}\),

$$\begin{aligned} \psi (x) = x e^{\frac{x^2}{2}} \int _x^\infty e^{-\frac{t^2}{2}} \, dt \end{aligned}$$

is increasing. Indeed, its first derivative is

$$\begin{aligned} \psi '(x) = - x + (1+ x^2 ) e^{\frac{x^2}{2}} \int _x^\infty e^{-\frac{t^2}{2}} \, dt, \end{aligned}$$

and by a second order analysis it is easy to show that the quantity \(\psi '(x) e^{-\frac{x^2}{2}}\) is positive. The volume condition \(\gamma (D_{\omega ,s}) = \phi (s)\) can be written as

$$\begin{aligned} \frac{2}{a(s)} e^{-\frac{a(s)^2}{2}} \psi (a(s)) = \frac{1}{s} e^{-\frac{s^2}{2}} \psi (s). \end{aligned}$$

Since \(\psi \) is increasing and \(a(s) > s\) we deduce by the upper bound for a(s) that

$$\begin{aligned} 2 e^{-\frac{a(s)^2}{2}}< \frac{a(s)}{s} e^{-\frac{s^2}{2}} < \Bigl ( 1 + \frac{\ln 2}{s^2} \Bigr ) e^{-\frac{s^2}{2}} . \end{aligned}$$

Hence we have the right-hand inequality in (73).

To prove the left-hand inequality in (72) we use the above estimate to obtain

$$\begin{aligned} a(s)^2 > s^2 + 2 \ln \left( \frac{2}{ 1 + \ln 2/s^2} \right) \ge s^2 + 2 \ln \left( 2(1 - \ln 2/s^2) \right) . \end{aligned}$$

In order to prove the inequality we need to show that

$$\begin{aligned} \sqrt{ s^2 + 2 \ln \left( 2(1 - \ln 2/s^2) \right) } > s + \frac{1}{s} \ln \left( 2-2/s^2\right) . \end{aligned}$$

This is equivalent to

$$\begin{aligned} 2 \ln \left( \frac{1 - \ln 2/s^2}{1 -1/s^2} \right) > \frac{1}{s^2} \ln ^2 \left( 2-2/s^2\right) . \end{aligned}$$

Use the fact that for \(0<y<1/9\) it holds \(\ln (1 + y) \ge 9y/10\) to estimate

$$\begin{aligned} 2 \ln \left( \frac{1 - \ln 2/s^2}{1 -1/s^2} \right) = 2 \ln \left( 1 + \frac{(1 - \ln 2)}{s^2-1} \right) \ge \frac{9}{5}\frac{1 - \ln 2}{s^2-1} \ge \frac{9}{5} \frac{1 - \ln 2}{s^2} . \end{aligned}$$

The claim follows from the fact that \(\ln ^2 \left( 2-2/s^2\right) < 9(1 - \ln 2)/5\).

In order prove the left-hand inequality in (73) we first obtain, by integrating by parts twice, that

$$\begin{aligned} \begin{aligned} \int _x^\infty e^{-\frac{t^2}{2}} \, dt&= \left( \frac{1}{x} - \frac{1}{x^3} \right) e^{-\frac{x^2}{2}} + 3 \int _x^\infty \frac{1}{t^4} e^{-\frac{t^2}{2}} \, dt. \end{aligned} \end{aligned}$$
(74)

This implies

$$\begin{aligned} \left( \frac{1}{x} - \frac{1}{x^3} \right) e^{-\frac{x^2}{2}}< \int _x^\infty e^{-\frac{t^2}{2}} \, dt < \left( \frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5} \right) e^{-\frac{x^2}{2}}. \end{aligned}$$

Then the volume condition \(\gamma (D_{\omega ,s}) = \phi (s)\) yields

$$\begin{aligned} \left( \frac{1}{a(s)} - \frac{1}{a(s)^3} + \frac{3}{a(s)^5} \right) 2 e^{-\frac{a(s)^2}{2}}> 2\int _{a(s)}^\infty e^{-\frac{t^2}{2}} \, dt = \int _s^\infty e^{-\frac{t^2}{2}} \, dt > \left( \frac{1}{s} - \frac{1}{s^3} \right) e^{-\frac{s^2}{2}} \end{aligned}$$

and therefore we have by (72) that

$$\begin{aligned} \begin{aligned} P_\gamma (D_{\omega ,s}) = 2 e^{-\frac{a(s)^2}{2}}&> \frac{a(s)}{s} e^{-\frac{s^2}{2}} \left( 1 - \frac{1}{s^2} \right) \cdot \left( 1 - \frac{1}{a(s)^2} + \frac{3}{a(s)^4} \right) ^{-1} \\&\ge \frac{a(s)}{s} e^{-\frac{s^2}{2}} \left( 1 - \frac{1}{s^2} \right) \cdot \left( 1 + \frac{1}{a(s)^2} - \frac{3}{a(s)^4} \right) \\&\ge \left( 1 + \frac{\ln 2}{s^2} - \frac{8}{s^4} \right) e^{-\frac{s^2}{2}} . \end{aligned} \end{aligned}$$

\(\square \)

Finally we prove the perimeter bounds in (10).

Lemma 6

Let E be a minimizer of (7). Then it holds

$$\begin{aligned} \frac{5}{6}e^{-\frac{s^2}{2}}\le P_\gamma (E)\le \Bigl ( 1 + \frac{\ln 2}{s^2}\Bigr ) e^{-\frac{s^2}{2}}. \end{aligned}$$

Proof

The bound from above follows by the minimality and by (73):

$$\begin{aligned} P_\gamma (E)\le {\mathcal {F}}(E)\le {\mathcal {F}}(D_{\omega ,s})=P_\gamma (D_{\omega ,s})\le \Bigl ( 1 + \frac{\ln 2}{s^2}\Bigr ) e^{-\frac{s^2}{2}}. \end{aligned}$$

The proof of the lower bound is slightly more difficult. Let \({\bar{s}}\) be such that \(\gamma (E) = \phi ({\bar{s}})\). The value \({\bar{s}}\) has to be non-negative, otherwise \({\mathcal {F}}(E)>{\mathcal {F}}({\mathbb {R}}^n\setminus E)\). If \({\bar{s}}\le s\), then the claim follows easily by the Gaussian isoperimetric inequality. If instead \({\bar{s}}> s\), then again by the isoperimetric inequality we have

$$\begin{aligned} {\mathcal {F}}(E) \ge P_\gamma (E) + \Lambda \sqrt{2\pi }\, (\phi ({\bar{s}}) - \phi (s)) \ge e^{-\frac{{\bar{s}}^2}{2}} + (s + 1) \int _s^{{\bar{s}}} e^{-\frac{t^2}{2}} \, dt. \end{aligned}$$

Define function \(f:[s, \infty ) \rightarrow {\mathbb {R}}\), \(f(x) := e^{-\frac{x^2}{2}} + (s + 1) \int _s^{x} e^{-\frac{t^2}{2}} \, dt\). By differentiating we get

$$\begin{aligned} f'(x) = (-x + s + 1) e^{-\frac{x^2}{2}}. \end{aligned}$$

The function is thus increasing up to \(x = s + 1\) and then decreasing. Denote \({\hat{s}}=s+ \frac{1}{6s}\). Let us show that \(f(x) > {\mathcal {F}}(D_{\omega ,s})\) for every \(x \ge {\hat{s}}\).

Note that \(f'(x) \ge \frac{1}{2} e^{-\frac{s^2}{2}}\) for every \(x \in (s, {\hat{s}})\). Therefore since \(f(s) = e^{-\frac{s^2}{2}}\) we get

$$\begin{aligned} f({\hat{s}}) \ge \left( 1+ \frac{1}{12s}\right) e^{-\frac{s^2}{2}}. \end{aligned}$$

Moreover we have by (74) that

$$\begin{aligned} \lim _{x \rightarrow \infty } f(x) = (s+1)\int _s^\infty e^{-\frac{t^2}{2}} \, dt \ge \left( 1 + \frac{1}{2 s}\right) e^{-\frac{s^2}{2}}. \end{aligned}$$

By the earlier analysis we deduce that for every \(x \ge {\hat{s}}\) it holds

$$\begin{aligned} f(x) \ge \min \{f({\hat{s}}), \lim _{x \rightarrow \infty } f(x) \} > \left( 1 + \frac{\ln 2}{s^2}\right) e^{-\frac{s^2}{2}}. \end{aligned}$$

Hence we conclude by (73) that \(f(x) > P_\gamma (D_{\omega ,s})= {\mathcal {F}}(D_{\omega ,s})\) for every \(x \ge {\hat{s}}\). This in turn implies that necessarily \({\bar{s}} < {\hat{s}}\). By the isoperimetric inequality we then have that

$$\begin{aligned} P_\gamma (E) \ge e^{-\frac{{\hat{s}}^2}{2}}\ge \frac{5}{6} e^{-\frac{s^2}{2}}. \end{aligned}$$

\(\square \)

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Barchiesi, M., Julin, V. Symmetry of minimizers of a Gaussian isoperimetric problem. Probab. Theory Relat. Fields 177, 217–256 (2020). https://doi.org/10.1007/s00440-019-00947-9

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Mathematics Subject Classification

  • 49Q20
  • 60E15