November 03, 2025
The investigation of the volume, surface area, and other geometric properties of sections of convex bodies, and in particular cubes, has a long history and a rich literature. However, much less is known when the cube has a volume distribution that is different from the Lebesgue measure; for example, a Gaussian density. We study the probability densities in the standard cube \(B^n_\infty=[-1,1]^n\) of \(\mathbb{R}^n\) generated by \(e^{-b\|x\|^2}\), \(b> 0\). We prove that the limit of the induced Gaussian-type volume of hyperplane sections of \(B^n_\infty\) through the origin and orthogonal to a main diagonal is \[\sqrt{\frac{b}{\pi}}\left (1-4\frac{e^{-b}\sqrt{b}}{2\sqrt{\pi}\mathop{\mathrm{erf}}(\sqrt{b})}\right)^{-\frac{1}{2}},\] as \(n\to\infty\). This extends the well-known result of Hensley (1979) for the Lebesgue measure and continues the investigations initiated by Barthe, Guédon, Mendelson, Naor (2005), Zvavitch (2008), and König, Koldobski (2013).
Let \(B^n_\infty=[-1,1]^n\) be the standard \(n\)-dimensional cube of edge length \(2\) centred at the origin. We denote by \(\|\cdot\|\) the Euclidean norm in \(\mathbb{R}^n\). Let \(b\geq 0\) be fixed. For every \(x=(x_j)_{j=1}^n\in B_\infty^n\), let \[{\mathrm d}\gamma_n[b](x)=\frac{e^{-b\|x\|^2}}{\left(\int_{-1}^1 e^{-b\sigma^2}\, {\mathrm d}\sigma\right)^n}\, {\mathrm d}x=\frac{\prod_{j=1}^n e^{-bx_j^2}}{\left(\int_{-1}^1 e^{-b\sigma^2}\, {\mathrm d}\sigma\right)^n}\, {\mathrm d}x\] be the Gaussian-type probability density with parameter \(b\) in \(B^n_\infty\). Note that if \(b=0\), then \({\mathrm d}\gamma_n[b](x)=1/2^n{\mathrm d}x\); the uniform density (normalized Lebesgue measure) in \(B^n_\infty\). Let \(S^{n-1}\) be the origin-centred unit sphere, and let \(\langle\cdot,\cdot\rangle\) denote the Euclidean inner product in \(\mathbb{R}^n\). Following König and Koldobsky [@KK13], we introduce the induced \((n-1)\)-measure \(\tilde{\gamma}_n[b]\) of the intersection of \(B^n_\infty\) with the hyperplane \(H(u)=\{x\in \mathbb{R}^n\colon \langle x,u\rangle=0\}\) as follows. For \(u\in S^{n-1}\), let \[A(u, \gamma_n[b])= \widetilde{\gamma}_n[b](B^n_\infty\cap H(u)),\] where \[\widetilde{\gamma}_n[b](B^n_\infty\cap H(u))\colon =\lim_{t\to 0^+} \frac{1}{2t}\gamma_n[b](\{x\in B^n_\infty\colon |\langle x, u\rangle|\leq t\}).\] König and Koldobsky [@KK13]*Proposition 2.1 proved 1 that the \(\widetilde{\gamma}_n[b]\) measure of hyperplane sections orthogonal to a main diagonal \(a=\frac{1}{\sqrt n}(1,\dots,1)\) of \(B^n_\infty\) is given by \[\label{eq:A40a44gamma41} A(a,\gamma_n[b]) = \frac{2}{\pi} \int_0^\infty\left(\frac{\int_0^1\cos(\frac{r}{\sqrt{n}}s)e^{-bs^2}\, {\mathrm d}s}{\int_0^1e^{-b\sigma ^2}\, {\mathrm d}\sigma}\right)^n\, {\mathrm d}r.\tag{1}\]
This result generalizes Ball’s formula for the Lebesgue measure of central sections of the unit cube [@B86], which corresponds to the special case \(b=0\). The origins of Ball’s volume formula trace back to Pólya [@P], see also Bartha, Fodor and González Merino [@BFG21]*(1). Using the volume formula for central sections, Ball showed in [@B86] that the maximal \((n-1)\)-dimensional Lebesgue measure of a hyperplane section of a unit cube is attained precisely when the hyperplane is an \((n-1)\)-dimensional subspace that contains an \((n-2)\)-dimensional face of \(B^n_\infty\); that is, for example, it is parallel to the vector \((1,1,0,\dots,0)\). Ivanov and Tsiutsiurupa [@IvTs], Ambrus [@Am], and Ambrus and Gárgyán [@AmGa] studied different aspects of local maximizers of central sections of the cube, König and Rudelson [@KoRu] and Moody, Stone, Zach and Zvavitch [@MSZZ] investigated non-central sections, and König and Koldobsky [@KK19] dealt with the case of maximizing the surface area. Other aspects of sections of the cube and other convex bodies have recently attracted attention; see, for instance, Abel [@Ul18], Alonso–Gutiérrez, Brazitikos and Chasapis [@ABC25], de Loera, Lopez–Campos and Torres [@LLT24], König [@Kon21; @Kon25], König and Koldobsky [@KonK11], Lonke [@Lo00], Marichal and Mossinghoff [@MaMo08], Meyer and Pajor [@MePa88], Brandenburg and Meroni [@MeBr25], Nayar and Tkocz [@NaTk23], Pournin [@Po1-23; @Po2-23; @Po24].
The problem of finding maximal sections of Gaussian cubes is still open. Barthe, Guédon, Mendelson and Naor [@BGMN05] proved general upper bounds for the measure of central hyperplane sections that work for all \(b\). Zvavitch [@Zvavitch] pointed out that when \(b\) is large enough, the central section of the cube, orthogonal to a main diagonal, has a larger \(\gamma_n[b]\) measure than the section parallel to the vector \((1,1,0,\dots,0)\). König and Koldobsky [@KK13] quantified Zvavitch’s result and proved [@KK13]*Theorem 1.2 that the maximal central sections with respect to the measure \(\gamma_n[b]\) are parallel to the vector \((1,1,0,\dots,0)\) if and only if \(b<\lambda_0\approx 0.1962627\). Notice that when \(b\) is close to \(0\), \(\gamma_n[b]\) in \(B^n_\infty\) is near the Lebesgue measure.
Hensley [@H79] proved that the limit of the sequence of the \((n-1)\)-dimensional volume of central diagonal sections of \(B^n_\infty\) tends to \(\sqrt{6/\pi}\) as \(n\to\infty\); a result he attributed originally to Selberg. König and Koldobsky [@KK19]*Prop. 6(a) showed that the volume of central diagonal sections of \(B^n_\infty\) is upper bounded by \(\sqrt{6/\pi}\). Using Laplace’s method and numerical tools, it was established by Bartha, Fodor and González Merino [@BFG21] that the Lebesgue measure of central sections of \(B^n_\infty\), orthogonal to a main diagonal, form a monotonically increasing sequence for \(n\geq 3\). We refer to Aliev [@Ali20; @Ali08], Borwein, Borwein and Leonard [@BBL], Ron, Ol’hava and Spektor [@KOS] for various properties of the behavior of this sequence. We also note that the volume of central sections can be evaluated explicitly via a closed formula (see Goddard [@Go45], Grimsey [@Gr45], Butler [@But60], Frank and Riede [@FR]; see also [@BFG21]*(2)). For a detailed survey and history on sections of convex bodies, we refer to the paper by Nayar and Tkocz [@NaTk23].
Our main result, Theorem 1, is the exact value of the limit of \(A(a,\gamma_n[b])\) as \(n\) tends to infinity. In particular, ?? extends the result of Hensley regarding the volume of central diagonal sections of \(B^n_\infty\) mentioned above and can be considered a first step in the investigation of the behavior of the sequence \(A(a, \gamma_n[b])\) as \(n\to\infty\).
Let \(\mathop{\mathrm{erf}}(x)=\frac{2}{\sqrt\pi}\int_0^x e^{-t^2}\, {\mathrm d}t\) denote the Gaussian error function for \(x\in[0,\infty)\).
Theorem 1. Let \(b>0\). Then \[\lim_{n\rightarrow\infty}A(a,\gamma_n[b]) = 2\sqrt{\frac{b}{\pi}}\left (1-4\frac{e^{-b}\sqrt{b}}{2\sqrt{\pi}\mathop{\mathrm{erf}}(\sqrt{b})}\right)^{-\frac{1}{2}}.\label{eq:limit}\qquad{(1)}\]
Notice that the expression of \(A(a,\gamma_n[b])\) in ?? satisfies \(\lim_{n\rightarrow \infty }A(a,\gamma_n[0])=\lim_{b\rightarrow 0^+}\lim_{n\rightarrow\infty}A(a,\gamma_n[b])=\sqrt{\frac{6}{\pi}}\), coinciding with the Lebesgue case.
The outline of the proof of Theorem 1 is as follows. We first observe in Section 2 that over every compact interval \([0,R]\) the sequence of functions in the integrand of 1 converges uniformly to an integrable function. Second, in Section 3 we prove that the tails of the sequence of integrals of functions converge to \(0\) within intervals \([R,\infty)\) for sufficiently large \(R>0\). Combining the two estimates, in Section 4 we obtain the exact value of the limit of the Gaussian central diagonal sections of the cube \(B^n_\infty\).
For \(b>0\), \(n\geq 2\) and \(r\in [0,\infty)\), let \[f(r,n,b) := \frac{\int_0^1\cos(\frac{r}{\sqrt{n}}s)e^{-bs^2}\, {\mathrm d}s}{\int_0^1e^{-b\sigma^2}\, {\mathrm d}\sigma}=\frac{\int_0^1\cos(\frac{r}{\sqrt{n}}s)e^{-bs^2}\, {\mathrm d}s}{\frac{\sqrt{\pi } \text{erf}\left(\sqrt{b}\right)}{2 \sqrt{b}}}.\]
Theorem 2. For any fixed \(b>0\) and \(r\in [0,\infty)\), \[\lim_{n\to\infty} f^n(r,n,b) =\exp\left \{\frac{r^2}{4b} \left(\frac{2 \sqrt{b} e^{-b}}{\sqrt{\pi } \mathop{\mathrm{erf}}\left(\sqrt{b}\right)}-1\right)\right\},\] and convergence is uniform in any interval \([0,r_0]\) for any \(r_0>0\).
Proof. Consider the second order Taylor series of \(f(r,n,b)\) with respect to \(r\) centred at \(r=0\), with the Lagrange error term: \[f(r,n,b)=1+\frac{r^2 \left(\frac{2 \sqrt{b} e^{-b}}{\sqrt{\pi } \mathop{\mathrm{erf}}\left(\sqrt{b}\right)}-1\right)}{4 b n} +\frac{\partial^3 f}{\partial r^3}(\xi_{r,n},n,b)\frac{r^3}{3!},\] for some \(\xi_{r,n}\in [0, r]\). Observe that \(\xi_{r,n}\) also depends on \(b\), but since \(b\) is fixed in this argument, we suppress this dependence. By direct calculation, we obtain \[\begin{align} \frac{\partial^3 f}{\partial r^3}(r,n,b)& =\left (\frac{e^{-b} \left(4 \sqrt{b} \sqrt{n} \left(r^2-4 b (b+1) n\right) \sin \left(\frac{r}{\sqrt{n}}\right)-8 b^{3/2} n r \cos \left(\frac{r}{\sqrt{n}}\right)\right)}{\sqrt{\pi }16 b^3 \mathop{\mathrm{erf}}\left(\sqrt{b}\right)n^{\frac{3}{2}}}\right.\\ &\qquad-\left.\frac{r \left(r^2-6 b n\right)f(r,n,b)}{8 b^3 n^{\frac{3}{2}}}\right )\frac{1}{n^{\frac{3}{2}}}\\ &=(h_1(r,n,b)-h_2(r,n,b))\frac{1}{n^{\frac{3}{2}}}, \end{align}\] where \[\begin{align} h_1(r,n,b)&=\frac{e^{-b} \left(4 \sqrt{b} \sqrt{n} \left(r^2-4 b (b+1) n\right) \sin \left(\frac{r}{\sqrt{n}}\right)-8 b^{3/2} n r \cos \left(\frac{r}{\sqrt{n}}\right)\right)}{\sqrt{\pi }16 b^3 \mathop{\mathrm{erf}}\left(\sqrt{b}\right)n^{\frac{3}{2}}},\\ h_2(r,n,b)&=\frac{r \left(r^2-6 b n\right)f(r,n,b)}{8 b^3 n^{\frac{3}{2}}}. \end{align}\] Thus, by the triangle inequality. \[\begin{align} \left |\frac{\partial^3 f}{\partial r^3}(r,n,b)\right |=\left |(h_1(r,n,b)-h_2(r,n,b))\right | \frac{1}{n^{\frac{3}{2}}} \leq \left (\left |(h_1(r,n,b)\right |+\left |h_2(r,n,b))\right |\right ) \frac{1}{n^{\frac{3}{2}}}. \end{align}\]
Let \(r_0\in [0,\infty)\) be arbitrary but fixed. If \(b>0\), \(n\geq 2\) are fixed, then both \(h_1\) and \(h_2\) are bounded on \([0,r_0]\) as they are continuous functions. Furthermore, if \(b>0\) is fixed, then \[\begin{align} |h_1(r,n,b)|&\leq C_1(r_0,b), \quad \text{ and }\\ |h_2(r,n,b)|&\leq C_2(r_0,b) \end{align}\] for all \(n\geq 2\) and \(r\in [0,r_0]\) for some suitable \(C_1(r_0,b)\) and \(C_2(r_0,b)\), meaning that they are uniformly bounded on \([0,r_0]\). The first of these inequalities is clear as the \(\sin x\) and \(\cos x\) functions are bounded. For the second inequality, note that \(f\) is continuous on \([0,r_0]\) for any \(n\geq 2\), and thus bounded. Moreover, for any fixed \(b>0\), the functions \(f(r,n,b)\) are clearly bounded for all \(n\geq 2\) uniformly on \([0,\infty)\). Therefore, \(h_2\) are \(O(n^{-\frac{1}{2}})\) as \(n\to\infty\) on \([0,r_0]\).
In summary, it follows that for any fixed \(b>0\), \(r_0\in [0,\infty)\) and \(r\in [0,r_0]\), \[\left |\frac{\partial^3 f}{\partial r^3}(\xi_{r,n},n,b)\right |\leq (C_1(r_0,b)+C_2(r_0,b))\frac{1}{n^{\frac{3}{2}}}\\ =C(r_0,b)\frac{1}{n^{\frac{3}{2}}}\] for all \(n\geq 2\).
Therefore, for any fixed \(b\), \(r_0\in [0,\infty]\), and \(r\in [0,r_0]\), we obtain \[\begin{align} \lim_{n\to\infty} f^n(r,n,b)&=\lim_{n\to\infty}\left (1+\frac{r^2 \left(\frac{2 \sqrt{b} e^{-b}}{\sqrt{\pi } \mathop{\mathrm{erf}}\left(\sqrt{b}\right)}-1\right)}{4 b n} +\frac{\partial^3 f}{\partial r^3}(\xi_{r,n},n,b)\frac{r^3}{3!}\right )^n\\ &=\lim_{n\to\infty}\left (1+\frac{r^2 \left(\frac{2 \sqrt{b} e^{-b}}{\sqrt{\pi } \mathop{\mathrm{erf}}\left(\sqrt{b}\right)}-1\right)}{4 b n} +C(r_0,b)\frac{r_0^3}{n^{\frac{3}{2}}3!}\right )^n\\ &=\exp\left \{\frac{r^2 \left(\frac{2 \sqrt{b} e^{-b}}{\sqrt{\pi } \mathop{\mathrm{erf}}\left(\sqrt{b}\right)}-1\right)}{4 b }\right \}. \end{align}\] Now, we show that the convergence is uniform in any interval \([0,r_0]\). Let \[a_n(u)=\left (1+\frac{u}{n}+\frac{C}{n^{\frac{3}{2}}}\right )\] for some constant \(C\). Then \[\log a_n(u)=n\log \left (1+\frac{u}{n}+\frac{C}{n^{\frac{3}{2}}}\right ).\] Let \(x_n(u)=\frac{u}{n}+\frac{C}{n^{\frac{3}{2}}}\). Using the Taylor series of \(\log (1+x)\) at \(x=0\), we obtain \[\log a_n(u)=n\left (x_n(u)+\frac{x_n^2(u)}{2}+O\left (x_n^2(u)\right )\right )=u+\frac{C}{\sqrt n}+O\left (\frac{1}{n}\right ).\] Now, by exponentiation and using the Taylor series of \(e^x\) at \(x=0\), we obtain \[a_n(u)=e^u e^{\frac{C}{\sqrt n}+O\left (\frac{1}{n}\right )}=e^u \left (1+\frac{C}{\sqrt n}+O\left (\frac{1}{n}\right )\right ).\] Thus, \[\left |a_n(u)-e^u\right |=e^u \left |\frac{C}{\sqrt n}+O\left (\frac{1}{n}\right )\right |\leq e^{u_0} \left |\frac{C}{\sqrt n}+O\left (\frac{1}{n}\right )\right |,\] from which the uniform convergence on \([0,u_0]\) follows. ◻
Theorem 3. Let \(b>0\), \(t\geq 0\). Then for every \(\varepsilon>0\), there exist \(R(\varepsilon)>0\) and \(n(\varepsilon)\in\mathbb{N}\) such that \[\int_{R(\varepsilon)}^\infty |f(r,n,b)^n|\, {\mathrm d}r \leq \varepsilon\] for every \(n\geq n(\varepsilon)\).
For every \(b>0\) and \(t\geq 0\) let \[G_b(t):=f(t,1,b)=\frac{\int_0^1\cos(ts)e^{-bs^2}\, {\mathrm d}s}{\int_0^1e^{-b\sigma ^2}\, {\mathrm d}\sigma}=\int_0^1\cos(ts)\, {\mathrm d}\mu_b(s) =\frac{F_b(t)}{N_b},\] where \[{\mathrm d}\mu_b(s):= \frac{e^{-bs^2}\, {\mathrm d}s}{\int_0^1e^{-b\sigma^2}\, {\mathrm d}\sigma} \mathbf{1}_{[0,1]}(s),\quad F_b(t):=\int_0^1\cos(ts)e^{-bs^2}\, {\mathrm d}s,\] and \[N_b:=\int_0^1e^{-b\sigma^2}\, {\mathrm d}\sigma.\]
Notice that \(G_b(t)\) is a multiple of the real part of the Fourier transform of the Gaussian \(e^{-bs^2}\). The Riemann–Lebesgue Lemma ensures \(G_b(t)\leq O(\frac{1}{t})\). However, the actual constant also depends on \(b\), and we need a more precise estimate for our arguments.
Lemma 1. Let \(b>0\) and \(t\geq 0\). Then \[|G_b(t)| \leq \begin{cases} \frac{e}{t}, & \text{ if }b\in(0,1),\\ 1.34\frac{\sqrt{b}}{t}, & \text{ if }b\geq 1. \end{cases}\]
Proof. We start bounding \(F_b(t)\) from above. We integrate by parts with \(u=e^{-bs^2}\) and \({\mathrm d}v=\cos(ts){\mathrm d}s\). Thus, \({\mathrm d}u=-2bse^{-bs^2}{\mathrm d}s\), \(v=\frac{\sin(ts)}{t}\) and \[\begin{align} \left | F_b(t)\right |&= \left | \frac{1}{t} \left (e^{-bs^2}\sin (ts) \right)_0^1+\frac{2b}{t}\int_0^1 s e^{-bs^2}\sin (ts)\, {\mathrm d}s\right |\\ &\leq \left | \frac{1}{t} e^{-b}\sin (t) \right |+\frac{2b}{t}\left |\int_0^1 s e^{-bs^2}\sin (ts)\, {\mathrm d}s\right |\\ &\leq\frac{1}{t} e^{-b}+\frac{2b}{t}\int_0^1 s e^{-bs^2}\, {\mathrm d}s\\ &= \frac{1}{t} e^{-b}-\frac{2b}{t}\frac{e^{-b}-1}{2b}\, {\mathrm d}s=\frac{1}{t}. \end{align}\]
Next, we bound \(N_b\) from below. Notice that \(N_b\) is a strictly monotonically decreasing function of \(b\).
If \(b\geq 1\), then using the substitution \(u=\sqrt b \sigma\), \({\mathrm d}u=\sqrt b\, {\mathrm d}\sigma\), we get \[\begin{align} N_b=\frac{1}{\sqrt b}\int_0^{\sqrt b} e^{-u^2}\, {\mathrm d}u\geq \frac{1}{\sqrt b}\int_0^{1} e^{-u^2}\, {\mathrm d}u=\frac{1}{\sqrt b}N_1\geq \frac{0.7468}{\sqrt b}, \end{align}\] where we used \(b\geq 1\). Thus, \[N_b\geq \begin{cases} 0.7468>e^{-1}, & \text{ if } b\in (0,1),\\ \frac{0.7468}{\sqrt b}, & \text{ if } b\geq 1. \end{cases}\] Since \(G_b(t)=F_b(t)/N_b\), the statement of the lemma follows. ◻
Lemma 2. Let \(t\geq 0\). Then \[|G_b(t)|\leq \begin{cases} 1-\frac{1}{16}t^2, & \text{ if } b\in (0,1) \text{ and } t\leq 2.75\\ 1-\frac{1}{25b}t^2, & \text{ if } b\geq 1 \text{ and } t\leq 1.4\sqrt b. \end{cases}\]
Proof. Notice that \(\cos x\leq 1-\frac{x^2}{4}\) for \(x\in [0,2.75]\). If \(s\in (0,1)\) and \(t\leq 2.75\), or \(b\geq 1\), \(t\leq 1.4\sqrt b\) and \(s\leq 1/\sqrt b\), then \[\cos (ts)\leq 1-\frac{t^2s^2}{4}.\] Now we split the domain of integration in \(G_b(t)\) at \(1/\sqrt{b}\) (note that if \(b\geq 1\) then the second term is void as \(\mu_b(s)\) is supported on \([0,1]\)) \[\begin{align} |G_b(t)|&=\left |\int_0^{\frac{1}{\sqrt b}} \cos(ts)\, {\mathrm d}\mu_b(s)+\int_{\frac{1}{\sqrt b}}^1 \cos(ts)\, {\mathrm d}\mu_b(s) \right |\\ &\leq \left |\int_0^{\frac{1}{\sqrt b}} 1-\frac{t^2s^2}{4}\, {\mathrm d}\mu_b(s) \right |+\int_{\frac{1}{\sqrt b}}^1 \left |\cos(ts)\right |\, {\mathrm d}\mu_b(s)\\ &\leq 1-\frac{t^2}{4}\int_0^{\frac{1}{\sqrt b}} s^2\, {\mathrm d}\mu_b(s). \end{align}\]
If \(b\in (0,1)\), then \[|G_b(t)|\leq 1-\frac{t^2}{4}\int_0^1 s^2\, {\mathrm d}\mu_b(s)\] We now use that the moment of second order \[\int_0^1 s^2\, {\mathrm d}\mu_b(s) = \frac{\int_0^1s^2e^{-bs^2}\,{\mathrm d}s}{\int_0^1e^{-b\sigma^2}\,{\mathrm d}\sigma}\] of \(e^{-bs^2}\) is decreasing in \(b\). Indeed, let \[B(b):=\int_0^1s^2e^{-bs^2}\, {\mathrm d}s \quad\text{and}\quad R(b):=\frac{B(b)}{N_b}.\] Differentiating \(R(b)\) with respect to \(b\) we obtain \[R'(b) = \frac{B(b)^2-N_bC(b)}{N_b^2},\] where \(C(b):=\int_0^1s^4e^{-bs^2}\, {\mathrm d}s\). The Cauchy-Schwarz inequality yields \[C(b)^2 = \left(\int_0^1e^{-\frac{bx^2}{2}}\cdot (x^2e^{-\frac{bx^2}{2}})\, {\mathrm d}x\right)^2 \leq \int_0^1 e^{-bx^2}\, {\mathrm d}x \int_0^1 x^4e^{-bx^2}\, {\mathrm d}x = N_bC(b).\] Hence \(R'(b)<0\) and therefore \(R(b)\) is decreasing in \(b\). Thus, we obtain \[|G_b(t)| \leq 1-\frac{t^2}{4} \int_0^1 s^2\, {\mathrm d}\mu_1(s) \leq 1-\frac{0.25}{4}t^2.\]
If \(b\geq 1\), then \(1/\sqrt{b}\leq 1\) and we consider \[\int_0^\frac{1}{\sqrt{b}} s^2\, {\mathrm d}\mu_b(s) = \frac{\int_0^\frac{1}{\sqrt{b}}s^2e^{-bs^2}\, {\mathrm d}s}{\int_0^1e^{-bs^2}\, {\mathrm d}s}.\] We bound the numerator from below and the denominator from above. Regarding the former, since \(s\leq1/\sqrt{b}\), then \(e^{-bs^2} \geq e^{-1}\), and thus \[\int_0^\frac{1}{\sqrt{b}}s^2e^{-bs^2}\, {\mathrm d}s \geq e^{-1}\int_0^\frac{1}{\sqrt{b}}s^2\, {\mathrm d}s = \frac{e^{-1}}{3b^{\frac{3}{2}}}.\] For the denominator, we obtain \[\int_0^1e^{-b\sigma^2}\, {\mathrm d}\sigma \leq \int_0^\infty e^{-b\sigma^2}\, {\mathrm d}\sigma = \frac{\sqrt{\pi}}{2\sqrt{b}}.\] Therefore, \[\int_0^\frac{1}{\sqrt{b}} s^2\, {\mathrm d}\mu_b(s) \geq \frac{2e^{-1}}{3\sqrt{\pi}b},\] and hence \[\begin{align} |G_b(t)| & \leq 1-\frac{t^2}{4}\frac{2e^{-1}}{3\sqrt{\pi}b} \leq 1-\frac{1}{25}\frac{t^2}{b}. \end{align}\] ◻
Proof of Theorem 3. We start with proving the case \(b\in(0,1)\). Let \(R>0\) and assume that \(2.73\sqrt n\geq R\). We split the expression 1 into \[\label{eq:b01Integral} \int_R^\infty \left|f(r,n,b)^n\right|\, {\mathrm d}r = \int_R^{2.73\sqrt{n}} \left|f(r,n,b)^n\right|\, {\mathrm d}r + \int_{2.73\sqrt{n}}^\infty \left|f(r,n,b)^n\right|\, {\mathrm d}r.\tag{2}\] We start with bounding the first term. Since \(t=\frac{r}{\sqrt{n}}\leq 2.73\), by Lemma 2 we have that \[\begin{align} \int_R^{2.73\sqrt{n}} \left|f(r,n,b)^n\right|\, {\mathrm d}r & \leq \int_R^{2.73\sqrt{n}} \left(1-\frac{1}{16}\frac{r^2}{n}\right)^n\, {\mathrm d}r \\ & \leq \int_R^{2.73\sqrt{n}} e^{-\frac{1}{16}r^2}\, {\mathrm d}r. \end{align}\] Evidently, for every \(\varepsilon>0\), we can choose \(R(\varepsilon)>0\) large enough so that \[\int_{R(\varepsilon)}^{\infty} e^{-\frac{1}{16}r^2}\, {\mathrm d}r \leq \frac{\varepsilon}{2}.\] We now bound the second term. Since \(t=\frac{r}{\sqrt{n}} \geq 0\), Lemma 1 yields \[\int_{2.73\sqrt{n}}^\infty \left|f(r,n,b)^n\right|\, {\mathrm d}r \leq \int_{2.73\sqrt{n}}^\infty \left(\frac{e}{r/\sqrt{n}}\right)^n\, {\mathrm d}r = \frac{2.73\sqrt{n}}{n-1}\left(\frac{e}{2.73}\right)^n.\] Again, since \(e/2.73<1\) there exists \(n(\varepsilon)\in\mathbb{N}\) such that \[\frac{2.73\sqrt{n}}{n-1}\left(\frac{e}{2.73}\right)^n\leq \frac{\varepsilon}{2}\] for every \(n\geq n(\varepsilon)\). Joining the left and the right estimates tells us that 2 is upper bounded by \(\varepsilon\) when choosing \(R:=R(\varepsilon)\) and for every \(n\geq n(\varepsilon)\) with \(n(\varepsilon)\geq \frac{R(\varepsilon)^2}{2.73^2}\), concluding the case \(b\in(0,1)\).
We now finish with the case \(b \geq 1\). Let \(R>0\) be the fixed lower value of the integral domain and assume that \(1.37\sqrt b\sqrt n\geq R\). We then split the expression 1 onto \[\label{eq:b1InftyIntegral} \int_R^\infty \left|f(r,n,b)^n\right|\, {\mathrm d}r = \int_R^{1.37\sqrt{b}\sqrt{n}} \left|f(r,n,b)^n\right|\, {\mathrm d}r + \int_{1.37\sqrt{b}\sqrt{n}}^\infty \left|f(r,n,b)^n\right|\, {\mathrm d}r.\tag{3}\] We start bounding the left hand side. Since \(t=\frac{r}{\sqrt{n}}\leq 1.37\sqrt{b} < 1.4\sqrt{b}\), by Lemma 2 we have that \[\begin{align} \int_R^{1.37\sqrt{b}\sqrt{n}} \left|f(r,n,b)^n\right|\, {\mathrm d}r & \leq \int_R^{1.37\sqrt{b}\sqrt{n}} \left(1-\frac{1}{25}\frac{r^2}{bn}\right)^n\, {\mathrm d}r \\ & \leq \int_R^{1.37\sqrt{b}\sqrt{n}} e^{-\frac{1}{25b}r^2}\, {\mathrm d}r. \end{align}\] Evidently, for every \(\varepsilon>0\) and fixed \(b\geq 1\), we can choose \(R(\varepsilon)>0\) large enough so that \[\int_{R(\varepsilon)}^{\infty} e^{-\frac{1}{25b}r^2}\, {\mathrm d}r \leq \frac{\varepsilon}{2}.\] We now proceed with the second term. Notice that \(t=\frac{r}{\sqrt{n}} \geq 1.37\sqrt{b} \geq\sqrt{b}\) and thus Lemma 1 we get that \[\int_{1.37\sqrt{b}\sqrt{n}}^\infty \left|f(r,n,b)^n\right|\, {\mathrm d}r \leq \int_{1.37\sqrt{b}\sqrt{n}}^\infty (1.34\sqrt{b})^n\frac{\sqrt{n}^n}{r^n}\, {\mathrm d}r = \frac{1.37\sqrt{b}\sqrt{n}}{n-1}\left(\frac{1.34}{1.37}\right)^n.\] Again, since \(1.34/1.37<1\) and \(b\geq 1\) is fixed, there exists \(n(\varepsilon)\in\mathbb{N}\) such that \[\frac{1.37\sqrt{b}\sqrt{n}}{n-1}\left(\frac{1.34}{1.37}\right)^n\leq \frac{\varepsilon}{2}\] for every \(n\geq n(\varepsilon)\). Joining the left and the right estimates tells us that 3 is upper bounded by \(\varepsilon\) when choosing \(R:=R(\varepsilon)\) and for every \(n\geq n(\varepsilon)\) with \(n(\varepsilon)\geq \frac{R(\varepsilon)^2}{1.37^2b}\), concluding the case \(b\geq 1\) and the theorem. ◻
We now show Theorem 1. Let us define \[g_b(r):=\exp\left \{\frac{r^2}{4b} \left(\frac{2 \sqrt{b} e^{-b}}{\sqrt{\pi } \text{erf}\left(\sqrt{b}\right)}-1\right)\right\}.\] For a suitable \(R>0\), let us write \[\begin{align} & \left|\int_0^\infty f^n(r,n,b)\, {\mathrm d}r - \int_0^\infty g_b(r)\, {\mathrm d}r\right| \\ & = \left|\int_0^R f^n(r,n,b)\, {\mathrm d}r + \int_R^\infty f^n(r,n,b)\, {\mathrm d}r - \int_0^R g_b(r)\, {\mathrm d}r-\int_R^\infty g_b(r)\, {\mathrm d}r \right| \\ & \leq \int_0^R \left|f^n(r,n,b)- g_b(r)\right|\, {\mathrm d}r + \int_R^\infty \left|f^n(r,n,b)\right|\, {\mathrm d}r+\left|\int_R^\infty g_b(r)\, {\mathrm d}r\right|. \end{align}\] First, notice that by Theorem 3, there exists \(R_1(\varepsilon)>0\) and \(n_1(\varepsilon)>0\) such that the second term is not larger than \(\frac{\varepsilon}{3}\) when choosing \(R:=R_1(\varepsilon)\) and every \(n\geq n_1(\varepsilon)\). Second, it is clear that \[\int_R^\infty g_b(r) dr= \int_R^\infty \exp\left \{\frac{r^2}{4b} \left(\frac{2 \sqrt{b} e^{-b}}{\sqrt{\pi } \text{erf}\left(\sqrt{b}\right)}-1\right)\right\}\, {\mathrm d}r \leq \frac{\varepsilon}{3}\] for some large enough \(R=R_2(\varepsilon)\). Take \(R:=\max\{R_1(\varepsilon),R_2(\varepsilon)\}\), so that the second and third terms are not larger than \(\frac{\varepsilon}{3}\) each whenever \(n\geq n_1(\varepsilon)\) (notice that the third term do not depend on \(n\) anyways). Third, notice that Theorem 2 ensures the uniform convergence of \(f^n(r,n,b)\) to \(g_b(r)\) on \(r\in[0,R]\), for every \(R>0\) when \(n\rightarrow\infty\). Thus, for our choice of \(R\), there exists \(n_2(\varepsilon)\in\mathbb{N}\) such that for every \(n\geq n_2(\varepsilon)\) then the first term in the equation above is not larger than \(\frac{\varepsilon}{3}\). Letting \(n_0:=\max\{n_1(\varepsilon),n_2(\varepsilon)\}\), we conclude that for every \(n\geq n_0\) then \[\left|\int_0^\infty f^n(r,n,b)\, {\mathrm d}r - \int_0^\infty g_b(r)\, {\mathrm d}r\right|\leq\varepsilon,\] concluding the proof.
F. Fodor’s research was supported by NKFIH project no. 150151. Project no. 150151 has been implemented with the support provided by the Ministry of Culture and Innovation of Hungary from the National Research, Development and Innovation Fund, financed under the ADVANCED_24 funding scheme.
B. González Merino was partially supported by Ministerio de Ciencia, Innovación y Universidades project PID2022-136320NB-I00/AEI/10.13039/501100011033/FEDER, UE.
We note that the product measure \(\gamma_n[b]\) in [@KK13] misses a factor of \(2\) and, as a result, the formula in Proposition 2.1 there also misses a factor of \(2\).↩︎