# Average measure of intersection of a convex region with its translate

Let $\lambda$ denote the Lebesgue-measure on $\mathbb{R}^n$, and let $C\subset\mathbb{R}^n$ be a convex region.

My question is about $$f(C):=\int_{C} \lambda(C \cap (x + C) ) \mathrm{d} x.$$ How large can $f(C)$ be? Of course, there is the trivial bound $f(C)\leq \lambda(C)^2$ but I would expect more something like $f(C) = O( \lambda(C) )$ at least. This question has probably been answered somewhere, and I suspect that the following might be very well known: for regions $C$ of constant measure, the value of $f(C)$ is maximal if $C$ is an $n$-ball (or something similar). I couldn't find any useful references regarding this question. If someone could help me out with some reference, or even a simple solution, I would be very grateful.

• $O(\lambda(C))$ estimate can not hold and it has not even physical meaning, since two expressions are homogeneous with different exponents. Oct 2, 2017 at 13:53
• @FedorPetrov Yes, you're right. I completely missed that. It would be nice to know what the extremal case looks like still.
– zref
Oct 2, 2017 at 15:49
• For $C$ a circle, I calculate $f(C)/\lambda(C)^2=1-3\sqrt{3}/4\pi$, roughly 0.5865 Oct 3, 2017 at 10:44
• For the ball in $R^n$, the corresponding formula seems to be $$\frac{\int_{0}^1 s^{n-1} V_n^{cap}(1,s/2)ds}{V_n^{cap}(1,0)/n}$$ where $V$ is as in math.stackexchange.com/a/162873/210229 Oct 3, 2017 at 12:37
• Looks like the optimal shapes are balls centered at the origin. Indeed, for 3 intervals $I,J,K\subset\mathbb R$ of given lengths, the integral $\int_I|J\cap(x+K)|\,dx$ is maximized when $I,J,K$ are all centered at the origin, so the Steiner symmetrization should do the job. On the other hand, I can easily miss something at 3AM, so it would be nice if someone else could verify (or refute) this claim :-) Oct 4, 2017 at 7:06

The following proposition answers OP's question regarding the upper bound of $$\tau(C) \Doteq f(C)/\lambda^2(C).$$

Let $$B_n$$ be the closed Euclidean unit ball of $$\mathbb{R^n}$$ centred at $$0$$, that is $$B_n = \{ (x_1,\dots, x_n) \in \mathbb{R^n} \, \vert \,\, x_1^2 + \cdots + x_n^2 \le 1\},$$ and let $$\tau_n = \tau(B_n)$$.

Proposition. The Euclidean unit ball $$B_n \subset \mathbb{R}^n$$ maximizes $$\tau(C)$$ among all convex subsets of finite and positive Lebesgue measure, i.e., $$\tau(C) \le \tau_n$$ holds for any such $$C$$.

Claim 3 below gives a way to compute efficiently $$\tau_n$$ for all $$n > 0$$.

The proof of the proposition follows Fedja's idea. The first part consists in proving that $$\tau$$ doesn't decrease after [Steiner symmetrization][2]. This is Claim 4 below. The second part uses [1, Theorem 7.1.10] showing that some Euclidean ball centred at $$0$$ can be obtained as the limit of Steiner symmetrizations of $$C$$ for any compact convex $$C$$ of positive Lebesgue measure (equivalently, with non-empty interior).

When $$n = 1$$, the result is slightly stronger.

Claim 1. Let $$C \subset \mathbb{R}$$ be a convex subset of finite Lebesgue measure $$\lambda(C) > 0$$. Then $$\tau(C) \le \frac{3}{4}$$ and equality holds if and only if $$C$$ is centred at $$0$$.

So, the above claim settles OP's question when $$n = 1$$. Unsurprisingly, the result generalizes to $$\mathbb{R}^n$$ in the following way.

Claim 2. Let $$C \subset \mathbb{R}^n$$ be a parallelotope of positive Lebesgue measure. Then $$\tau(C) \le (\frac{3}{4})^n$$ and equality holds if and only if $$C$$ is centred at $$0$$.

Computing the maximum $$\tau_n$$ of $$\tau$$ can be achieved via the formula

Claim 3. We have $$\tau_n = \frac{3}{2} \frac{\int^{\pi/3}_0 \sin^n(\theta) d\theta}{\int^{\pi/2}_0 \sin^n(\theta) d\theta}.$$ In particular, $$\tau_1 = \frac{3}{4}, \, \tau_2 = 1 - \frac{3 \sqrt{3}}{4 \pi}, \tau_3 = \frac{15}{32}, \, \tau_4 = 1 - \frac{9 \sqrt{3}}{8 \pi}, \tau_5 = \frac{159}{512}, \, \tau_6 = 1 - \frac{27 \sqrt{3}}{20 \pi}$$.

The above formula can be rephrased as $$\tau_n = \frac{3}{2}\frac{W_n'}{W_n}$$ where $$W_n$$ is the $$n$$-th [Wallis' integral][3] and $$W_n'$$ is recursively defined by $$W'_0 = \frac{\pi}{3}, \,W_1' = \frac{1}{2}$$ and $$W_n' = - \frac{1}{2n}(\frac{\sqrt{3}}{2})^{n - 1} + \frac{n - 1}{n}W'_{n - 2}$$. The latter formula allows an efficient computation of $$\tau_n$$ for any $$n > 0$$. As $$W_n' \le \frac{\pi}{3}(\frac{\sqrt{3}}{2})^n$$ and $$W_n \sim \sqrt{\frac{\pi}{2n}}$$, the sequence $$(\tau_n)_n$$ converges to $$0$$ exponentially fast (we have for instance $$10^{-7} < \tau_{91} < 10^{-6}$$).

Before proving the claims, let us state some obvious, but important, remarks.

Remark 1. Let $$A \in GL_n(\mathbb{R})$$ and $$d \in \mathbb{R}^n$$. Let $$C$$ be any subset of $$\mathbb{R}^n$$. Then we have $$f(AC) = \vert \det(A) \vert^2f(C), \quad f(C + d) = \int_C \lambda\left(C \cap (C + d + x) \right)d\lambda(x).$$

In particular the ratio $$\tau(C)$$ is $$GL_n(\mathbb{R})$$-invariant but it is not invariant under translation. Indeed, if $$C$$ is a convex subset of finite positive Lebesgue measure whose interior contains $$0$$, then the function $$f_C(d): d \mapsto f(C + d)$$ has a non-empty and bounded support.

The following tightly relates to OP's question.

Question. Let $$C \subset \mathbb{R}^n$$ be a convex subset of finite positive Lebesgue measure. For which $$d$$ is $$f_C(d)$$ maximal?

The following relates $$\tau$$ to Steiner symmetrization.

Claim 4. Let $$C \subset \mathbb{R}^n$$ be a convex subset of finite positive Lebesgue measure. Let $$H =\{0\} \times \mathbb{R}^{n -1}$$ and let $$C' \Doteq \text{St}_H(C)$$ be the [Steiner symmetrization][2] of $$C$$ with respect to $$H$$. Then we have $$\tau(C) \le \tau(C')$$.

We now turn to the proofs of the above claims.

Proof of Claim 1. Because of Remark 1, we can restrict, without loss of generality, to intervals of the form $$[-1, 1] + d$$ with $$d \ge 0$$. A direct computation yields $$f([-1, 1] + d) = \int_{-1}^1 (2 - \vert d + x \vert)^{+} dx = \left\{\begin{array}{cc} 3 -d^2 & \text{ if } 0 \le d \le 1, \\ \frac{1}{2}(3 - d)^2 & \text{ if } 1 \le d \le 3, \\ 0 & \text{ if } d \ge 3. \end{array}\right.$$ where $$g^{+} \Doteq \max(g ,0)$$ for $$g$$ a real-valued function. The result follows.

Proof of Claim 2. By Remark 1, we can restrict to the convex sets of the form $$C = [-1, 1]^n + d$$ with $$d = (d_1, \dots, d_n) \in \mathbb{R}^n$$. We have $$f(C) = \int_{[-1, 1]^n} (2 - \vert x_1 + d_1 \vert)^{+} \cdots (2 - \vert x_n + d_n\vert)^{+}dx_1 \cdots dx_n$$. It follows from Claim 1, that $$f(C) \le \int_{[-1, 1]^n} (2 - \vert x_1 \vert)^{+} \cdots (2 - \vert x_n \vert)^{+}dx_1 \cdots dx_n = 3^n$$ and the inequality is strict whenever one of the $$d_i$$'s is not zero. The result follows.

We denote by $$V_n(r)$$ the Lebesgue measure of the Euclidean ball of radius $$r \ge 0$$, that is, $$V_n(r) = r^n\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2} + 1)}$$, and let $$\nu_n = V_n(1) = \lambda(B_n)$$. Given a real number $$h$$ such that $$-1 \le h \le 1$$, we denote by $$B_n^{\text{cap}}(h)$$ the hyperspherical cap of height $$h$$ along $$x_n$$ , that is the set of points $$x = (x_1, \dots, x_n) \in B_n$$ such that $$x_n \ge h$$. We also set $$V_n^{\text{cap}}(h) = \lambda(B_n^{\text{cap}}(h))$$.

Proof of Claim 3. By definition, we have $$f(B_n) = \int_{B_n} \lambda(B_n \cap (B_n + x)) d\lambda(x) = 2 \int_{B_n} V_n^{\text{cap}}(\frac{\vert x \vert}{2}) d\lambda(x)$$ where $$\vert \cdot \vert$$ stands for the Euclidean norm of $$\mathbb{R}^n$$. Changing to [$$n$$-dimensional spherical coordinates][4], we get $$f(B_n) = 2 \int_{D_n} V_n^{\text{cap}}(\frac{r}{2})r^{n - 1} \sin^{n - 2}(\phi_1)\sin^{n - 3}(\phi_2) \cdots \sin(\phi_{n - 1}) dr d\phi_1 d\phi_2 \cdots d \phi_{n - 1}$$ where $$D_n = [0,2\pi] \times [0, \pi] \times \cdots \times [0, \pi] \times [0, 1]$$. Using the classical computation of $$\nu_n = \lambda(B_n)$$, we deduce that $$f(B_n) = 2n\nu_n \int_{[0, 1]} V_n^{\text{cap}}(\frac{r}{2})r^{n - 1}dr = 2^{n + 1}n\nu_n \int_{[0, \frac{1}{2}]} V_n^{\text{cap}}(s)s^{n - 1}ds .$$ Since $$V_n^{\text{cap}}(s) = \int_s^1 V_{n - 1}(\sqrt{1 - h^2})dh = \nu_{n - 1} \int_0^{\text{arcos}(s)} \sin^n(\theta)d \theta$$, we obtain $$f(B_n) = 2^{n + 1}n\nu_n \nu_{n - 1} \int_{[0, \frac{1}{2}]} J_n(\text{arcos}(s))s^{n - 1}ds$$ where $$J_n(\phi) = \int_0^{\phi}\sin^n(\theta)d\theta$$. Using the substitution $$\theta = \text{arcos}(s)$$, we get $$\begin{eqnarray} \frac{f(B_n)}{2^{n + 1}n\nu_n \nu_{n - 1}} &=& \int_{[\frac{\pi}{3}, \frac{\pi}{2}]} J_n(\theta)\cos(\theta)^{n - 1} \sin(\theta)d\theta \\ &=& \frac{[J_n(\theta) \cos^n(\theta)]^{\frac{\pi}{3}}_{\frac{\pi}{2}} + \int^{\frac{\pi}{2}}_{\frac{\pi}{3}} (\cos(\theta)\sin(\theta))^n d\theta}{n} \\ &=& \frac{2J_n(\frac{\pi}{3}) + \int^{\pi}_{\frac{2\pi}{3}} \sin(\theta)^n d\theta}{2^{n + 1} n} \\ &=& \frac{3J_n(\frac{\pi}{3})}{2^{n + 1} n}. \end{eqnarray}$$ Now the result follows from the identity $$\frac{\nu_{n - 1}}{\nu_n} = \frac{1}{2 W_n}$$.

Claim 4 is a specialization of

Fedja's Lemma. Let $$\mathcal{I}(C, D, E) \Doteq \int_C \lambda(D \cap (E + x)) d\lambda(x)$$ where $$C$$, $$D$$ and $$E$$ are convex subsets of $$\mathbb{R}^n$$ of positive finite Lebesgue measure. Then $$\mathcal{I}(C, D, E) \le \mathcal{I}(C', D', E')$$, where $$X'$$ denotes the Steiner symmetrization of $$X$$ with respect to $$H = \{0\} \times \mathbb{R}^{n - 1}$$.

The proof of Fedja's Lemma can be reduced to the case $$n = 1$$. The proof of the case $$n = 1$$ is eased off by

Lemma 1. Let $$g: \mathbb{R} \rightarrow \mathbb{R}^{+} \cup \{0\}$$ be an even non-negative function which is concave on its support and non-decreasing on the negative real numbers. Then we have $$\int_I g d\lambda \le \int_{I'} g d\lambda$$ for every convex subset $$I$$ of $$\mathbb{R}$$ where $$I'$$ is the closed interval of measure $$\lambda(I)$$ centred at $$0$$.

Proof of Lemma 1. The non-negative number $$\int_I g d\lambda$$ is the area of convex subset $$C(I) \subset \mathbb{R}^2$$ located between the $$x$$-axis and the graph of $$g$$. By hypothesis, we have $$C(I)' \subset C(I') = \int_{I'} g d\lambda$$, where $$C(I)'$$ is Steiner symmetrization of $$C(I)$$ with respect to $$H = \{0\} \times \mathbb{R}$$. As $$C(I)$$ and $$C(I)'$$ have the same area, the result follows.

We are now in position to prove Fedja's Lemma.

Proof of Fedja's Lemma. Let us assume first that $$n = 1$$. As $$\mathcal{I}(C + x, D, E) = \mathcal{I}(C, D, E + x)$$ and $$\mathcal{I}(C, D + x, E + x) = \mathcal{I}(C, D, E)$$, we can assume, without loss of generality, that $$C$$ and $$D$$ are centred at zero. Let us write $$E = E’ + \eta$$ and let $$g(x) \Doteq \lambda(D \cap (E' + x))$$. Then we have $$g(x) = \left\lbrace \begin{array}{ccc} \min(d, e) & \text{ if } \vert x \vert \le \frac{ \vert d - e \vert }{2}, \\ \frac{d + e}{2} - \vert x \vert & \text{ if } \frac{ \vert d - e \vert }{2} \le \vert x \vert \le \frac{d + e}{2}, \\ 0 & \text{ if } \vert x \vert \ge \frac{d + e}{2}, \end{array}\right.$$ where $$d = \lambda(D) > 0$$ and $$e = \lambda(E) > 0$$. The function $$g$$ satisfies the hypothesis of Lemma $$1$$, thus $$\mathcal{I}(C, D, E) = \mathcal{I}(C + \eta, D, E’) = \int_{C + \eta} g(x) d\lambda(x) \le \int_{C} g(x) d\lambda(x) = \mathcal{I}(C, D, E’)$$ which proves the result for $$n = 1$$. Assume now that $$n > 1$$. We denote by $$\lambda_{n - 1}$$ the Lebesgue measure on $$\mathbb{R}^{n - 1}$$ and we identify $$H$$ with $$\mathbb{R}^{n - 1}$$. For $$\xi \in \mathbb{R^n}$$, we write $$\xi = (\xi_1, \zeta)$$, with $$\xi_1 \in \mathbb{R}$$ and $$\zeta \in H$$. Given $$A \subset \mathbb{R}^n$$ and $$\zeta \in H$$, we set $$A^{\zeta} \Doteq \{ \xi_1 \, \vert \, (\xi_1, \zeta) \in A\} \subset \mathbb{R}$$. Let $$x = (x_1, z) \in \mathbb{R} \times H$$. Observe that $$(A^{\zeta})’ = (A’)^{\zeta}$$ and $$(A + x)^{\zeta} = A^{\zeta - z} + x_1$$. By the Fubini-Tonelli Theorem, we have $$\lambda_n(D \cap (E + x)) = \int_H \lambda_1(D^{\zeta} \cap (E^{\zeta - z} + x_1)) d \lambda_{n - 1}(\zeta)$$. We shall use the abbreviations $$dz = d \lambda_{n - 1}(z)$$ and $$d\zeta = d \lambda_{n - 1}(\zeta)$$. Using Fubini-Tonelli’s Theorem and the case $$n = 1$$, we obtain $$\begin{eqnarray} \mathcal{I}(C, D, E) & = & \int_{H^2} \mathcal{I}(C^{z}, D^{\zeta}, E^{\zeta - z})d\zeta dz \\ & \le & \int_{H^2} \mathcal{I}((C’)^{z}, (D’)^{\zeta}, (E’)^{\zeta - z}) d\zeta dz \\ & = & \mathcal{I}(C’, D’, E’). \end{eqnarray}$$

Eventually, we can prove the proposition.

Proposition's proof. By [1, Theorem 7.1.10], there is a Euclidean ball $$B$$ centred at zero which is the limit in the Hausdorff distance topology of Steiner symmetrizations of $$C$$ with respect to linear hyperplanes. As $$\lambda$$, $$f$$ and hence $$\tau$$ are continuous with respect to the Hausdorff distance topology (we use that $$C$$ has non-empty interior), we have $$\lambda(B) = \lambda(C) > 0$$ and we infer from Claim 4 that $$\tau(C) \le \tau(B)$$. By Remark 1, we have $$\tau(B) = \tau_n$$, which completes the proof.

[1] S. G. Krantz, H. R. Parks, "The Geometry of Domains in Space", 1999.
[2]: https://en.wikipedia.org/wiki/Symmetrization_methods [3]: https://en.wikipedia.org/wiki/Wallis%27_integrals [4]: https://en.wikipedia.org/wiki/N-sphere

• I'ved added a post which gives an analytic formula for $\tau_m$ in terms of incomplete beta function. I've also derived a exponential bound which is essentially tight. May 10, 2020 at 15:30
• It seems convexity is not used at all in the proof of "Fedja's Lemma", since in the 1D case, you don't need the sets to be intervals. Also, Fubini-Tonelli still carries over $1 \rightarrow n$. So, I guess compactness + convexity of D and E can be weakened to demanding that $D$ and $E$ be measurable with finite volume. Then, the limitting argument needed to complete the proof of the ultimate result can be replaced with a weaker form of convergence: almost sure convergence to a ball, of Steiner symmetrizations w.r.t random hyperplanes link.springer.com/article/10.1007/s00526-012-0493-4. May 28, 2020 at 10:44
• @dohmatob This is a very interesting remark, thank you. I'll check the details. (By the way, you didn't answer my question about tightness: what does "tight" mean for your?) May 30, 2020 at 12:56
• Fixed (see comment below). For my remark on Steiner symmetrization compact nonconvex sets, actually you get convergence in a nonprobabilistic sense. See Theorem 1.1 of researchgate.net/profile/Aljosa_Volcic/publication/… Jun 2, 2020 at 1:37

How about a short solution to a much general problem, and without convexity assumption (which turns out to be really just a distraction here) ?

So, more generally, for nonempty compact subsets $$A$$, $$B$$, and $$C$$ of $$\mathbb R^n$$, define $$F(A,B,C)$$ by

$$F(A,B,C) := \int_{A}\lambda(B \cap (x + C))dx.$$

In particular, the OP's question is concerned with $$f(C) := F(C,C,C)$$.

We shall proceed in 3 short steps.

• Step 1. The first step in analyzing $$F$$ is to realize that we can rewrite it as $$F(A,B,C) = G(1_A,1_B,1_C)$$, where

\begin{align} G(u,v,w) = \int_{\mathbb R^n}\int_{\mathbb R^n} u(x)w(x-y)v(y)\,dxdy. \end{align}

• Step 2. Now, by the Riez rearrangement inequality, we know that $$G(u,v,w) \le G(u^\star,v^\star,w^\star),$$ where $$u^\star$$ is the Symmetric decreasing rearrangement of the function $$u$$. On the other hand, it is clear that $$(1_A)^{\star} = 1_{K(A)},$$ where $$K(A)$$ is the centered ball of same volume as $$A$$. We deduce that $$F(A,B,C) \le F(K(A),K(B),K(C)),$$

and in particular, $$f(C) \le f(K(C))$$. The solution to the OP's question then follows from computation for the ball, in the begining of the accepted answer https://mathoverflow.net/a/282941/78539.

## Analytic formula and an exponential bound

For $$r \in [0, 1]$$, define $$\tau_m(r) \in [0, 1]$$ by $$\tau_m(r) := \frac{1}{\lambda(B_m(r))}\int_{B_m(r)} \frac{\lambda(B_m(1) \cap (B_m(1) + x))}{\lambda(B_m(1))}dx.$$

The case $$r=1$$ has been analysed in accepted answer. Extending user @Matt's comment, one can establish that (for the core idea, see this post)

$$\begin{split} \tau_m(r) &= \frac{1}{V_m^{cap}(r, 0)}\int_{0}^r ms^{m-1}V_m^{cap}(1, s/2)ds = \frac{1}{r^mV_m^{cap}(1, 0)}\int_{0}^r ms^{m-1}V_m^{cap}(1, s/2)ds\\ &= \frac{1}{V_m^{cap}(1, 0)}\int_{0}^1 ms^{m-1}V_m^{cap}(1, rs/2)ds, \end{split}$$

where $$V_m(r; h)$$ is the volume (i.e Lebesgue measure) of the half-lens $$\{x \in B_m(r) \mid x_1 \ge h\}$$.

Preamble. I propose to obtain an analytic formula for $$\tau_m(r)$$ in terms of special functions (mostly hypergeometric functions). We will also obtain a good upper bound for $$\tau_m(r)$$.

The end result will be the following exponential decay of $$\tau_m(r)$$ as a function on the dimension $$m$$.

Theorem. For every $$r \in [0, 1]$$ and large integer $$m$$, we have the upper bound $$\begin{split} \tau_m(r) &\lesssim \exp\left(-\frac{mr^2}{8-r^2}\right) + \frac{r(1-r^2/4)^{m/2}}{\sqrt{m\pi}}\\ &\lesssim \exp\left(-\frac{mr^2}{8}\right) + \frac{r(1-r^2/4)^{m/2}}{\sqrt{m\pi}}. \end{split}$$

Proof. Now, it is a classical computation that for $$h \ge 0$$ we have $$V_m^{cap}(1, h) = \frac{\lambda(B_{m - 1}(1))}{2}\beta_{1-h^2}\left(\frac{m+1}{2},\frac{1}{2}\right)$$, where $$\beta_x(a, b) := \int_0^x t^{a-1}(1-t)^{b-1}dt$$ defines the incomplete beta function. In particular, $$V_m^{cap}(1, 0) = \frac{\lambda(B_{m - 1}(1))}{2}\beta_{1}\left(\frac{m+1}{2},\frac{1}{2}\right)$$.

Thus, integrating by parts, we get

$$\begin{split} V_m^{cap}(1,0)\tau_m(r) &= \int_{0}^1 ms^{m-1}V_m^{cap}(rs/2)ds = \int_{0}^1 \beta_{1-(rs)^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right)ds^m\\ &= \left[s^m\beta_{1-(rs)^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right)\right]_0^1 + r\int_0^1 s^m(1-(rs)^2/4)^{\frac{m-1}{2}}ds \\ &= \beta_{1-(r/2)^2}\left(\frac{m+1}{2},\frac{1}{2}\right) + R_m, \end{split} \tag{1}$$ where $$R_m := r\int_0^1 s^m(1-(rs/2)^2)^{\frac{m-1}{2}}ds$$. Now, consider the change of variable $$s = 2\sqrt{t}$$. This gives $$s^m(1-s^2/4)^{(m-1)/2}ds=2^mt^{m/2}\cdot(1-t)^{(m-1)/2}\cdot t^{-1/2}dt = 2^mt^{(m-1)/2}(1-t)^{(m-1)/2}dt$$. Thus,

$$\begin{split} R_m &= r\int_0^1 s^m(1-(rs)^2/4)^{\frac{m-1}{2}}ds = \frac{2^m}{r^m}\int_0^{(r/2)^2}t^{(m-1)/2}(1-t)^{(m-1)/2}dt\\ &=: \frac{2^m}{r^m}\beta_{(r/2)^2}\left(\frac{m+1}{2},\frac{m+1}{2}\right). \end{split} \tag{2}$$

Combining (1) and (2) then gives the analytic formula,

Analytic formula. For any $$r \in [0, 1]$$ and positive integer $$m$$, it holds that $$\tau_m(r) = I_{1-(r/2)^2}\left(\frac{m+1}{2},\frac{1}{2}\right) + \frac{(2/r)^m\beta_{(r/2)^2}\left(\frac{m+1}{2},\frac{m+1}{2}\right)}{\beta_1\left(\frac{m+1}{2},\frac{1}{2}\right)},$$ where $$u \mapsto I_u(a, b) := \beta_u(a,b) / \beta_1(a, b)$$ is the CDF of $$(a,b)$$-beta distribution.

Bounding the first term. Let $$\delta := (r/2)^2$$ and $$p = I_{1-\delta}((m+1)/2,1/2)$$. It is known since Temme (1992) that for $$p \in (0, 1)$$ and large $$a > 0$$, the solution of the equation $$p = I(t; a,b)$$ is given (approximately) by

$$t=t_p(a, b) \approx e^{-(1/a)Q_{1-p}(\Gamma(b,1))}, \tag{4}$$

where $$Q_{1-p}(\Gamma(b,1))$$ is the $$1-p$$ quantile of the unit-scale gamma distribution with shape parameter $$b$$. Now by standard concentration results (e.g see Boucheron et al. textbook),

$$Q_{1-p}(\Gamma(b,1)) \le \log(1/p) + \sqrt{2b\log(1/p)}. \tag{5}$$

In particular, for $$a=(m+1)/2$$ and $$b=1/2$$ we get

$$Q_{1-p}(\Gamma(1/2,1)) \le \log(1/p) + \sqrt{\log(1/p)}. \tag{6}$$

Putting (2), (4), and (6) together and using the basic inequality $$1-e^{-z} \le 2z/(2+z)\;\forall z \ge 0$$, we see that

$$\begin{split} \delta &= 1 - t_{p}\left((m+1)/2,1/2\right) \approx 1-e^{-\frac{2Q_{1-p}(\Gamma(1/2,1))}{m+1}} \le 1-e^{-\frac{2}{m+1}\left(\log\left(\frac{1}{p}\right) + \sqrt{\log\left(\frac{1}{p}\right)}\right)} \le \frac{2\alpha}{2+\alpha}, \end{split}$$

where $$\alpha = \frac{2}{m+1}(\log(1/p)+\sqrt{\log(1/p)})$$. Thus, $$\alpha \ge \delta/(1 - \delta/2)$$, and this wolfram-alpha query tells us that we must have $$p \lesssim \exp(-mr^2/(8-r^2))$$. Putting things together then gives

$$\begin{split} I_{1-(r/2)^2}((m+1)/2,1/2) = p \lesssim \exp(-mr^2/(8-r^2)) \le \exp(-mr^2/8). \end{split}$$

Bounding the second term. It is a well-known result that

$$\beta_z(a,b) \equiv a^{-1}z^a{}_2F_1(a,1-b,a+1;z),$$

where $$z \mapsto {}_2F_1(u, v, w; z)$$ is the hypergeometric function. Thus, one computes

$$\begin{split} (2/r)^m\beta_{(r/2)^2}\left(\frac{m+1}{2},\frac{m+1}{2}\right) &= \frac{r}{m+1}{}_2F_1\left(\frac{m+1}{2},\frac{1-m}{2},\frac{m+3}{2},\frac{r^2}{4}\right) \\ &= \frac{r}{m+1}{}_2F_1\left(\frac{1-m}{2},\frac{m+1}{2},\frac{m+3}{2},\frac{r^2}{4}\right)\\ &\approx \frac{r}{m+1}{}_2F_1\left(\frac{1-m}{2},\frac{m+1}{2},\frac{m+1}{2},\frac{r^2}{4}\right)\\ &= \frac{r}{m+1}(1-r^2/4)^{(m-1)/2}, \end{split}$$

where we have used the powerful identity $${}_2F_1(u, v, v; z) \equiv (1 - z)^{-u}$$. On the other hand, Stirling's formula gives

$$\beta_1\left(\frac{m+1}{2},\frac{1}{2}\right) \approx \Gamma(1/2)\left(\frac{m+1}{2}\right)^{-1/2} = \sqrt{\frac{\pi}{m+1}}.$$

Thus,

$$\frac{(2/r)^m\beta_{(r/2)^2}\left(\frac{m+1}{2},\frac{m+1}{2}\right)}{\beta_1\left(\frac{m+1}{2},\frac{1}{2}\right)} \approx \frac{r(1-(r/2)^2)^{(m-1)/2}}{\sqrt{\pi(m+1)}}$$

Putting every together then completes the proof of the theorem. $$\quad\quad\quad\Box$$

• @LucGuyot There were a bunch of typos in what I wrote, leading to the inconsistent inequalities you mentioned. Fixed. Let me know if you still have any trouble following. May 13, 2020 at 7:51
• I was hasty in simplifying (6) to $\ldots \le 2\log(1/p)$, which is only true for $p \le 1/e$. Actually, I can do without this simplification and still get the $\exp(-mr^2/(8-r^2))$ term of my bound. See update. May 13, 2020 at 10:13
• Sorry, I wasn't very careful in the calculations. I meant $1-e^{-z} \le 2z / (2+z)$ for all $z \ge 0$. Also, indeed the next inequalities were reversed. Everything should be fine now. Thanks for spotting the errors. May 16, 2020 at 14:19
• Because you are not using the big-O nor the small-o notation, and because there are at least two approximations in your computations, I would expect a constant factor $C$ (as close to $1$ as one would desire) to appear on the right side of the theorem's inequality. You also claim that your upper bound is tight. In which sense? Consider $r = 1$. We have $e^{-1/7} \approx 0.8668 > 0.8660 \approx \frac{\sqrt{3}}{2}$. So $(\frac{\sqrt{3}}{2})^m$ would be a better asymptotic upper bound of $\tau_m(1)$. May 21, 2020 at 13:12
• Ah, the $\le$ in my theorem was meant to be be implicitly read as $\lesssim$. I'll make it more explicit, if that helps. BTW, glad to know that my bounds match (upto within dust) the quantity you computed using a completely different strategy. May 21, 2020 at 14:22