Shallow sections of the hypercube

Denote by $H^{-}$ the half-space of $\mathbb{R}^{d}$ made up of the points $x$ such that $a\mathord{\cdot}x\leq{b}$. It is proven in [2] (see equation (2) therein) that the $d$-dimensional volume of $H^{-}\cap[0,1]^{d}$ is given by the expression

where the sum is over the vertices $v$ of $[0,1]^{d}$ such that $a\mathord{\cdot}v\leq{b}$. Now observe that the distance $t$ between $H$ and the center of $[0,1]^{d}$ is

Assume for a moment that $b$ is at least $\sigma(a)/2$ or, equivalently, that the center of $[0,1]^{d}$ is contained in $H^{-}$. In this case,

Replacing $b$ in (2) by the right-hand side of this equality and differentiating with respect to $t$ results in the $(d-1)$-dimensional volume of $H\cap[0,1]^{d}$. The expression that results from the differentiation is (1), as desired.

Now if $b$ is less than $\sigma(a)/2$, then

As above, replacing $b$ in (2) by the right-hand side of this equality and differentiating with respect to $-t$ results in (1). ∎

Consider the volume formula provided by Theorem 2.1 and observe the $(d-1)$-dimensional volume of $H\cap[0,1]^{d}$ can be obtained from it by a straightforward change of variables. Further observe that $\|a\|$ is a continuously derivable function of $a$ on $]0,+\infty[^{d}$. Hence, by the symmetry of this formula with respect to the coordinates of $a$, it is sufficient to show that the partial derivative

exists and is a continuous function of $a$ on $]0,+\infty[^{d}$. It is a consequence of (3) that the cosine term is no longer dependent of $a$ after the change of variables (see also Proposition 1 from [11]). The differentiability is obtained by Leibniz’s rule for differentiation under the integral sign whereby

However, as the integral is improper, Leibniz’s rule requires that

converges uniformly when $n$ goes to infinity, in a neighborhood of each point $a$ from $]0,+\infty[^{d}$. The uniform convergence also implies the desired continuity of the partial derivative. Observe that any point contained in $]0,+\infty[^{d}$ admits, as a neighborhood, the interior of a closed ball contained in the open orthant $]0,+\infty[^{d}$. Consider such a closed ball $B$ and assume that $a$ belongs to the interior of $B$. Note that, when $u$ is non-zero,

By a straightforward calculation,

As a consequence,

where $m$ is the smallest possible value of the product of the coordinates of a point from $B$. Under the assumption that $d$ is at least $3$, the desired uniform convergence property therefore follows from Cauchy’s criterion. ∎

Consider a point $x$ from $[0,+\infty[^{d}$ with at least two non-zero coordinates. Assume, without loss of generality, that the first two coordinates of $x$ are non-zero. Consider a closed $d$-dimensional ball $B$ centered at $x$, whose interior is non-empty. Choose the radius of $B$ small enough so that the first two coordinates of all the points it contains are non-zero. Observe that

is a continuous function of $a$ for any positive integer $n$ as a definite integral of a continuous function of $a$ and $u$ (under the convention that $\sin(0)/0=1$). In fact, that function converges uniformly on $B$ when $n$ goes to infinity. Indeed, for any non-zero real number $u$ and any vector $a$ from $B$,

where $m$ is the smallest value for the product of the first two coordinates of a point from $B$. Hence, by the Cauchy criterion, the desired uniform convergence property holds. As a consequence,

is a continuous function of $a$ on $B$. It remains to remark that, according to Theorem 2.2, the value of this function at any point $a$ from $B$ (still under the convention that $\sin(0)/0=1$, and by the same change of variables as in the proof of Corollary 2.4) is precisely the volume of $H\cap[0,1]^{d}$. ∎

Assume that $t>1/2$. By that assumption, $H$ is disjoint from $[0,1]^{d}$ when $a$ has a unique non-zero coordinate. Now observe that $H$ remains disjoint from $[0,1]^{d}$ in a neighborhood of any such vector $a$. As a consequence, the $(d-1)$-dimensional volume of $H\cap[0,1]^{d}$ is equal to $0$ in neighborhood of $a$. This, together with Corollary 2.6, proves that the $(d-1)$-dimensional volume of $H\cap[0,1]^{d}$ is continuous on the whole of $\mathbb{S}^{d-1}\cap[0,+\infty[^{d}$. The result therefore follows from the compactness of $\mathbb{S}^{d-1}\cap[0,+\infty[^{d}$. ∎

Under the assumption that (9) holds, $V$ must be positive at $a$. Indeed, according to (9), the distance between the center $c$ of $[0,1]^{d}$ and $H$ is less than the distance between $c$ and a vertex of $[0,1]^{d}$ and, therefore, $V$ is sometimes positive (as, for instance when all the coordinates of $a$ coincide). Since $V$ attains its maximum on $\mathbb{S}^{d-1}\cap[0,+\infty[^{d}$ at $a$, it is necessarily positive at $a$.

Now observe that (9) also implies that the distance between $c$ and $H$ is greater than the distance between $c$ and the midpoint of the edges of $[0,1]^{d}$. Hence, if a coordinate of $a$ were equal to $0$, then $H$ and $[0,1]^{d}$ would be disjoint and $V$ would be equal to $0$. Therefore $a$ necessarily belongs to $]0,+\infty[^{d}$. In turn, $b$ must be positive at $a$. Indeed, otherwise $H$ and $[0,1]^{d}$ would either be disjoint (in the case when $b<0$), or intersect precisely at the origin of $\mathbb{R}^{d}$ (in the case when $b=0$). In both cases, $V$ would be equal to $0$ at $a$.

Finally, observe that the gradient of

is non-zero in $\mathbb{R}^{d}\mathord{\setminus}\{0\}$. Hence, as $a$ belongs to $]0,+\infty[^{d}$, it must be a critical point of $L_{\lambda}$ for some real number $\lambda$ by the Lagrange multipliers theorem. ∎

Assume that the coordinates of $a$ are all at least $b$ and that $b$ is positive. In this case, the only vertex $v$ of $[0,1]^{d}$ that satisfies $a\mathord{\cdot}v<b$ is the origin of $\mathbb{R}^{d}$. It therefore follows from Theorem 2.2 that

In turn, according to (3) and (7),

For any two integers $j$ and $k$ such that $1\leq{j}<k\leq{d}$, (8) implies that

Using the expression for the partial derivatives from (10) and because of the assumption that $b$ is positive, this can be rewritten as

Assume, for contradiction, that $a_{j}$ and $a_{k}$ do not coincide. In this case, one can divide (11) by $a_{k}-a_{j}$, which yields

Since both $a_{j}$ and $a_{k}$ are at least $b$, this equality only possibly holds when $d$ is at most $5$. If $d$ is equal to $5$, it implies that both $a_{j}$ and $a_{k}$ coincide with $b$, contradicting the assumption that $a_{j}$ and $a_{k}$ are distinct. If $3\leq{d}\leq 4$, then (12) implies that $a_{j}/2$ and $a_{k}/2$ cannot both be greater than $b$. Geometrically, this means that one of the closed half-spaces of $\mathbb{R}^{d}$ bounded by $H$ contains both the origin of $\mathbb{R}^{d}$ and the midpoint of an edge of $[0,1]^{d}$ which, in this case, is forbidden by the assumption that $t>\sqrt{d-1}/2$. ∎

Assume that $d$ is at least $3$. It follows from this and from (13) that $t>1/2$. In turn, by Proposition 2.8, $V$ admits a maximum on $\mathbb{S}^{d-1}\cap[0,+\infty[^{d}$. Thanks to the symmetries of the hypercube, it suffices to show that the maximum of $V$ is uniquely attained at the vector in $\mathbb{S}^{d-1}\cap]0,+\infty[^{d}$ whose all coordinates coincide. Indeed, the coordinates of that vector are $1/\sqrt{d}$ and it then follows from Theorem 2.2 and from (3) that $V$ is equal to the announced volume. As (13) holds, one just needs to combine Lemmas 3.1 and 3.2 in order to prove that the coordinates of any vector from $\mathbb{S}^{d-1}\cap]0,+\infty[^{d}$ where $V$ attains its maximum on $\mathbb{S}^{d-1}\cap]0,+\infty[^{d}$ necessarily coincide. ∎

Observe that the desired inequality can be rewritten as

When $d\geq 5$, the right-hand side of this inequality is a decreasing function of $t$ on the considered interval. Indeed, its derivative with respect to $t$ is

where $A=(d-1)\sqrt{d-1}-(d-2)\sqrt{d}-2t$. In particular, that derivative has the sign of $A$ on the considered interval for $t$. When $t$ is in that interval,

Rearranging the terms of the right-hand side, and simultaneously multiplying and dividing it by $(\sqrt{d}+\sqrt{d-1})(\sqrt{d-1}+\sqrt{d-2})$ yields

which, as $\sqrt{d-2}<\sqrt{d-1}$ implies the negativity of $A$ when $d\geq 5$. Hence, it suffices to show that the desired inequality holds for the smallest $t$ in the considered interval. For this reason, it is assumed that $t=\sqrt{d-2}/2$ in the remainder of the proof. Observe that when $d=5$, the left-hand side of (15) is then greater than $0.7$ and its right-hand side less than $0.7$, proving the inequality when $d=5$. Assume that $d\geq 6$, and observe that

As a consequence, it is sufficient to show that

Now observe that, since $d\geq 6$,

Finally, as $(4/\sqrt{5})^{d}-8d+16$ is positive when $d=6$ and an increasing function of $d$ on $[6,+\infty[$ (because its derivative is positive), (15) holds. ∎

Recall that, if all the coordinates of $a$ are equal, then $V$ is

As $V$ attains its maximum on $\mathbb{S}^{d-1}\cap[0,+\infty[^{d}$ at $a$, then $V$ must be at least (17). According to (16), the distance between the center of $[0,1]^{d}$ and $H$ is greater than the distance between the center of $[0,1]^{d}$ and the center of a square face of $[0,1]^{d}$. In particular, if two of the coordinates of $a$ were equal to $0$, then $H$ would be disjoint from $[0,1]^{d}$. Since $V$ is positive, this shows that all the coordinates of $a$ are non-zero except possibly one.

Assume for contradiction that a unique coordinate of $a$ is equal to zero and identify for a moment $\mathbb{R}^{d-1}$ with the subspace of $\mathbb{R}^{d}$ spanned by the other $d-1$ coordinates. In this case, $V$ is equal to the $(d-2)$-dimensional volume of $H\cap[0,1]^{d-1}$. Now observe that (16) can be rewritten as

If $t\geq\sqrt{d-1}/2$, then either $H$ and $[0,1]^{d-1}$ are disjoint or their intersection is a vertex of $[0,1]^{d-1}$. In both of these cases, $V$ would be equal to zero, which is impossible. If $t<\sqrt{d-1}/2$, then by Theorem 3.3, $V$ is at most

and according to Proposition 3.4, this quantity is less than (17), contradicting the above observation that $V$ is at least (17). This proves that $a$ belongs to $]0,+\infty[^{d}$. One obtains that $b$ is positive at $a$ and that, for some number $\lambda$, $a$ is a critical point of $L_{\lambda}$ by the same argument as in the proof of Lemma 3.1. ∎

Assume that $b$ is positive. It will be shown that, under the assumptions made on $d$ and $t$, none of the coordinates of $a$ can be less than $b$. Observe that, in this case, the result then follows from Lemma 3.2. Assume, for contradiction that $a$ has a unique coordinate less than $b$. It can be required without loss of generality that this coordinate is the first. In this case, the only two vertices $v$ of $[0,1]^{d}$ such that $a\mathord{\cdot}v<b$ are the origin of $\mathbb{R}^{d}$ and the vertex whose only non-zero coordinate is the first. Hence, by Theorem 2.2,

Therefore, it follows from (7) that,

and that, when $i\geq 2$,

Consider an integer $j$ such that $2\leq{j}\leq{d}$. According to (8),

Multiplying this equality by $-2a_{1}a_{j}/b^{d+1}$, and using the above expressions for the partial derivatives, it can be rewritten as

where $x=a_{j}/b$, $y=1-a_{1}/b$, and

Note that $y$ belongs to the interval $]0,1[$ because $0<a_{1}<b$. By Proposition 4.1, $x$ cannot belong to $[1,+\infty[$ when $d\geq 6$, thereby contradicting the assumption that $a_{j}\geq{b}$. It remains to reach a contradiction when $d=5$. In this case, the two solutions of the quadratic equation (18) are $y+1$ and $(y^{2}+1)/(y+1)$. Only the former belongs to $[1,+\infty[$, hence

In particular, all the coordinates of $a$ but the first are equal to $2b-a_{1}$. Now recall that $\|a\|^{2}=1$ and that (3) holds. These equalities yield

Replacing $2b-a_{1}$ by $2t/3$ in the first of these equalities and remembering the assumption that $t>\sqrt{d-2}/2$, one obtains $a_{1}^{2}<-1/3$, a contradiction. ∎

The proof proceeds just as that of Theorem 3.3, except that Lemmas 3.5 and 4.2 are used instead of Lemmas 3.1 and 3.2. ∎