## Laplacian spectrum of small graphs

This is a collection of entirely unoriginal remarks about Laplacian spectrum of graphs. For an accessible overview of the subject I recommend the M.S. thesis The Laplacian Spectrum of Graphs by Michael William Newman. It also includes a large table of graphs with their spectra. Here I will avoid introducing matrices and enumerating vertices.

Let ${V}$ be the vertex set of a graph. Write ${u\sim v}$ if ${u, v}$ are adjacent vertices. Given a function ${f\colon V\to \mathbb R}$, define ${L f(v) = \sum_{u\colon u\sim v}(f(v)-f(u))}$.
This is a linear operator (the graph Laplacian) on the Euclidean space ${\ell^2(V)}$ of all functions ${f\colon V\to \mathbb R}$ with the norm ${\|f\|^2 = \sum_{v\in V} f(v)^2}$. It is symmetric: ${\langle L f, g\rangle = \langle f, L g\rangle }$ and positive semidefinite: ${\langle L f, f\rangle = \frac12 \sum_{u\sim v}(f(u)-f(v))^2\ge 0}$. Since equality is attained for constant ${f}$, 0 is always an eigenvalue of ${L}$.

This is the standard setup, but I prefer to change things a little and replace ${\ell^2(V)}$ by the smaller space ${\ell^2_0(V)}$ of functions with zero mean: ${\sum_{v\in V}f(v)=0}$. Indeed, ${L}$ maps ${\ell^2(V)}$ to ${\ell^2_0(V)}$ anyway, and since it kills the constants, it makes sense to focus on ${\ell^2_0(V)}$. It is a vector space of dimension ${n-1}$ where ${n=|V|}$.

One advantage is that the smallest eigenvalue is 0 if and only if the graph is disconnected: indeed, ${\langle L f, f\rangle=0}$ is equivalent to ${f}$ being constant on each connected component. We also gain better symmetry between ${L}$ and the Laplacian of the graph complement, denoted ${L'}$. Indeed, since ${L' f(v) = \sum_{u\colon u\not \sim v}(f(v)-f(u))}$, it follows that ${(L+L')f(v) = \sum_{u\colon u\ne v} (f(v)-f(u)) = n f(v)}$ for every ${f\in \ell^2_0(V)}$. So, the identity ${L+L' = nI}$ holds on ${\ell^2_0(V)}$ (it does not hold on ${\ell^2(V)}$). Hence the eigenvalues of ${L'}$ are obtained by subtracting the eigenvalues of ${L}$ from ${n}$. As a corollary, the largest eigenvalue of ${L}$ is at most ${n}$, with equality if and only if the graph complement is disconnected. More precisely, the multiplicity of eigenvalue ${n}$ is one less than the number of connected components of the graph complement.

Let ${D}$ denote the diameter of the graph. Then the number of distinct Laplacian eigenvalues is at least ${D}$. Indeed, let ${u, v}$ be two vertices at distance ${D}$ from each other. Define ${f_0(u) = 1}$ and ${f_0=0}$ elsewhere. Also let ${f_k=L^k f_0}$ for ${k=1, 2, \dots}$. Note that ${f_k\in \ell_0^2(V)}$ for all ${k\ge 1}$. One can prove by induction that ${f_k(w)=0}$ when the distance from ${w}$ to ${u}$ is greater than ${k}$, and ${(-1)^k f_k(w) > 0}$ when the distance from ${w}$ to ${u}$ is equal to ${k}$. In particular, ${f_k(v) = 0}$ when ${k and ${f_D(v)\ne 0}$. This shows that ${f_D}$ is not a linear combination of ${f_1, \dots, f_{D-1}}$. Since ${f_k = L^{k-1}f_1}$, it follows that ${L^{D-1}}$ is not a linear combination of ${L^0, L^1, \dots, L^{D-2}}$. Hence the minimal polynomial of ${L}$ has degree at least ${D}$, which implies the claim.

Let’s consider a few examples of connected graphs.

## 3 vertices

There are two connected graphs: the 3-path (D=2) and the 3-cycle (D=1). In both cases we get D distinct eigenvalues. The spectra are [1, 3] and [3, 3], respectively.

## 4 vertices

• One graph of diameter 3, the path. Its spectrum is ${[2-\sqrt{2}, 2, 2+\sqrt{2}]}$.
• One graph of diameter 1, the complete graph. Its spectrum is ${[4, 4, 4]}$. This pattern continues for other complete graphs: since the complement is the empty graph (${n}$ components), all ${n-1}$ eigenvalues are equal to ${n}$.
• Four graphs of diameter 2, which are shown below, with each caption being the spectrum.

Remarks:

• The graph [1, 3, 4] has more distinct eigenvalues than its diameter.
• The graph [2, 2, 4] is regular (all vertices have the same degree).
• The smallest eigenvalue of graphs [1, 1, 4] and [2, 2, 4] is multiple, due to the graphs having a large group of automorphisms (here rotations); applying some of these automorphisms to an eigenfunctions for the smallest eigenvalue yields another eigenfunction.
• [1, 3, 4] and [2, 4, 4] also have automorphisms, but their automorphisms preserve the eigenfunction for the lowest eigenvalue, up to a constant factor.

## 5 vertices

• One graph of diameter 4, the path. Its spectrum is related to the golden ratio: it consists of ${(3\pm \sqrt{5})/2, (5\pm \sqrt{5})/2}$.
• One graph of diameter 1, the complete one: [5, 5, 5, 5]
• Five graphs of diameter 3. All have connected complement, with the highest eigenvalue strictly between 4 and 5. None are regular. Each has 4 distinct eigenvalues.
• 14 graphs of diameter 2. Some of these are noted below.

Two have connected complement, so their eigenvalues are less than 5 (spectrum shown on hover):

One has both integers and non-integers in its spectrum, the smallest such graph. Its eigenvalues are ${3\pm \sqrt{2}, 3, 5}$.

Two have eigenvalues of multiplicity 3, indicating a high degree of symmetry (spectrum shown on hover).

Two have all eigenvalues integer and distinct:

The 5-cycle and the complete graph are the only regular graphs on 5 vertices.

## 6 vertices

This is where we first encounter isospectral graphs: the Laplacian spectrum cannot tell them apart.

Both of these have spectrum ${3\pm \sqrt{5}, 2, 3, 3}$ but they are obviously non-isomorphic (consider the vertex degrees):

Both have these have spectrum ${3\pm \sqrt{5}, 3, 3, 4}$ and are non-isomorphic.

Indeed, the second pair is obtained from the first by taking graph complement.

Also notable are regular graphs on 6 vertices, all of which have integer spectrum.

Here [3, 3, 3, 3, 6] (complete bipartite) and [2, 3, 3, 5, 5] (prism) are both regular of degree 3, but the spectrum allows us to tell them apart.

The prism is the smallest regular graph for which the first eigenvalue is a simple one. It has plenty of automorphisms, but the relevant eigenfunction (1 on one face of the prism, -1 on the other face) is compatible with all of them.

## 7 vertices

There are four regular graphs on 7 vertices. Two of them are by now familiar: 7-cycle and complete graph. Here are the other two, both regular of degree 4 but with different spectra.

There are lots of isospectral pairs of graphs on 7 vertices, so I will list only the isospectral triples, of which there are five.

Spectrum 0.676596, 2, 3, 3.642074, 5, 5.681331:

Spectrum 0.726927, 2, 3.140435, 4, 4, 6.132637:

Spectrum 0.867363, 3, 3, 3.859565, 5, 6.273073:

Spectrum 1.318669, 2, 3.357926, 4, 5, 6.323404:

All of the triples mentioned so far have connected complement: for example, taking the complement of the triple with the spectrum [0.676596, 2, 3, 3.642074, 5, 5.681331] turns it into the triple with the spectrum [1.318669, 2, 3.357926, 4, 5, 6.323404].

Last but not least, an isospectral triple with an integer spectrum: 3, 4, 4, 6, 6, 7. This one has no counterpart since the complement of each of these graphs is disconnected.

## 8 vertices

Regular graphs, excluding the cycle (spectrum 0.585786, 0.585786, 2, 2, 3.414214, 3.414214, 4) and the complete one.

#### Degree 6 regular

Credits: the list of graphs by Brendan McKay and NetworkX library specifically the methods read_graph6 (to read the files provided by Prof. McKay), laplacian_spectrum, diameter, degree, and draw.

## Extremal Taylor polynomials

Suppose ${f(z)=a_0+a_1z+a_2z^2+\cdots}$ is a holomorphic function in the unit disk ${|z|<1}$ such that ${|f|\le 1}$ in the disk. How large can its Taylor polynomial ${T_n(z)=a_0+a_1z+\cdots +a_n z^n}$ be in the disk?

We should not expect ${T_n}$ to be bounded by 1 as well. Indeed, the Möbius transformation ${f(z)=(z+1/2)/(1+z/2)}$ has Taylor expansion ${(z+1/2)(1-z/2+O(z^2)) = 1/2 + (3/4)z + O(z^2)}$, so ${T_1(1)=5/4}$ in this case. This turns out to be the worst case: in general ${T_1}$ is bounded by 5/4 in the disk.

For the second-degree polynomial ${T_2}$ the sharp bound is ${89/64}$, attained when ${f(z) = (8z^2 + 4z + 3)/(3z^2 + 4z + 8)}$; the image of the unit circle under the extremal ${T_2}$ is shown below. Clearly, there is something nontrivial going on.

Edmund Landau established the sharp bound for ${|T_n|}$ in his paper Abschätzung der Koeffizientensumme einer Potenzreihe, published in Archiv der Mathematik und Physik (3) 21 in 1913. Confusingly, there are two papers with the same title in the same issue of the journal: one on pages 42-50, the other on pages 250-255, and they appear in different volumes of Landau’s Collected Works. The sharp bound is in the second paper.

### First steps

By rotation, it suffices to bound ${|T_n(1)|}$, which is ${|a_0+\cdots +a_n|}$. As is often done, we rescale ${f}$ a bit so that it’s holomorphic in a slightly larger disk, enabling the use of the Cauchy integral formula on the unit circle ${\mathbb T}$. The Cauchy formula says ${2\pi i a_k = \int_{\mathbb T} z^{-k-1} f(z) \,dz}$. Hence

${\displaystyle 2\pi |T_n(1)| = \left| \int_{\mathbb T} z^{-n-1}(1+z+\dots+z^n) f(z) \,dz \right|}$

It is natural to use ${|f(z)|\le 1}$ now, which leads to

${\displaystyle 2\pi |T_n(1)| \le \int_{\mathbb T} |1+z+\dots+z^n|\, |dz| }$

Here we can use the geometric sum formula and try to estimate the integral of ${|(1-z^{n+1})/(1-z)|}$ on the unit circle. This is what Landau does in the first of two papers; the result is ${O(\log n)}$ which is the correct rate of growth (this is essentially the Dirichlet kernel estimate from the theory of Fourier series). But there is a way to do better and get the sharp bound.

### Key ideas

First idea: the factor ${1+z+\dots+z^n}$ could be replaced by any polynomial ${Q}$ as long as the coefficients of powers up to ${n}$ stay the same. Higher powers contribute nothing to the integral that evaluates ${T_n(1)}$, but they might reduce the integral of ${|Q|}$.

Second idea: we should choose ${Q}$ to be the square of some polynomial, ${Q=P^2}$, because ${(2\pi)^{-1}\int_{\mathbb T} |P(z)|^2\, |dz|}$ can be computed exactly: it is just the sum of squares of the coefficients of ${P}$, by Parseval’s formula.

### Implementation

Since ${1+z+\dots+z^n}$ is the ${n}$-th degree Taylor polynomial of ${(1-z)^{-1}}$, it is natural to choose ${P}$ to be the ${n}$-th degree Taylor polynomial of ${(1-z)^{-1/2}}$. Indeed, if ${P_n(z) = (1-z)^{-1/2} + O(z^{n+1})}$, then ${P_n(z)^2 = (1-z)^{-1} + O(z^{n+1}) = 1+z+\dots+z^n + O(z^{n+1})}$ as desired (asymptotics as ${z\to 0}$). The binomial formula tells us that
${\displaystyle P_n(z)=\sum_{k=0}^n (-1)^k\binom{-1/2}{k}z^k }$

The coefficient of ${z^k}$ here can be written out as ${(2k-1)!!/(2k)!!}$ or rewritten as ${4^{-k}\binom{2k}{k}}$ which shows that in lowest terms, its denominator is a power of 2. To summarize, ${|T_n(1)|}$ is bounded by the sum of squares of the coefficients of ${P_n}$. Such sums are referred to as the Landau constants,

${\displaystyle G_n = 1+ \left(\frac{1}{2}\right)^2 + \left(\frac{1\cdot 3}{2\cdot 4}\right)^2 + \cdots + \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 }$

A number of asymptotic and non-asymptotic formulas have been derived for ${G_n}$, for example Brutman (1982) shows that ${G_n - (1/\pi)\log(n+1)}$ is between 1 and 1.0663.

### Sharpness

To demonstrate the sharpness of the bound ${|T_n|\le G_n}$, we want ${|f|\equiv 1}$ and ${P_n(z)^2f(z)/z^n\ge 0}$ on the unit circle. Both are arranged by taking ${f(z) = z^n P_n(1/z) / P_n(z)}$ which is a Blaschke product of degree ${n}$. Note that the term ${P_n(1/z)}$ can also be written as ${\overline{P_n(1/\bar z)}}$. Hence ${P_n(z)^2f(z)/z^n = P_n(z) \overline{P_n(1/\bar z)}}$ which is simply ${|P_n(z)|^2}$ when ${|z|=1}$. Equality holds in all the estimates above, so they are sharp.

Here are the images of the unit circle under extremal Taylor polynomials ${T_5}$ and ${T_{20}}$.

These polynomials attain large values only on a short subarc of the circle; most of the time they oscillate at levels less than 1. Indeed, the mean value of ${|T_n|^2}$ cannot exceed the mean of ${|f|^2}$ which is at most 1. Here is the plot of the roots of extremal ${T_n}$:  they are nearly uniform around the circle, except for a gap near 1.

### But we are not done…

Wait a moment. Does ${f(z) = z^n P_n(1/z) / P_n(z)}$ define a holomorphic function in the unit disk? We are dividing by ${P_n}$ here. Fortunately, ${P_n}$ has no zeros in the unit disk, because its coefficients are positive and decreasing as the exponent ${k}$ increases. Indeed, if ${p(z)=c_0+c_1z+\cdots + c_nz^n}$ with ${c_0>c_1>\dots>c_n > 0}$, then ${(1-z)p(z)}$ has constant term ${c_0}$ and other coefficients ${c_1-c_0}$, ${c_2-c_1}$, … ${c_n-c_{n-1}}$, ${-c_n}$. Summing the absolute values of the coefficients of nonconstant terms we get ${c_0}$. So, when these coefficients are attached to ${z^k}$ with ${|z|<1}$, the sum of nonconstant terms is strictly less than ${c_0}$ in absolute value. This proves ${P_n\ne 0}$ in the unit disk. Landau credits Adolf Hurwitz with this proof.

In fact, the zeros of ${P_n}$ (Taylor polynomials of ${(1-z)^{-1/2}}$) lie just outside of the unit disk.

The zeros of the Blaschke products formed from ${P_n}$ are the reciprocals of the zeros of  ${P_n}$, so they lie just inside the unit circle, much like the zeros of ${T_n}$ (though they are different).

## The distribution of pow(2, n, n)

For a positive integer ${n}$ let ${f(n)}$ be the remainder of the division of ${2^n}$ by ${n}$. This is conveniently computed in Python as pow(2, n, n). For example, the first 20 values are returned by

[pow(2, n, n) for n in range(1, 21)]

and they are:

[0, 0, 2, 0, 2, 4, 2, 0, 8, 4, 2, 4, 2, 4, 8, 0, 2, 10, 2, 16]

Obviously, ${f(n)=0}$ if and only if ${n}$ is a power of ${2}$. It also looks like ${f}$ is always even, with powers of 2 dominating the list of its values. But ${f}$ does take on odd values, although this does not happen often: only 6 times in the first 100 integers.

[(n, pow(2, n, n)) for n in range(1, 101) if pow(2, n, n) % 2]

returns

[(25, 7), (45, 17), (55, 43), (91, 37), (95, 13), (99, 17)]

It is conjectured that the range of ${f}$ consists of all nonnegative integers except 1. Here is a proof that ${f(n)\ne 1}$, provided by Max Alexeyev on the OEIS page for A036236

Suppose ${2^n \equiv 1 \bmod n}$ for some ${n>1}$. Let ${p}$ be the smallest prime divisor of ${n}$. Then ${2^n\equiv 1 \bmod p}$. This means that the order of ${2}$ in the multiplicative group ${(\mathbb Z/ p \mathbb Z)^*}$ is a divisor of ${n}$. But this group has ${p-1}$ elements, and the only divisor of ${n}$ that is smaller than ${p}$ is ${1}$. Thus, the order of ${2}$ is ${1}$, which means ${2\equiv 1 \bmod p}$, which is absurd.

The fact that the smallest ${n}$ with ${f(n) = 3}$ is 4700063497 deserves to be mentioned here. But I would like to consider the most frequent values of ${f}$ instead. Experimentally, they are found like this:

from collections import Counter
freq = Counter(pow(2, n, n) for n in range(1, 100000000 + 1))
print(freq.most_common(20))

which prints 20 pairs (value, frequency):

(2, 5763518),
(4, 3004047),
(8, 2054167),
(16, 1569725),
(64, 1076293),
(256, 824438),
(512, 737450),
(1024, 668970),
(32, 638063),
(4096, 569705),
(128, 466015),
(65536, 435306),
(262144, 389305),
(1048576, 351723),
(2097152, 326926),
(16384, 246533),
(16777216, 243841),
(32768, 232037),
(2048, 153537),
(8192, 131614)

These are all powers of 2… but not exactly in the order one might expect. I repeated the experiment for the first ${10^8k}$ integers with ${k=1, 2, 3, 4, 5, 6}$. The frequency order remained stable at the top. These are the most common 11 values, with their frequency (in %) listed for the aforementioned six ranges.

2, [5.8, 5.5, 5.4, 5.3, 5.3, 5.2]
4, [3.0, 2.9, 2.8, 2.8, 2.7, 2.7]
8, [2.1, 2.0, 1.9, 1.9, 1.9, 1.8]
16, [1.6, 1.5, 1.5, 1.4, 1.4, 1.4]
64, [1.1, 1.0, 1.0, 1.0, 1.0, 1.0]
256, [0.8, 0.8, 0.8, 0.8, 0.7, 0.7]
512, [0.7, 0.7, 0.7, 0.7, 0.7, 0.7]
1024, [0.7, 0.6, 0.6, 0.6, 0.6, 0.6]
32, [0.6, 0.6, 0.6, 0.6, 0.6, 0.6]
4096, [0.6, 0.5, 0.5, 0.5, 0.5, 0.5]
128, [0.5, 0.4, 0.4, 0.4, 0.4, 0.4]

It seems that 32 and 128 are consistently less common than one might expect.

The most common value, 2, is contributed by primes and pseudoprimes to base 2. The value of 4 appears when ${n = 2p}$ with ${p}$ prime (but not only then). Still, the primes having zero density makes it difficult to explain the frequency pattern based on primality. A more convincing reason could be: when ${n=2k}$ is even, we are computing ${4^k \bmod 2k}$ and that is likely to be a power of ${4}$. This boosts the frequency of the even (generally, composite) powers of 2 in the sequence.

## Magic angles

The function ${u(x, y, z) = 2z^2 - x^2 - y^2}$ is notable for the following combination of properties:

1. It is a harmonic function: ${\Delta u = -2 - 2 + 4 = 0}$
2. It vanishes at the origin ${(0, 0, 0)}$ together with its gradient.
3. It is positive on the cone ${C=\{(x, y, z) : z > \sqrt{(x^2+y^2)/2}\}}$

The cone C has the opening angle ${\theta_3 = \cos^{-1}(1/\sqrt{3}) \approx 57.4^\circ}$ which is known as the magic angle in the context of NMR spectroscopy. Let us consider the mathematical side of its magic.

If C is replaced by any larger cone, the properties 1-2-3 cannot be satisfied by a harmonic function in a neighborhood of the origin. That is, C is the largest cone such that a harmonic function can have a critical point at its vertex which is also a point of its extremum on the cone. Why is that?

Let ${u}$ be a harmonic function in some neighborhood of ${(0,0,0)}$ and suppose ${u(0,0,0)=0}$, ${\nabla u(0,0,0) = 0}$, and ${u>0}$ on some cone ${C = \{(x, y, z) \colon z>c \sqrt{x^2+y^2+z^2} \}}$. Expand ${u}$ into a sum of polynomials ${p_2+p_3+\cdots }$ where each ${p_k}$ is a harmonic homogeneous polynomial of degree ${k}$. Let ${m}$ be the smallest integer such that ${p_m}$ is not identically zero. Then ${p_m}$ has the same properties as ${u}$ itself, since it dominates the other terms near the origin. We may as well replace ${u}$ by ${p_m}$: that is, ${u}$ is a spherical harmonic from now on.

Rotating ${u}$ around the ${z}$-axis preserves all the properties of interest: harmonic, positive on the cone, zero gradient at the origin. Averaging over all these rotations we get a rotationally symmetric function known as a zonal spherical harmonic. Up to a constant factor, such a function is given by ${P_m(\cos \phi)}$ where ${\phi}$ is a spherical coordinate (angle with the ${z}$-axis) and ${P_m}$ is the Legendre polynomial of degree ${m}$.

The positivity condition requires ${P_m(t) > 0}$ for ${t>c}$. In other words, the bound on ${\theta}$ comes from the greatest zero of the Legendre polynomial. As is true for orthogonal polynomials in general, the zeros are interlaced: that is, the zeros of ${P_m=0}$ appear strictly between any two consecutive zeros of ${P_{m+1}}$. It follows that the value of the greatest zero grows with ${m}$. Thus, it is smallest when ${m=2}$. Since ${P_2(t) = (3t^2-1)/2}$, the zero of interest is ${1/\sqrt{3}}$, and we conclude that ${c \ge 1/\sqrt{3}}$. Hence the magic angle.

The magic angle is easy to visualize: it is formed by an edge of a cube and its space diagonal. So, the magic cone with vertex at (0,0,0) is the one that passes through (1, 1, 1), as shown above.

In other dimensions the zonal harmonics are expressed in terms of Gegenbauer polynomials (which reduce to Chebyshev polynomials in dimensions 2 and 4). The above argument applies to them just as well. The relevant Gegenbauer polynomial of degree ${2}$ is ${nx^2-1}$ up to a constant. Thus, in ${\mathbb R^n}$ the magic angle is ${\cos^{-1}(1/\sqrt{n})}$, illustrated by the harmonic function ${u(x)=nx_n^2 - |x|^2}$.

This analysis is reminiscent of the Hopf Lemma which asserts that if a positive harmonic function in a smooth domain has boundary value 0 at some point, the normal derivative cannot vanish at that point. The smoothness requirement is typically expressed as the interior ball condition: one can touch every boundary point by a ball contained in the domain. The consideration of magic angles shows that if the function is also harmonic in a larger domain, the interior ball condition can be replaced by the interior cone condition, provided that the opening angle of the cone is greater than ${\cos^{-1}(1/\sqrt{n})}$.

## Measurability of Banach space valued functions

There is only an indirect proof of the existence of a function ${f\colon [0, 1]\to \mathbb R}$ that is not Lebesgue measurable. But it’s easy to give an explicit example when the codomain of ${f}$ is a Banach space: just let ${b(t)}$ be the sequence of the binary digits of ${t}$, considered as an element of the sequence space ${\ell_\infty}$.

Why is ${b}$ not measurable? Recall that a Banach space-valued function ${f}$ is (Bochner) measurable iff there is a sequence of simple functions ${\sum v_k \chi_{A_k}}$ (finite sum, measurable ${A_k}$, arbitrary vectors ${v_k}$) that converges to ${f}$ almost everywhere. This property implies that, with an exception of a null set, the range of ${f}$ lies in the separable subspace spanned by all the vectors ${v_k}$ used in the sequence of simple functions. But ${b}$ has the property ${\|b(t)-b(s)\|=1}$ whenever ${t\ne s}$, so the image of any uncountable set under ${b}$ is nonseparable.

Another way to look at this: on the interval [0, 1) the function ${b}$ is injective and its range has discrete topology, which implies that every subset of [0, 1) is the preimage of some open subset of ${\ell_\infty}$ under ${b}$.

The binary-digits functions can also be used to illustrate an issue with the duality of Lebesgue-Bochner spaces ${L_p(0, 1; X)}$ where ${X}$ is a Banach space and ${1\le p<\infty}$. (So, ${f}$ belongs to this space iff it is Bochner measurable and the ${L^p}$ norm of ${\|f\|\colon [0, 1]\to [0, \infty)}$ is finite.) In general we do not have the expected relation ${L_p(0, 1; X)^* = L_q(0, 1; X^*)}$ with ${1/p+1/q=1}$. The natural isometric embedding of ${L_q(0, 1; X^*)}$ into ${L_p(0, 1; X)^*}$ is still there: any ${g\in L_q(0, 1; X^*)}$ acts on ${L_p(0, 1; X)}$ by ${f\mapsto \int \langle f(t), g(t) \rangle\, dt}$. But unless ${X^*}$ has the Radon–Nikodym property, these are more bounded linear functionals on ${L_p(0, 1; X)}$.

To construct such a functional, let ${b_n(t)}$ be the ${n}$-th binary digit of ${t}$. Given ${f\in L_1(0, 1; \ell_1)}$, write it in coordinates as ${(f_1, f_2, \dots)}$ and define ${\varphi(f) = \sum_n \int_0^1 f_n b_n}$. This is a bounded linear functional, since ${|\varphi(f)|\le \sum_n \int_0^1 |f_n| = \|f\|}$. But there is no function ${g\in L_\infty(0, 1; \ell_\infty)}$ that represents it, i.e., ${\varphi(f) = \int_0^1 \langle f(t), g(t)\rangle \,dt = \sum_n \int_0^1 f_n g_n }$. Indeed, if such ${g}$ existed then by considering ${f}$ with only one nonzero coordinate, we find that ${g_n}$ must be ${b_n}$, using the duality ${L_1^* = L_\infty}$ in the scalar case. But the function ${[0, 1]\to \ell_\infty}$ with the components ${(b_1, b_2, \dots)}$ is not measurable, as shown above.

This example, which applies to all ${1\le p<\infty}$, also serves as a reminder that the duality relation ${L_p(0, 1; X)^* = L_q(0, 1; X^*)}$ depends on the dual space ${X^*}$ having the Radon-Nikodym property (RNP), not ${X}$ itself. Indeed, ${X=\ell_1}$ has the RNP; its dual does not.

The importance of ${X^*}$ having the RNP becomes clear once one tries to follow the usual proof of ${L_p^*=L_q}$. Given ${\varphi\in L_p(0,1;X)^*}$, we can define an ${X^*}$-valued measure ${\tau}$ on ${[0, 1]}$ by ${\tau(A)(v) = \varphi( v\chi_A)}$ where ${A\subset [0, 1]}$ is Lebesgue measurable and ${v\in X}$. This measure has reasonable finiteness and continuity properties coming from ${\varphi}$ being a bounded functional. Still, the existence of a density ${g\colon [0, 1]\to X^*}$ of the measure ${\tau}$ depends on the structure of the Banach space ${X^*}$.

## Immaculate perfection

— I am so lucky. I was born on April 4th 1944, that’s 4/4/44. If you add that up, it comes to 16, one six. One plus six is seven: luckiest number of all.
— It’s more than math, Mike, it’s… immaculate perfection.

The value of 7 for the the digital root indeed seems preferable, at least on this blog. But looking at the dates formed by repeated digits is too anthropocentric (not to mention xkcd1179). Besides, mathematics already has a concept of perfect numbers: being equal to the sum of proper divisors.

So, are there any perfect numbers with digital root equal to 7, equivalently ${n\equiv 7\bmod 9}$? For even perfect numbers, one can expect that the Euclid-Euler theorem ${n = 2^{p-1} (2^p-1)}$ will help with the answer, but even so, the solution turns out to be surprisingly simple.

A lucky coincidence is that ${2^6\equiv 1\bmod 9}$ (${64 = 1 + 7\cdot 9}$). Since all primes greater than 3 have remainder 1 or 5 mod 6, there are only a few cases to check:

• ${p\equiv 1\bmod 6}$, hence ${2^{p-1}\equiv 2^{1-1} \equiv 1 \bmod 9}$ and ${2^p-1\equiv 2^1-1 \equiv 1\bmod 9}$. In this case ${n\equiv 1\cdot 1 \equiv 1 \bmod 9}$
• ${p\equiv 5\bmod 6}$, hence ${2^{p-1}\equiv 2^{5-1} \equiv 7 \bmod 9}$ and ${2^p-1\equiv 2^5-1 \equiv 4\bmod 9}$. In this case ${n\equiv 7\cdot 4 \equiv 1 \bmod 9}$
• Exceptional cases p=2, 3 correspond to n = 6, 28. The latter is congruent to 1 mod 9, matching the preceding cases.

So, every even perfect number except 6 is congruent to 1 mod 9. Should a perfect number be congruent to 7 mod 9, it has to be odd. An odd perfect number, should it exist, must be either divisible by 9 (not good for us) or congruent to 1 mod 12. (Reference: On the form of an odd perfect number). The condition ${n\equiv 1 \bmod 12}$ is consistent with ${n\equiv 7 \bmod 9}$: for example, 25 satisfies both. Come to think of it, the sum of proper divisors of 25 is… not 25. But there might still be a 7 mod 9 perfect number out there.

What numbers k have the property that the sequence {n mod k : n is perfect} stabilizes? Assuming odd perfect numbers do not exist, every power of 2 has this property. So do 3 and 9 by the above (and therefore, their products with powers of 2). I don’t know any other such k, in particular n mod 27 does not appear to stabilize. Of course, it is impossible to prove any negative result here since we don’t know if there are infinitely many perfect numbers.

## Bipartite metric spaces and positive kernels

Bipartite metric spaces provide simple examples of highly non-Euclidean metric spaces. Let ${A}$ and ${B}$ be two disjoint abstract sets, say, with ${m}$ and ${n}$ points respectively. Define a metric on the union ${A\cup B}$ as follows, where ${a, b, c}$ are some positive numbers:

• The distance between distinct elements of ${A}$ is ${a}$.
• The distance between distinct elements of ${B}$ is ${b}$.
• The distance between two elements from different sets is ${c}$.

The triangle inequality is satisfied provided that ${a, b\le 2c}$. When ${c}$ is large, the space is not very interesting: it is approximately the union of two simplexes placed at some large distance from each other. On the other hand, when ${a=b=2c}$ we encounter a bizarre situation: every point of ${B}$ is a “midpoint” between any two distinct points of ${A}$, and vice versa.

A function ${K\colon X\times X \to \mathbb R}$ is called a positive kernel if the matrix ${K(x_i, x_j)_{i, j=1}^n}$ is positive semidefinite for any choice of ${n}$ and ${x_1, \dots, x_n\in X}$. A classical theorem of Schoenberg says that a metric space ${(X, d)}$ is isometric to a subset of a Hilbert space if and only if ${\exp(-\gamma d^2)}$ is a positive kernel for all ${\gamma >0}$. One may ask: is there any function ${f\colon [0, \infty)\to\mathbb R}$ such that ${f\circ d}$ is a positive kernel on ${X}$ for every metric space ${(X, d)}$?

Suppose ${f}$ is such a function. Applying it to the bipartite metric defined above, we find that the following matrix must be positive semidefinite (the case ${m=3}$, ${n=2}$ shown; in general the diagonal blocks have sizes ${m\times m}$ and ${n\times n}$):

${\displaystyle \begin{pmatrix} f(0) & f(a) & f(a) & f(c) & f(c) \\ f(a) & f(0) & f(a) & f(c) & f(c) \\ f(a) & f(a) & f(0) & f(c) & f(c) \\ f(c) & f(c) & f(c) & f(0) & f(b) \\ f(c) & f(c) & f(c) & f(b) & f(0) \end{pmatrix} }$

Test the PSD property against vectors of the form ${(s, s, s, t, t)}$, that is ${m}$ entries equal to some number ${s}$ followed by ${n}$ entries equal to some number ${t}$. The result is the quadratic form ${As^2 + Bt^2 + 2C^2st }$ where ${A = (m^2-m) f(a)+mf(0)}$, ${B = (n^2-n)f(b)+nf(0)}$, and ${C = mnf(c)}$.

For this quadratic form to be positive semidefinite, we need ${A\ge 0}$ and ${C^2\le AB}$ to hold. The inequality ${A\ge 0}$ amounts to ${f(a)\ge 0}$ and ${f(0)\ge 0}$, because ${m}$ can be any positive integer. Simply put, ${f}$ must be nonnegative.

The inequality ${C^2\le AB}$ simplifies to

${ mn f(c)^2 \le ( (m-1) f(a)+f(0)) \cdot ((n-1)f(b)+ f(0)) }$

When ${m=n=1}$ and ${a=b}$, we obtain ${f(c)\le f(0)}$; so, ${f}$ is maximized at ${0}$. Letting ${m, n\to \infty}$ yields ${f(c)^2 \le f(a)f(b)}$. What does this mean?

If ${c=(a+b)/2}$, the inequality ${f(c)^2 \le f(a)f(b)}$ amounts to log-convexity of ${f}$, and there are lots of log-convex functions. But crucially, our only constraint on ${c}$ in the construction of a bipartite metric space is ${c\ge \max(a, b)/2}$. In the special case ${a=b}$ we find that ${f(c)\le f(a)}$ whenever ${c\ge a/2}$ (and ${a>0}$). On one hand, ${f}$ must be nonincreasing because all ${c\ge a}$ are allowed. On the other hand, ${f(a/2)\le f(a)}$ implies that the sequence ${k\mapsto f(2^k)}$ is nondecreasing for ${k\in\mathbb Z}$. These conditions can be satisfied together only if ${f}$ is constant on ${(0, \infty)}$.

And there you have it, a complete description of functions ${f\colon [0, \infty)\to\mathbb R}$ that transform every distance matrix into a positive semidefinite matrix: ${f\equiv c\ge 0}$ on ${(0, \infty)}$, and ${f(0)\ge c}$. It is immediate that such ${f}$ indeed works for all metric spaces, not just bipartite ones.