## Apropos of et al

I have 111 MathSciNet reviews posted, and there are three more articles on my desk that I should be reviewing instead of blogging. Even though I think of canceling my AMS membership, I don’t mind helping the society pay their bills (MathSciNet brings about 37% of the AMS revenue, according to their 2010-11 report.)

Sure, reviews need to be edited, especially when written by non-native English speakers like myself. Still, I’m unhappy with the edited version of my recent review:

This was the approach taken in the foundational paper by J. Heinonen et al. [J. Anal. Math. 85 (2001), 87-139]

The paper was written by J. Heinonen, P. Koskela, N. Shanmugalingam, and J. T. Tyson. Yes, it’s four names. Yes, the 14-letter name is not easy to pronounce without practice. But does the saving of 45 bytes justify omitting the names of people who spent many months, if not years, working on the paper? Absolutely not. The tradition of using “et al” for papers with more than 3 authors belongs to the age of typewriters.

P.S. I don’t think MathSciNet editors read my blog, so I emailed them.

P.P.S. The names are now restored. In the future I’ll be sure to add in “comments to the editor” that names should not be replaced by et al.

## This ain’t like dusting crops, boy

The hyperspace is a set of sets equipped with a metric or at least with a topology. Given a metric space $X$, let $\mathcal{H}(X)$ be the set of all nonempty closed subsets of $X$ with the Hausdorff metric: $d(A,B) if no matter where you are in one set, you can jump into the other by traveling less than $r$. So, the distance between letters S and U is the length of the longer green arrow.

The requirement of closedness ensures $d(A,B)>0$ for $A\ne B$. If $X$ is unbounded, then $d(A,B)$ will be infinite for some pairs of sets, which is natural: the hyperspace contains infinitely many parallel universes which do not interact, being at infinite distance from one another.

Every continuous surjection $f\colon X\to Y$ has an inverse $f^{-1}\colon Y\to \mathcal{H}(X)$ defined in the obvious way: $f^{-1}(y)=f^{-1}(y)$. Yay ambiguous notation! The subset of $\mathcal{H}(X)$ that consists of the singletons is naturally identified with $X$, so for bijective maps we recover the usual inverse.

Exercise: what conditions on $f$ guarantee that $f^{-1}$ is (a) continuous; (b) Lipschitz? After the previous post it should not be surprising that

• Even if $f$ is open and continuous, $f^{-1}$ may be discontinuous.
• If $f$ is a Lipschitz quotient, then $f^{-1}$ is Lipschitz.

Proofs are not like dusting crops—they are easier.

## Continuous:Lipschitz :: Open:?

A map is continuous if the preimage of every open set is open. If the topology is defined by a metric, we can reformulate this as: the inverse image of an open ball $B_R(f(x))$ contains an open ball $B_r(x)$. Like this:

But bringing these radii $R$ and $r$ into the picture will not serve any purpose unless we use them to quantify continuity. For example, if we insist that $r\ge cR$ for a fixed constant $c>0$, we arrive at the definition of a Lipschitz map.

But why do we look at the inverse image; what happens if we take the direct image instead? Then we get the definition of an open map: the image of every open set is open. Recasting this in metric terms: the image of an open ball $B_R(x)$ contains an open ball $B_r(f(x))$. Like this:

If we quantify openness by requiring $r\ge cR$ for a fixed $c>0$, we arrive at the definition of a co-Lipschitz map. [Caution: some people use “co-Lipschitz” to mean $|f(a)-f(b)|\ge c|a-b|$, which is a different condition. They coincide if $f$ is bijective.]

I don’t know if openness without continuity is good for anything other than torturing students with exercises such as: “Construct an open discontinuous map from $\mathbb R$ to $\mathbb R$.” We probably want both. At first one can hope that open continuous maps will have reasonable fibers $f^{-1}(x)$: something $(m-n)$-dimensional when going from $m$ dimensions to $n$, with $m\ge n$. The hope is futile: an open continuous map $f\colon \mathbb R^2\to\mathbb R^2$ can squeeze a line segment to a point (construction left as an exercise).

A map that is both Lipschitz and co-Lipschitz is called a Lipschitz quotient; this is a quantitative analog of “open continuous”. It turns out that for any Lipschitz quotient $f\colon \mathbb R^2\to\mathbb R^2$ the preimage of every point is a finite set. Moreover, $f$ factors as $f=g\circ h$ where $g$ is a complex polynomial and $h$ is a homeomorphism.

This is encouraging… but going even one dimension higher, it remains unknown whether a Lipschitz quotient $f\colon \mathbb R^3\to\mathbb R^3$ must have discrete fibers. For an overview of the subject, see Bill Johnson’s slides.

## Sweetened and flavored dessert made from gelatinous or starchy ingredients and milk

Takagi (高木) curves are fractals that are somehow less known than Cantor sets and Sierpinski carpets, yet they can also be useful as (counter-)examples. The general 高木 curve is the graph $y=f(x)$ of a function $f$ that is built from triangular waves. The $n$th generation wave has equation $y=2^{-n} \lbrace 2^n x \rbrace$ where $\lbrace\cdot\rbrace$ means the distance to the nearest integer. Six of these waves are pictured below.

Summation over $n$ creates the standard 高木 curve $T$, also known as the blancmange curve:

$\displaystyle y=\sum_{n=0}^{\infty} 2^{-n} \lbrace 2^n x\rbrace$

Note the prominent cusps at dyadic rationals: more on this later.

General 高木 curves are obtained by attaching coefficients $c_n$ to the terms of the above series. The simplest of these, and the one of most interest to me, is the alternating 高木 curve $T_{alt}$:

$\displaystyle y=\sum_{n=0}^{\infty} (-2)^{-n} \lbrace 2^n x\rbrace$

The alternation of signs destroys the cusps that are so prominent in $T$. Quantitatively speaking, the diameter of any subarc of $T_{alt}$ is bounded by the distance between its endpoints times a fixed constant. The curves with this property are called quasiarcs, and they are precisely the quasiconformal images of line segments.

Both $T$ and $T_{alt}$ have infinite length. More precisely, the length of the $n$th generation of either curve is between $\sqrt{(n+1)/2}$ and $\sqrt{n+1}+1$. Indeed, the derivative of $x\mapsto 2^{-k}\lbrace 2^k x\rbrace$ is just the Rademacher function $r_k$. Therefore, the total variation of the sum $\sum_{k=0}^n c_k 2^{-k}\lbrace 2^k x\rbrace$ is the $L^1$ norm of $\sum_{k=0}^n c_k r_k$. With $c_k=\pm 1$ the sharp form of the Хинчин inequality from the previous post yields

$\displaystyle 2^{-1/2}\sqrt{n+1} \le \left\|\sum_{k=0}^n c_k r_k\right\|_{L^1} \le \sqrt{n+1}$

For the upper bound I added 1 to account for the horizontal direction. Of course, the bound of real interest is the lower one, which proves unrectifiability. So far, a construction involving these curves shed a tiny bit of light on the following questions:

Which sets $K\subset \mathbb R^n$ have the property that any quasiconformal image of $K$ contains a rectifiable curve?

I won’t go (yet) into the reasons why this question arose. Any set with nonempty interior has the above property, since quasiconformal maps are homeomorphisms. A countable union of lines in the plane does not; this is what 高木 curves helped to show. The wide gap between these results remains to be filled.

## The Khintchine inequality

Today’s technology should make it possible to use the native transcription of names like Хинчин without inventing numerous ugly transliterations. The inequality is extremely useful in both analysis and probability: it says that the $L^p$ norm of a linear combination of Rademacher functions (see my post on the Walsh basis) can be computed from its coefficients, up to a multiplicative error that depends only on $p$. Best of all, this works even for the troublesome $p=1$; in fact for all $0. Formally stated, the inequality is

$\displaystyle A_p\sqrt{\sum c_n^2} \le \left\|\sum c_n r_n\right\|_{L^p} \le B_p\sqrt{\sum c_n^2}$

where the constants $A_p,B_p$ depend only on $p$. The orthogonality of Rademacher functions tells us that $A_2=B_2=1$, but what are the other constants? They were not found until almost 60 years after the inequality was proved. The precise values, established by Haagerup in 1982, behave in a somewhat unexpected way. Actually, only $A_p$ does. The upper bound is reasonably simple:

$\displaystyle B_p=\begin{cases} 1, \qquad 0

The lower bound takes an unexpected turn:

$\displaystyle A_p=\begin{cases} 2^{\frac{1}{2}-\frac{1}{p}},\qquad 0

The value of $p_0$ is determined by the continuity of $A_p$, and is not far from $2$: precisely, $p_0\approx 1.84742$. Looks like a bug in the design of the Universe.

For a concrete example, I took random coefficients $c_0...c_4$ and formed the linear combination shown above. Then computed its $L^p$ norm and the bounds in the Khintchine inequality. The norm is in red, the lower bound is green, the upper bound is yellow.

It’s a tight squeeze near $p=2$

## Another orthonormal basis: Hermite functions

This is an orthonormal basis for $L^2(\mathbb R)$. Since the measure of $\mathbb R$ is infinite, functions will have to decay at infinity in order to be in $L^2$. The Hermite functions are
$\displaystyle \Phi_n(x)=(2^n n! \sqrt{\pi})^{-1/2} H_n(x)e^{-x^2/2}$
where $H_n$ is the nth Hermite polynomial, defined by
$\displaystyle H_n(x)=(-1)^n e^{x^2} \left(\frac{d}{dx}\right)^n e^{-x^2}$.
The goal is to prove that the functions $\Phi_n$ can be obtained from $x^n e^{-x^2/2}$ via the Gram-Schmidt process. (They actually form a basis, but I won’t prove that.)

One can observe that the term $e^{-x^2/2}$ would be unnecessary if we considered the weighted space $L^2(\mathbb R, w)$ with weight $w(x)=e^{-x^2}$ and the inner product $\langle f,g\rangle=\int_{\mathbb R} fg\,w\,dx$. In this language, we orthogonalize the sequence of monomials $\lbrace x^n\rbrace\subset L^2(\mathbb R, w)$ and get the ON basis of polynomials $\{c_n H_n\}$ with $c_n = (2^n n! \sqrt{\pi})^{-1/2}$ being a normalizing constant. But since weighted spaces were never introduced in class, I’ll proceed with the original formulation. First, an unnecessary graph of $\Phi_0,\dots,\Phi_4$; the order is red, green, yellow, blue, magenta.

Claim 1. $H_n$ is a polynomial of degree $n$ with the leading term $2^n x^n$. Proof by induction, starting with $H_0=1$. Observe that

$\displaystyle H_{n+1}=- e^{x^2} \frac{d}{dx}\left(e^{-x^2} H_n\right) =2x H_n - H_n'$

where the first term has degree $n+1$ and the second $n-1$. So, their sum has degree exactly $n+1$, and the leading coefficient is $2^{n+1}$. Claim 1 is proved.

In particular, Claim 1 tells us that the span of the $\Phi_0,\dots,\Phi_n$ is the same as the span of $\lbrace x^k e^{-x^2/2}\colon 0\le k\le n\rbrace$.

Claim 2. $\Phi_m\perp \Phi_n$ for $m\ne n$. We may assume $m. Must show $\int_{\mathbb R} H_m(x) H_n(x) e^{-x^2}\,dx=0$. Since $H_m$ is a polynomial of degree $m, it suffices to prove

(*) $\displaystyle \int_{\mathbb R} x^k H_n(x) e^{-x^2}\,dx=0$ for integers $0\le k.

Rewrite (*) as $\int_{\mathbb R} x^k \left(\frac{d}{dx}\right)^n e^{-x^2} \,dx=0$ and integrate by parts repeatedly, throwing the derivatives onto $x^k$ until the poor guy can't handle it anymore and dies. No boundary terms appear because $e^{-x^2}$ decays superexponentially at infinity, easily beating any polynomial factors. Claim 2 is proved.

Combining Claim 1 and Claim 2, we see that $\Phi_n$ belongs to the $(n+1)$-dimensional space $\mathrm{span}\,\lbrace x^k e^{-x^2/2}\colon 0\le k\le n\rbrace$, and is orthogonal to the $n$-dimensional subspace $\mathrm{span}\,\lbrace x^k e^{-x^2/2}\colon 0\le k\le n-1\rbrace$. Since the “Gram-Schmidtization'' of $x^n e^{-x^2/2}$ has the same properties, we conclude that $\Phi_n$ agrees with this “Gram-Schmidtization'' up to a scalar factor.

It remains to prove that the scalar factor is unimodular ($\pm 1$ since we are over reals).

Claim 3. $\langle \Phi_n, \Phi_n\rangle=1$ for all $n$. To this end we must show $\int_{\mathbb R} H_n(x)H_n(x)e^{-x^2}\,dx =2^n n! \sqrt{\pi}$. Expand the first factor $H_n$ into monomials, use (*) to kill the degrees less than n, and recall Claim 1 to obtain
$\int_{\mathbb R} H_n(x)H_n(x)e^{-x^2}\,dx = 2^n \int_{\mathbb R} x^n H_n(x)e^{-x^2}\,dx = (-1)^n 2^n\int_{\mathbb R} x^n \left(\frac{d}{dx}\right)^n e^{-x^2} \,dx$.
As in the proof of Claim 2, we integrate by parts throwing the derivatives onto $x^n$. After n integrations the result is
$2^n \int_{\mathbb R} n! e^{-x^2} \,dx = 2^n n! \sqrt{\pi}$, as claimed.

P.S. According to Wikipedia, these are the “physicists’ Hermite polynomials”. The “probabilists’ Hermite polynomials” are normalized to have the leading coefficient 1.

## The Walsh basis

If you ask a random passerby to give you an orthonormal basis for $L^2[0,1]$, they will probably respond with $e_n(t)=\exp(2\pi i nt)$, $n\in \mathbb Z$. There is a lot to like about this exponential basis: most importantly, it diagonalizes the $\frac{d}{dt}$ operator: $\frac{d}{dt}e_n=2\pi i n e_n$. This property makes the exponential basis indispensable in the studies of differential equations. However, I prefer to describe the Walsh basis, which has several advantages:

• the basis functions take just two values $\pm 1$, which simplifies the computation of coefficients
• the proof of the basis property is easier than for the exponential basis
• there is a strong connection to probability: the Walsh expansion can be seen as conditional expectation, and the partial sums form a Doob martingale
• partial sums converge a.e. for any $L^1$ function, which is not the case for the exponential basis.

First, introduce the Rademacher functions $r_n=\mathrm{sign}\, \sin (2^{n+1} \pi t)$, $n=0,1,2,\dots$ (The enumeration is slightly different from what I used in class.) These are $r_0,r_1,r_2,r_3$:

Alternatively, one can define $r_n$ as the function which takes the values $+1,-1$ alternatively on the dyadic intervals $\displaystyle \bigg[\frac{j}{2^{n+1}},\frac{j+1}{2^{n+1}}\bigg)$.

To define the $n$th Walsh function $W_n$, express the index $n$ as the sum of powers of 2, i.e., $n=2^{p_1}+2^{p_2}+\dots$ and let $W_n=r_{p_1}r_{p_2}\dots$. For example, $W_{13}=r_3r_2r_0$ because $13=2^3+2^2+2^0$. Since the binary representation is unique, the definition makes sense. We also have $W_0=1$ because the product of an empty set of numbers is 1.

In class I checked that the set $\lbrace W_n\colon n=0,1,2,\dots\rbrace$ is orthonormal. Also, for any integer $k\ge 0$ the linear span of $\lbrace W_n\colon 0\le n< 2^k \rbrace$ is the space $V_k$ of all functions that are constant on the dyadic intervals of length $2^{-k}$. This follows by observing that $\lbrace W_n\colon 0\le n< 2^k \rbrace\subset V_k$ and that the dimension of $V_k$ is $2^k$.

To prove that the Walsh basis is indeed a basis, suppose that $h\in L^2[0,1]$ is orthogonal to all $W_n$. Since $h\perp V_k$ for all $k$, the integral of $h$ over any dyadic interval is zero (note that the characteristic function of any dyadic interval belongs to some $V_k$). But any subinterval $I\subset [0,1]$ can be written as a disjoint countable union of dyadic intervals: just take all dyadic intervals that are contained in $I$. (You don't necessarily get the right type of endpoints, but as long as we work with integrals, the difference between open and closed intervals is immaterial.) Thus, the integral of $h$ over any subinterval of $[0,1]$ is zero. By the Lebesgue differentiation theorem, for a.e. $t$ we have $\displaystyle h(t)=\lim_{\delta\to 0}\frac{1}{2\delta}\int_{t-\delta}^{t+\delta} h =0$. Thus $h=0$ as required.

The proof is even simpler if we use the non-centered form of the Lebesgue differentiation theorem: for a.e. $t$ the average $\frac{1}{b-a}\int_a^b h$ approaches $h(t)$ as $a,b\to t$ in such a way that $a\le t\le b$. Armed with this theorem, we can consider the sequence of dyadic intervals containing $t$, and immediately obtain $h(t)=0$ a.e.

Having proved that $\lbrace W_n\rbrace$ is a basis, let’s expand something in it. For example, this moderately ugly function $f$:

I used Maple to compute the coefficients $c_n=\langle f, W_n\rangle$ and plotted the partial sums $\sum_{n=0}^N c_n W_n$ for $N=1,3,7,15$:

Such partials sums (those that use $2^k$ basis functions) are particularly nice: they are obtained simply by averaging $f$ over each dyadic interval of length $2^{-k}$. In probability theory this is known as conditional expectation. The conditional expectation is a contraction in any $L^p$ space, including $L^1$ which gives so much trouble to the exponential basis. The highly oscillatory parts of $f$ are killed by the dyadic averaging; in contrast, when integrated against the exponentials, they may cleverly accumulate and destroy the convergence of partial sums.