For the sake of completeness

Let’s prove the completeness of $\ell^p$. The argument consists of two steps.

Claim 1. Suppose $X$ is a normed space in which every absolutely convergent series converges; that is, $\sum_{n=1}^{\infty} x_n$ converges whenever $x_n\in X$ are such that $\sum_{n=1}^{\infty} \|x_n\|$ converges. Then the space is complete.

Proof. Take a Cauchy sequence $\{y_n\}\subset X$. For $j=1,2,\dots$ find an integer $n_j$ such that $\|y_n-y_m\|<2^{-j}$ as long as $n,m\ge n_j$. (This is possible because the sequence is Cauchy.) Also let $n_0=1$ and consider the series $\sum_{j=1}^{\infty} (y_{n_{j}}-y_{n_{j-1}})$. By the hypothesis this series converges. Its partial sums simplify (telescope) to $y_{n_j}-y_1$. Hence the subsequence $\{y_{n_j}\}$ has a limit. It remains to apply a general theorem about metric spaces: if a Cauchy sequence has a convergent subsequence, then the entire sequence converges. This proves Claim 1.

Claim 2. Every absolutely convergent series in $\ell^p$ converges.

Proof. The elements of $\ell^p$ are functions from $\mathbb N$ to $\mathbb C$, so let’s write them as such: $f_j\colon \mathbb N\to \mathbb C$. (This avoids confusion of indices.) Suppose the series $\sum_{j=1}^{\infty} \|f_j\|$ converges. Then for any $n$ the series $\sum_{j=1}^{\infty} |f_j(n)|$ also converges, by Comparison Test. Hence $\sum_{j=1}^{\infty} f_j(n)$ converges (absolutely convergent implies convergent for series of real or complex numbers). Let $f(n) = \sum_{j=1}^{\infty} f_j(n)$. So far the convergence is only pointwise, so we are not done. We still have to show that the series converges in $\ell^p$, that is, its tails have small $\ell^2$ norm: $\sum_{n=1}^\infty |\sum_{j=k}^{\infty} f_j(n)|^p \to 0$ as $k\to\infty$.

What we need now is a dominating function, so that we can apply the Dominated Convergence Theorem. Namely, we need a function $g\colon \mathbb N\to [0,\infty)$ such that
(1) $\sum_{n=1}^{\infty} g(n)<\infty$, and
(2) $|\sum_{j=k}^{\infty} f_j(n)|^p \le g(n)$ for all $k,n$.

Set $g=(\sum_{j=1}^{\infty} |f_j|)^p$. Then (2) follows from the triangle inequality. Also, $g$ is the increasing limit of functions $g_k =(\sum_{j=1}^k |f_j|)^p$, for which we have
$\sum_n g_k(n) \le (\sum_{j=1}^k \|f_j\|)^p \le (\sum_{j=1}^{\infty} \|f_j\|)^p<\infty$
using the triangle inequality in $\ell^p$. Therefore, $\sum_n g(n)<\infty$ by the Monotone Convergence Theorem.

Almost norming functionals, Part 2

Let $E$ be a real Banach space with the dual $E^*$. Fix $\delta\in (0,1)$ and call a linear functional $e^*\in E^*$ almost norming for $e$ if $|e|=|e^*|=1$ and $e^*(e)\ge \delta$. In Part 1 I showed that in any Banach space there exists a continuous selection of almost norming functionals. Here I will prove that there is no uniformly continuous selection in $\ell_1$.

Claim. Let $S$ be the unit sphere in $\ell_1^n$, the $n$-dimensional $\ell_1$-space.  Suppose that $f\colon S\to \ell_{\infty}^n$ is a map such that $f(e)$ is almost norming $e$ in the above sense. Then the modulus of continuity $\omega_f$ satisfies $\omega_f(2/n)\ge 2\delta$.

(If an uniformly continuous selection was available in $\ell_1$, it would yield selections in $\ell_1^n$ with a modulus of continuity independent of $n$.)

Proof. Write $f=(f_1,\dots,f_n)$. For any $\epsilon\in \{-1,1\}^n$ we have $n^{-1}\epsilon \in S$, hence

$\sum\limits_{i=1}^n \epsilon_i f_i(n^{-1}\epsilon)\ge n\delta$ for all $\epsilon\in \{-1,1\}^n$. Sum over all $\epsilon$ and change the order of summation:

$\sum\limits_{i=1}^n \sum\limits_{\epsilon}\epsilon_i f_i(n^{-1}\epsilon)\ge n2^n\delta$

There exists $i\in\{1,2,\dots,n\}$ such that

$\sum\limits_{\epsilon}\epsilon_i f_i(n^{-1}\epsilon) \ge 2^n \delta$

Fix this $i$ from now on. Define $\tilde \epsilon$ to be the same $\pm$ vector as $\epsilon$, but with the $i$th component flipped. Rewrite the previous sum as

$\sum\limits_{\epsilon} -\epsilon_i f_i(n^{-1}\tilde \epsilon)\ge 2^n\delta$

$\sum\limits_{\epsilon}\epsilon_i [f_i(n^{-1}\epsilon)-f_i(n^{-1}\tilde \epsilon)]\ge 2^{n+1}\delta$

Since $\|n^{-1}\epsilon-n^{-1}\tilde \epsilon\|=2/n$, it follows that $2^n \omega_f(2/n) \ge 2^{n+1}\delta$, as claimed.

A relation between polynomials

This is a brief foray into algebra from a 2006 REU project at Texas A&M.

Given two polynomials $P,Q \in \mathbb C[z_1,\dots,z_n]$, we write $Q\preccurlyeq P$ if there is a differential operator $T\in \mathbb C[\frac{\partial}{\partial z_1},\dots, \frac{\partial}{\partial z_n}]$ such that $Q=T P$.

The relation $\preccurlyeq$ is reflexive and transitive, but is not antisymmetric. If both $Q\preccurlyeq P$ and $Q\preccurlyeq P$ hold, we say that $P$ and $Q$ are $\partial$-equivalent, denoted $P\thicksim Q$.

A polynomial is $\partial$-homogeneous if it is $\partial$-equivalent to a homogeneous polynomial. Obviously, any polynomial in one variable has this property. Polynomials in more than one variable usually do not have it.

The interesting thing about $\partial$-homogeneous polynomials is that they are refinable, meaning that one has a nontrivial identity of the form $P(z)=\sum_{j\in\mathbb Z^n} c_{j} P(\lambda z-j)$ where $c_{j}\in \mathbb C$, $j\in \mathbb Z^n$, and only finitely many of the coefficients $c_j$ are nonzero. The value of $\lambda$ does not matter as long as $|\lambda|\ne 0,1$. Conversely, every $\lambda$-refinable polynomial is $\partial$-homogeneous.

Rigidity of functions

Suppose that $u\colon \mathbb R^2\to\mathbb R$ is a continuously differentiable function such that at every point of the plane at least one of the partial derivatives $u_x, u_y$ is zero.

Prove that $u$ depends on just one variable (either $x$ or $y$). In other words, either $u_x\equiv 0$ or $u_y\equiv 0$.

Controlled bilipschitz extension

A map $f\colon X\to Y$ is $L$-bilipschitz if $L^{-1} |a-b| \le |f(a)-f(b)| \le L |a-b|$ for all $a,b\in X$. This definition makes sense if X and Y are general metric spaces, but let’s suppose they are subsets on the plane $\mathbb R^2$.

Definition 1. A set $A\subset \mathbb R^2$ has the BL extension property if any bilipschitz map $f\colon A\to\mathbb R^2$ can be extended to a bilipschitz map $F\colon \mathbb R^2\to\mathbb R^2$. (Extension means that $F$ is required to agree with $f$ on $A$.)

Lines and circles have the BL extension property. This was proved in early 1980s independently by Tukia, Jerison and Kenig, and Latfullin.

Definition 2. A set $A\subset \mathbb R^2$ has the controlled BL extension property if there exists a constant $C$ such that any $L$-bilipschitz map $f\colon A\to\mathbb R^2$ can be extended to a $C L$-bilipschitz map $F\colon \mathbb R^2\to\mathbb R^2$.

Clearly, Definition 2 asks for more than Definition 1. I can prove that a line has the controlled BL extension property, even with a modest constant such as $C=2000$. (Incidentally, one cannot take $C=1$.) I still can’t prove the controlled BL extension property for a circle.

Update: extension from line is done in this paper.

WeBWork class roster import

One way to import classroster into WeBWork (at SU):

2. The first four columns A,B,C,D will be Last Name, First Name, UserName, Student ID.
3. Append the column with the function
=CONCATENATE(D2,",";A2,",",B2,",C,,,,",C2,"@syr.edu,",C2)

(second row shown). Or

=CONCATENATE(D2;",";A2;",";B2;",C,,,,";C2;"@syr.edu,";C2)

if using OpenOffice.

4. Using WeBWork file manager, create a file roster.lst and paste this new column into it.
5. Use the Import Users command under Classlist editor.

Almost norming functionals, Part 1

Let $E$ be a real Banach space with the dual $E^*$. By the Hahn-Banach theorem, for every unit vector $e\in E$ there exists a functional $e^*\in E^*$ of unit norm such that $e^*(e)=1$. One says that $e^*$ is a norming functional for $e$. In general, one cannot choose $e^*$ so that it depends continuously on $e$. For example, the 2-dimensional space with $\ell_1$ norm does not allow such a continuous selection.

Fix $\delta\in (0,1)$ and call a linear functional $e^*$ almost norming for $e$ if $|e|=|e^*|=1$ and $e^*(e)\ge \delta$. In any Banach space there exists a continuous selection of almost norming functionals.