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 ith component flipped. Rewrite the previous sum as

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

and add them together:

\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.

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):

  1. Download the roster from Blackboard Grade Center and import it into a spreadsheet
  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.

Continue reading “Almost norming functionals, Part 1”