User bruce blackadar - MathOverflow

Non-analytic function with convergent Taylor series everywhere

2011-11-22T14:49:44Z

Is there a smooth function on an interval in $\mathbb R$, not analytic on any subinterval, whose Taylor series at every point has positive radius of convergence? The Fabius function might be an example, but this is questionable since the n'th derivative has maximum $2^{\sigma(n)}$, where $\sigma(n)=\frac{n(n+1)}{2}$, which is not quite good enough using a crude estimate for radius of convergence if there are points where many derivatives are close to the maximum.

Does local strict contractibility imply ANR?

2012-07-19T22:04:21Z

Say that a space (= compact metrizable space) $X$ is locally strictly contractible if, for every $p\in X$ and neighborhood $U$ of $p$, there is a neighborhood $V$ of $p$ which can be contracted to $p$ within $U$ relative to $p$, i.e. by a homotopy leaving $p$ fixed. (This may be bad terminology, and I welcome alternatives.)

Every ANR is locally strictly contractible. What about the converse? Borsuk's example of a locally contractible space which is not an ANR is not locally strictly contractible.

Incidentally, "locally contractible" is defined in different ways in various references. What is the currently accepted definition?

Answer by Bruce Blackadar for C*-algebras with bizzarre structure of projections

2012-01-26T23:44:49Z

Here is perhaps the simplest example. Let $A$ be the C*-algebra of all sequences of $2 \times 2$ matrices converging to a scalar multiple of diag(1,0). Let $p$ be the constant sequence diag(1,0), and $q$ a sequence of rank 1 projections converging to diag(1,0) but never exactly equal. Then $p$ and $q$ have no upper bound at all. This example can be tweaked to make it unital by allowing any limit matrix at infinity and taking $q$ to alternate diag(1,0) and nearby but unequal projections. Then $p$ and $q$ have no least upper bound.

Decomposability of Hausdorff measure

2012-01-24T23:25:22Z

Consider $s$-dimensional Hausdorff measure $\mathcal{H}^s$ on the Borel sets in $\mathbb{R}^n$.
$\mathcal{H}^s$ is not $\sigma$-finite if $s < n$, but it is semifinite (on Borel sets!)
Is it known whether $\mathcal{H}^s$ can be decomposable, i.e. can there be a partition of $\mathbb{R}^n$ into disjoint Borel sets ${X_i:i\in I}$ ($I$ necessarily uncountable) such that $\mathcal{H}^s(X_i)<\infty$ for all $i$ and, for every Borel set $E$, $\mathcal{H}^s(E)=\sum\limits_{i\in I}\mathcal{H}^s(E\cap X_i)$? Does the answer depend on any set-theoretic assumptions?

Answer by Bruce Blackadar for Existence of a smooth function with nowhere converging Taylor series at every point

2011-11-22T05:59:35Z

Good. I thought something like this might work.

One more question before I get off this subject, which might be harder (or easier!): what about the other extreme? Is there a smooth function on an interval in $\mathbb R$, not analytic on any subinterval, whose Taylor series at every point has positive radius of convergence? The Fabius function might be an example, but this is questionable since the $n$'th derivative has maximum $2^{\sigma(n)}$, where $\sigma(n)=\frac{n(n+1)}{2}$, which is not quite good enough using a crude estimate for radius of convergence if there are points where many derivatives are close to the maximum.

Answer by Bruce Blackadar for Existence of a smooth function with nowhere converging Taylor series at every point

2011-11-20T23:12:27Z

Neither of the responses exactly answers the question asked. The question asks whether the Taylor series of a smooth function at every point can have radius of convergence 0. This is more restrictive than not being analytic anywhere. The responses just treat this weaker question. The Taylor series of the Fabius function at any dyadic rational actually has infinite radius of convergence (only finitely many terms are nonzero) but does not represent the function on any interval.

It should be noted that the proof in the article of Kim and Kwon is incorrect, and although it is highly likely the function they consider is not analytic anywhere, I don't see how to prove it. (Their "proof" assumes that the sum of a tail of the series is analytic, which essentially contradicts the conclusion they are trying to prove!)

In fact, I do not know an example of what the questioner is asking, although the example of Kim and Kwon is a candidate. There is an example in Big Rudin (Chapter 19, problem 13) of a complex-valued smooth function on R whose Taylor series has radius of convergence zero at every point, but the argument there does not seem to be adaptable to getting a real-valued example; in particular, it is not at all clear that the real and imaginary parts of this example have the desired property. The question asked seems like a very interesting one, perhaps quite difficult and possibly open.

Comment by Bruce Blackadar

2012-07-23T01:00:33Z

This is a tricky (but worthwhile) example to understand. I'm still not sure I could write out a rigorous proof of what I claimed, but I'm pretty sure it's right.

Comment by Bruce Blackadar

2012-07-22T13:18:20Z

This space is a sort of multidimensional version of the infinite comb space (or, more accurately, the infinite ladder), where to contract everything must be moved through a corner point.

Comment by Bruce Blackadar

2012-07-22T02:16:03Z

Our notation seems to be different. I was using Borsuk's notation where the Hilbert cube is the product over all $n$ of $[0,1/n]$. In your notation the point $x$ has first coordinate 0 and all other coordinates 1/2. By the two ends of $X_m$ I meant the end with $x_m=0$ and the end with $x_m=1/m$ (Borsuk's notation).

Comment by Bruce Blackadar

2012-07-21T16:44:16Z

The problem is that to contract a neighborhood, you have to be able to work in a coordinate $m$ for which the neighborhood doesn't hit both ends of $X_m$, so points can be moved around one end. But for any neighborhood of the point $x$, there are only finitely many such coordinates to work with, so the end you move around can't be passed off to infinity. (Hope this cryptic description makes sense.)

Comment by Bruce Blackadar

2012-07-21T14:06:09Z

I guess I still don't see how to contract Borsuk's example to, say, the point with $x_1=0$ and $x_n=1/2n$ for $n>1$. It seems that any contraction of a neighborhood has to send this point via a point with $n$'th coordinate 0 or $1/n$ for some $n>1$.

Comment by Bruce Blackadar

2012-07-21T12:47:05Z

OK, I see my mistake in analyzing Borsuk's example. But the concept of local equiconnectedness seems to be the right one for the work I am currently doing, rather than pointed local contractibility. Thanks for pointing me in the right direction.

Comment by Bruce Blackadar

2012-07-20T19:34:56Z

Thanks. Not really being a topologist (although we operator algebra people sometimes like to think our subject includes topology as a special case), I'm not always up on correct topology terminology or the topology literature. What name do you suggest for the property I describe?

Comment by Bruce Blackadar

2012-07-20T01:55:18Z

Borsuk's definition of "locally contractible" is the same as the one I gave of "locally strictly contractible" without the "relative to $p$" part. A stronger definition sometimes used is that $X$ is locally contractible if every point has a neighborhood base of contractible open sets. By this stronger definition even an AR is not necessarily locally contractible.

Comment by Bruce Blackadar

2012-01-27T21:12:05Z

I carelessly misspoke slightly in the previous comment. The argument only works for semifinite measure spaces where each point is contained in a measurable set of finite measure. This of course holds for Hausdorff measure, but not in general; for a simple counterexample, see IX.1.9.3 and IX.1.9.6 in the Real Analysis manuscript on my website <wolfweb.unr.edu/homepage/bruceb/>; .

Comment by Bruce Blackadar

2012-01-26T03:45:07Z

This statement is true about Borel sets of infinite Hausdorff measure. See the article by J. Howroyd (Proc. London Math. Soc. (3) 70 (1995) 581-6O4) for the most general results known along this line. However, the result is false in general for $\mathcal{H}^s$-measurable sets; see section 439H of Fremlin's book. This is why I phrased the question in terms of Borel sets. What your argument shows is that a semifinite measure space whose $\sigma$-algebra has cardinality $\aleph_1$ is decomposable. It answers the question under the assumption of the continuum hypothesis.

Comment by Bruce Blackadar

2012-01-25T21:16:11Z

My first instinct is also that it is not possible if $0 < s < n$. But this might not be provable; it might be possible using some set-theoretic assumptions like maybe the continuum hypothesis. I have not found any discussion of this question in the literature.

Comment by Bruce Blackadar

2011-11-23T15:32:56Z

After I finally read some definitive history on this subject such as Dave's survey, what strikes me is how consistently many mathematicians (now myself included) have been ignorant of previous work in this area over a rather long time. I don't offhand recall any topic, at least in analysis, where basic examples and results have been rediscovered, reproved, and republished so often, and I would guess that even today many mathematicians would not know about this work.

Comment by Bruce Blackadar

2011-11-22T21:22:35Z

OK, thanks, I guess I didn't do my homework on this one.

Comment by Bruce Blackadar

2011-11-22T14:15:19Z

Sorry, I'll do this. I'm new to this site.

Comment by Bruce Blackadar

2011-11-21T16:29:08Z

Nice example! I assume this is original, but if not, can you give a reference? If $I$ is an interval in $\mathbb R$, does the set of functions with the proerty that all Taylor series have radius of convergence zero form a residual set in $C^\infty(I)$ in its usual topology?