Challenge 14: Smooth is not enough

[Office_Shredder]

A function [itex]f:\mathbb{R} \to \mathbb{R}[/itex] is called "smooth" if its k-th derivative exists for all k. A function is called analytic at a if its Taylor series
[tex] \sum_{n\geq 0} \frac{f^{(n)}(a)}{n!} (x-a)^n [/tex]
converges and is equal to f(x) in a small neighborhood around a.

The challenge: Construct or otherwise prove the existence of a function which is smooth, but which is not analytic on as large a set as possible.

 

[hilbert2]

I remember seeing an example of that kind of a function in a textbook years ago. It was like this:

Spoiler

Define the function ##f(x)## piecewise as follows:

##f(x)=\begin{cases}
0, & \text{if }x\leq 0\\
e^{-1/x^{2}}, & \text{if }x>0
\end{cases}##

Now ##f(x)## is smooth everywhere, including at the point ##x=0##, but its Taylor series about the point ##x=0## is just the identically zero function because all derivatives are zero at that point. Therefore the Taylor series does not converge to ##f(x)##, which has nonzero value if ##x>0##.

 

[mfb]

To extend hilbert2's solution,

Spoiler

we can modify the function a bit:$$f_t(x)=\begin{cases}
1, & \text{if }x= 0\\
1-e^{-1/(x-t)^{2}}, & \text{if }x \neq 0
\end{cases}$$
The Taylor series around x=t looks the same (apart from the constant term of course), all derivatives exist everywhere, the derivatives at 0 are x=t, but the Taylor series around x=t does not converge to the function.

Note that ##0<f_t(x)\leq 1##.
All derivatives have the shape ##f^{(n)}_0(x) = \frac{f(x)}{x^{3n}}p(x)## with a polynomial p(x) of degree <3n. This is smooth and goes to zero for ##x \to \pm \infty##, therefore the derivatives are bounded.

f_t, seen as function ##\mathbb{C} \to \mathbb{C}##, is analytic everywhere apart from x=t. Therefore, the taylor series around any other point converges to the function in some finite range.


This allows to find functions with "smooth is not enough" ("SINE") points at all integers:

$$g(x) =\sum_{i \in \mathbb{Z}} \frac{1}{1+i^4} f_i(x)$$

To verify that this sum is well-defined, note that ##|g(x)| < \sum_{i \in \mathbb{Z}} \frac{1}{1+i^4}## and the second sum is finite. In the same way, all derivatives are well-defined.

Around all integer values q, the taylor series of this sum converges to the function minus the contribution coming from f_q, therefore it does not converge to the function.

In the same way, we can add more similar terms with prefactors. As long as the sum of the prefactors is finite, the function is well-defined and SINE at all values r with an f_r-contribution:

$$g(x) = \sum_{p \in \mathbb{N}^+} \frac{1}{1+p^4} \sum_{i \in \mathbb{Z}} \frac{1}{1+i^4} f_{i/p}(x)$$

This is SINE at all rational numbers.

 

[hilbert2]

In mathematical physics, one sometimes encounters so-called "test functions" that are smooth at all points but have compact support, i.e. have zero value outside some interval. Obviously such functions can't be analytic at all points of the real axis. See http://en.wikipedia.org/wiki/Bump_function .

 

[jbunniii]

[edit] Found several mistakes - will work on an amended version when I get a chance.

 

[Office_Shredder]

Nice solutions guys! I'm not sure why this week's challenge gets the spoiler treatment though...

mfb, your first g(x) solution has that every function in the sum is analytic in a neighborhood of length 1/2 (or any other small number of your choice) around an integer point except for one, which show that it is not analytic at that point easily. But for your last example there seems to be a subtlety: if I pick a rational number p/q, and some interval I containing p/q, infinitely many of the functions in your sum are not analytic at the point p/q for the entire interval I. I think I can see an easy work around to this problem for your base function f_t(x) but it's not clear (and I'm interested if anyone knows) that it is necessarily the case that if you sum up any non-analytic function like that you will necessarily get a non-analytic function. My gut says that there must exist some stupid counterexample because there always exists some stupid counterexample.

 

[mfb]

Hmm. We are summing up infinitely many analytic functions (but with their non-analytic point arbitrarily close to the point we are looking at) and one non-analytic function. So we have to find out if this sum can be analytic. I would not expect this, but I don't have a proof.

 

[bpet]

A kth antiderivative of the Weierstrass function would be smooth but nowhere analytic.