This will be the last of this series of posts on my love affair with functional equations (here are links to parts one, two and three).
1. A simple solution of the functional equation
In the previous posts, I told of how I came to know of the functional equations
(*) $latex f(t) = fleft(frac{t+1}{2}right) + f left( frac{t-1}{2}right) ,, , ,, t in [-1,1]$
and more generally
(**) $latex f(t) = f(delta_1(t)) + f(delta_2(t)) ,, , ,, t in [-1,1]$
(where $latex delta_1$ and $latex delta_2$ satisfy some additional conditions) and my long journey to discover that these equations have, and now I will give it away…
… so I told of my journey to discover that equation (*) (and more generally (**)) always has continuous solutions which are not of the form $latex f(t) = ct$. After a very long time in which I considered this an interesting question, I discovered a proof of this fact which I thought (and I still think) is very surprising and interesting. But it turned out in the end, that this problem, for the special case (*), is much easier than I had thought. In fact, I even give a similar problem (motivated by chapter 6 in Emil Artin’s “The Gamma Function”) as a homework exercise in my Calculus class; the Hebrew speaking readers can check out Question 2 (d) in the “for submission” part, and question 2 in the “not for submission” part in this problems sheet (to be honest, I have been told by some students that none of the students solved this problem).
Eliahu Levy, my friend and collaborator from the Technion, and who is one of the rare mathematicians that always has his eyes and ears and mind and heart open, kindly brought to my attention the following simple solution to (*). Let $latex f$ be a (say) continuous function that satisfies (*). Then one may compute the Fourier coefficients $latex {hat{f}(n)}_{n in mathbb{Z}}$ of $latex f$ and use the functional equation to obtain a constraint on the coefficients:
$latex hat{f}(n) = frac{1}{2} int_{-1}^1 f(t) e^{-npi i t} dt = frac{1}{2} int_{-1}^{1} fleft(frac{t+1}{2}right)e^{-npi i t}dt + frac{1}{2} int_{-1}^{1} fleft(frac{t-1}{2}right)e^{-npi i t} dt = $
$latex = int_{0}^1 f(u) e^{n pi i 2u} e^{-n pi i} du + int_{-1}^0 f(u) e^{n pi i 2u} e^{n pi i} du =$
$latex = 2 (-1)^n hat{f}(2n),$
in short: if $latex f$ is a solution to (*) then
(C) $latex hat{f}(2n) = frac{(-1)^n}{2} hat{f}(n)$
for all $latex n$. This leads one to the guess that if $latex f$ has a Fourier series that converges uniformly, and if the Fourier coefficients of $latex f$ satisfy (C) above, then $latex f$ is a continuous solution to (*). This can be verified, and in any case, one may write down infinitely many linearly independent continuous solutions to (*) which are not linear, such as
$latex f(t) = e^{mpi i t} – sum_{n=1}^infty frac{1}{2^n} e^{2^n m pi i t} ,$
where $latex m$ is an odd integer (to see that they are linearly independent (and not linear), recall the uniqueness of Fourier coefficients).
This solution is by far more straightforward and natural then the proof that I eventually found (and linked in the previous post). It does not answer the question for any functional equation of the form (**). However, since (*) serves as a guiding example it is interesting enough, and in any case it shows that to obtain uniqueness of solutions to (**) (in the sense that there is only a one dimensional space of solutions) one does need, in general, to impose an additional conditions.
Note that the solutions found this way are not necessarily increasing (the solutions I found in my paper were homeomorphisms). Eliahu Levy has also told me of a way to find increasing solutions, but I will not write about that now (perhaps I will write in a later post, and perhaps Eliahu will comment about it 🙂 ).
2. Some things that I have learned
Here are some things that I learned from this story.
- As I first found, you never can tell what will lead you where. I found the solution to a problem I was interested in by surprise, while thinking about something different (not completely different, but still not obviously connected).
- On the other hand, as I later found, there are some basic tricks that one has to try first, before the fancy stuff. Of course, I did try to use Fourier series to solve this problem, but something went wrong with the computations and I looked for a different way. An experienced mathematician should understand that (*) has immediate consequences for the Fourier coefficients, and this understanding should be robust enough to survive miscalculations.
- I still ask myself how I overlooked the solutions given above for so long. If I know myself, or at least believe that the person that I used to be a few years ago has the same faults as the person writing these lines, then I must come to the conclusion that I wanted the functional equation (*) to have no non-linear continuous solutions. This happens to me often enough, that when working on a problem that I begin to have an opinion what the answer should be. This is a good thing as well as a bad thing. I would not be hasty to give up having my opinions, but one should be careful: in some cases, my opinion has guided me the wrong way.
- On the other hand, the wrong way is sometimes a very good way to take. The wrong way is at times more interesting, by far (to be honest, I am not working in mathematics because I am searching for the Truth, but because I am searching for something interesting).
- Perhaps the most important lesson I have learned from this story is that one should not work in isolation and on isolated problems. One should strive to be part of a large and diverse community and talk to people about problems and thoughts. This is important not only because it is much more fun to be part of a community; it is important because it helps one form a map of the landscape: which problems are easy and which are hard, which are important and which are esoteric.
Interesting. To understand better the history of the story, perhaps you can tell the readers when exactly Eliahu Levy told you about his solution?
Another thing: I think you somewhat reduced the contribution of your paper, since actually you solved (**) but this fact is not so evident from this sequence of posts. However, it will be interesting to check whether Levy’s ideas enable one to solve (**).
As for your lessons: well, I guess that you also learned that there exist some happy (mathematical) moments that worth a lot to the one who experiences them. Anyway, regarding item 5, note that, as implicit from item 1, sometimes actually these esoteric problems can lead to a progress elsewhere, so they do have some value. Sometimes they just lead one to have a degree, which is something valuable by itself even if no real impact is made elsewhere. In general, what seems as good/bad in one moment or one stage in one’s career, can be evaluated differently in other stages. As is somewhat implied from item 4, for you this problem was attractive and valuable in certain important stages in your career, although now one can underestimate this in retrospective or by looking at some lists of publications, etc. And besides, there is always the chance that somewhere in the future such problems will be considered as important. The history of science is full of such examples. Something related was written in a comment to another post in your blog:
http://noncommutativeanalysis.wordpress.com/2012/09/17/functional-analysis-a-preface-to-the-introduction/comment-page-1/#comment-2
Hello Daniel,
Thanks for you comment!
Regarding your last paragraph, do not worry: I do not regret spending so much time on the problem. Thinking about it, I do not regret anything I did in mathematics (that makes it sound like a very safe profession).
Incidentally, Eliahu told me his solution five years after I solved it – that’s as long as it took me to solve the problem!
Regarding your second paragraph: I did spend most of the time thinking directly about (*), so I kicked myself for not finding the Fourier series solution to the equation (*) by myself. And using Fourier series to solve functional equations was under my nose, see the following paper which I was aware of:
http://www.jstor.org/discover/10.4169/amer.math.monthly.119.03.245?uid=3738240&uid=2&uid=4&sid=21102300624811
arxiv version:
http://arxiv.org/pdf/1104.1201v2.pdf
Additional comments:
1. I still think that there is something misleading in the way your paper “CONJUGACY OF P-CONFIGURATIONS AND NONLINEAR
SOLUTIONS TO A CERTAIN CONDITIONAL CAUCHY
EQUATION”, Banach J. Math. Anal. 3 (2009), no. 1, 28–35,
http://www.emis.de/journals/BJMA/tex_v3_n1_a3.pdf
is regarded: after all, you solved (**) which is definitely more general than (*), and you also did other things. However, this is not so apparent from this post and the previous ones. One may obtain a wrong impression. Perhaps you may considering updating some of these posts only because of certain people sitting in certain committees (at least you may want to update certain relevant adjectives).
2. Regarding Fourier series: they were under my nose too when I thought about (*) about 9-10 years ago and definitely also in the year you solved the problem [this is somewhat evident from the bottom of the first page in v1 of the arXiv paper in your comment above 🙂 ].
Daniel, thanks for setting that straight. I had a humblilng experience, so if you say that I make my part look lesser than it really is, then I guess it shows through my writing.
As Orr has asked in the text, here is my comment.
I. Differentiable solutions to the functional equation.
As in the blog text, define the Fourier coefficients of $latex f$ as:
$latex F(n)=(1/2)int_{-1}^1exp(ipi nt)f(t),dt $.
(I put the 1/2 there only because the interval has length 2)
And one finds that for a general $latex L^1$ function $latex f$ , the functional equation holds (a.e.) iff
(*) $latex F(2n)=[(-1)^n](1/2)F(n)$ for all integer $latex n$.
For example, the function $latex f(t)equiv t$ has
$latex F(n)=[(-1)^n]/n ,, , ,, nne 0. F(0)=0,$
which clearly satisfies (*).
Now, if we want a solution $latex f$ to be absolutely continuous, i.e. be a.e. differentiable with an $latex L^1$ derivative, and be the integral of that derivative (note that, if we close the interval to a circle, $latex f$ is an integral of that $latex L^1$ function plus some multiple of the delta “function” at (-1, identified with 1)), then $latex nF(n)$ must be a multiple of $latex (-1)^n$ plus something tending to 0 when $latex ntoinfty$, thus, by (*), must be just that multiple of $latex (-1)^n$, making $latex f(t)= a$ multiple of $latex t$.
II. Of course, the Fourier coefficients approach refers to solutions not necessarily non-decreasing.
If one insists on non-decreasing $latex f(t)$, one can get many different solutions in a natural way as follows:
If $latex f$ is continuous non-decreasing, it defines a measure on the interval $latex [-1,1]$, giving to $latex [a,b]$ the mass $latex f(b)-f(a)$. The equation just says that this measure is invariant w.r.t. the two-to-one self-map of the
interval which maps $latex (t+1)/2$ and $latex (t-1)/2$ to $latex t$.
Now, if we work in $latex [0,1]$ instead of $latex [-1,1]$ and view each number in $latex [0,1]$ as the sequence of its binary digits (forget about fractions with a finite number of 0’s or of 1’s), this self-map is just the shift – erasing the first digit.
Thus we are asking about shift-invariant measures on the 0-1 sequences (a.k.a a stationary random 0-1 sequence) and there are many:
Choosing each digit independently with probability $latex 1/2$ will give the solution $latex f(t)=t$, but we can choose independently 0 with probability $latex p$ and 1 with probability $latex 1-p$, or not independently (the system has memory), yet still stationarily.
Note that as long as any initial finite sequence has positive probability, and there are no atoms (which, by the way, can only be ultimately periodic sequences, i.e. rational $latex t$), $latex f$ will be a continuous strictly increasing function, i.e. a homeomorphism.