Where have all the functional equations gone (the end of the story and the lessons I’ve learned)

by Orr Shalit

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

(*)  f(t) = f\left(\frac{t+1}{2}\right) + f \left( \frac{t-1}{2}\right) \,\, , \,\, t \in [-1,1]

and more generally

(**) f(t) = f(\delta_1(t)) + f(\delta_2(t)) \,\, , \,\, t \in [-1,1]

(where \delta_1 and \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 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 f be a (say) continuous function that satisfies (*). Then one may compute the Fourier coefficients \{\hat{f}(n)\}_{n \in \mathbb{Z}} of f and use the functional equation to obtain a constraint on the coefficients:

\hat{f}(n) = \frac{1}{2} \int_{-1}^1 f(t) e^{-n\pi i t} dt = \frac{1}{2} \int_{-1}^{1} f\left(\frac{t+1}{2}\right)e^{-n\pi i t}dt + \frac{1}{2} \int_{-1}^{1} f\left(\frac{t-1}{2}\right)e^{-n\pi i t} dt =

= \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 =

= 2 (-1)^n \hat{f}(2n),

in short: if f is a solution to (*) then

(C) \hat{f}(2n) = \frac{(-1)^n}{2} \hat{f}(n)

for all n. This leads one to the guess that if f has a Fourier series that converges uniformly, and if the Fourier coefficients of f satisfy (C) above, then 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

f(t) = e^{m\pi i t} - \sum_{n=1}^\infty \frac{1}{2^n} e^{2^n m \pi i t} ,

where 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.

  1. 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).
  2. 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.
  3. 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.
  4. 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).
  5. 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.
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