I plan in two post to comment on a type of problems, given at some IMO level competitions. We have a function defined on the lattice points of the plane. The value of at any lattice point is the average of its values taken at all or some of the neighbouring points. For example
Such kind of functions we call harmonic. Of course, there exist plenty of such functions, for example constant ones. We also can define at three neighbouring points and spread out the values of the function complying to the condition. An interesting fact is that if we impose to be bounded, the only possibility is it to be constant. Moreover, if is lower (or upper) bounded, the same is also true. Two methods will be presented. The first one – using elementary math, the second one is the fastest one but uses a convex analysis fact – the Krein-Milman theorem.
(USA TST #1 IMO 2018. p2.) Find all functions such that for any integers and ,
One may skip this rather long explanation and jump directly to the solution in the next section, which starts from a scratch.
So, we have a bounded function on the set of lattice points. The value of at any point is the average of the values at its bottom neighbour and its left one . To see roughly why should be a constant, suppose takes only finite number of values. Then, takes its maximal value at some point . By we get and in the same way Now, we take again the maximal value of on and repeating it infinitely many times obtain is constant on entire .
It seems, an argument like that may help in the continuous case. Oh, not at all! This problem is indeed difficult. For example the problem presented at the beginning (see ) was given at a Bulgarian TST for BMO 2003 (see ). As far as I know, no one solved it. Nevertheless, let’s try it.
Suppose, satisfy . Then any linear combination also satisfy . Any translation of where also comply to that condition. Particularly also satisfies . may be interpret as the velocity of changing of . Now, we cannot take the maximal value of , but we may take a value as close to the as we want. Let . Using , it follows . Applying it again several times, we get
It means, for any fixed , taking small enough we can ensure be also close to . It is of little help, but applying the same argument to we could ensure the velocity of changing of is not too small for as long interval as we want. But it would mean can deviate a large amplitude, which is not allowed since is bounded.
Solution of Problem 1.
Seeking for contradiction, suppose there exists satisfying which is not constant. Then, there exists a point for which either or . WLOG, suppose the former holds (otherwise we can flip and ). We also can assume , otherwise we consider . Let . Clearly, also satisfies and it is bounded. Denote , hence . Fix some and let satisfies . Condition gives us . Applying it again several times, we get
where Finally, we have
The LHS is bounded, but we can make the RHS as large as we want by taking sufficiently small (thus as large as we want), contradiction.
 Bulgarian Mathematical Competitions, 2003 – 2006, P. Boyvalenkov, O. Mushkarov, E. Kolev, N. Nikolov. GIL Publishing House 2007.