186 3 Real Analysis
But we like more the nonstandard functional equations. Here is one, which is a
simplified version of a short-listed problem from the 42nd International Mathematical
Olympiad. We liked about it the fact that the auxiliary functionhfrom the solution
mimics, in a discrete situation, harmonicity—a fundamental concept in mathematics.
The solution applies the maximum modulus principle, which states that ifhis a harmonic
function then the maximum of|h|is attained on the boundary of the domain of definition.
Harmonic functions, characterized by the fact that the value at one point is the average
of the values in a neighborhood of the point, play a fundamental role in geometry. For
example, they encode geometric properties of their domain, a fact made explicit in Hodge
theory.
Example.Find all functionsf:{ 0 , 1 , 2 ,...}×{ 0 , 1 , 2 ,...}→Rsatisfying
f(p,q)={ 1
2 (f (p+^1 ,q−^1 )+f(p−^1 ,q+^1 ))+1ifpq =^0 ,
0ifpq= 0.Solution.We see thatf( 1 , 1 )=1. The defining relation givesf( 1 , 2 )= 1 +f( 2 , 1 )/ 2
andf( 2 , 1 )= 1 +f( 1 , 2 )/2, and hencef( 2 , 1 ) =f( 1 , 2 )=2. Thenf( 3 , 1 ) =
1 +f( 2 , 2 )/2,f( 2 , 2 )= 1 +f( 3 , 1 )/ 2 +f( 1 , 3 )/2,f( 1 , 3 )= 1 +f( 2 , 2 )/2. So
f( 2 , 2 )=4,f( 3 , 1 )=3,f( 1 , 3 )=3. Repeating such computations, we eventually
guess the explicit formulaf(p,q)=pq,p, q≥0. And indeed, this function satisfies
the condition from the statement. Are there other solutions to the problem? The answer
is no, but we need to prove it.
Assume thatf 1 andf 2 are both solutions to the functional equation. Leth=f 1 −f 2.
Thenhsatisfies
h(p, q)={ 1
2 (h(p+^1 ,q−^1 )+h(p−^1 ,q+^1 )) ifpq =^0 ,
0ifpq= 0.Fix a linep+q=n, and on this line pick(p 0 ,q 0 )the point that maximizes the value of
h. Because
h(p 0 ,q 0 )=1
2
(h(p 0 + 1 ,q 0 − 1 )+h(p 0 − 1 ,q 0 + 1 )),it follows thath(p 0 + 1 ,q 0 − 1 )=h(p 0 − 1 ,q 0 + 1 )=h(p 0 ,q 0 ). Shifting the point,
we eventually conclude thathis constant on the linep+q=n, and its value is equal to
h(n, 0 )=0. Sincenwas arbitrary, we see thathis identically equal to 0. Therefore,f 1 =
f 2 , the problem has a unique solution, and this solution isf(p,q)=pq,p, q≥0.
And now an example of a problem about a multivariable function, from the same
short list, submitted by B. Enescu (Romania).