A problem with polynomial functions
by Dan Shved
Here is a problem that popped up in my head about a year ago.
Problem 1. A function is polynomial of each argument, i.e. for every the function can be represented by a polynomial with integer coefficients, and the same goes for for every . Is it true then that must be polynomial of both arguments simultaneously, i.e. representable by an element of ?
If you’ve ever been a freshman in an analysis course, you are bound to be aware of a common trap with the notion of differentiability. The trap is like this: you are asked to prove that a function is differentiable everywhere. You have several months of single-variable analysis behind your back. So your first instinct may be to prove that and exist at every point and be done with it. But this is not enough, as demonstrated by
Problem 1 asks if there is the same trap with polynomiality (not a word?) as there is with differentiability. When I came up with problem 1 I was busy with more “serious” math, so the question didn’t get any real thought. I always suspected that it would’t take long to solve, but somehow I was too lazy to do the job. That is, I was until this morning, when I finally put in the required several minutes and got this thing sorted out. If you’re curious, you might want to do the same before moving on to the next paragraph.
Without further ado, the answer to problem 1: no. A function can be polynomial of each argument, but not polynomial overall. Here is an example:
The sum is actually finite at every point : all the terms with are zero. Also, if we fix , then becomes a polynomial of with degree at most .
It remains to prove that is not polynomial itself. But if it were, then would too. (I wonder if this was grammatically correct). And it is clear that grows faster than any polynomial, so we are done.
Now, what if we replace with another ring? Something larger maybe, like all rational numbers, or all real numbers. Will the trick still work? In case of the rationals it will, with minor modifications.
Exercise. Problem 1 still has a negative answer when is replaced with .
But if we move on from to the uncountable , the trick stops working. In fact, the problem changes its answer.
Fact of life. If a function is polynomial of each argument, then it is polynomial.
Proof. Let’s introduce some new notation for convenience. For every , define by . Similarly, for every let , . We know that all the functions and are polynomial.
First, we prove that there is a natural and an infinite set such that for every . To do this, for every define . Clearly, . is uncountably infinite and is countable. It follows that there is a such that is infinite, which gives us the desired and .
OK, so for every we know that is a polynomial function with degree at most . This means that there are functions such that
for all , .
The next order of business is to prove that each is representable by a polynomial, i.e. that there is a such that for all . This is immediately obvious for , because . It is only a bit more difficult for an arbitrary . It is an easy exercise to show that coefficients of a polynomial with degree at most can be found as linear combinations of its values at distinct points. This means that for every there exist constants such that
for all , and thus is a linear combination of functions . Therefore, is polynomial.
And we are almost done. We have polynomials , , that have the same values as on set . Now build a new polynomial function defined on all of like this:
It is obvious that . If we fix an arbitrary , we see that functions and are both polynomial and have the same restriction to . Since is infinite, it follows that these two functions coincide everywhere, i.e. for every . But then and are the same function, and the proof is complete.