From: israel@math.ubc.ca (Robert Israel) Newsgroups: sci.math Subject: Re: IT IS A POLYNOMIAL Date: 5 May 1998 20:27:44 GMT Alexander Abian wrote: > Can you give as clear and as simple as possible proof of the following > in the Real Analysis? > > Let f be an infinitely differentiable function on [0, 1] such > that at every x in [0, 1] an n-th order derivative of f is zero, > where n depends on x. Prove that f is a polynomial. It brings back memories: this problem (the "polynomial problem") was going around U. of Chicago ca. 1970. Suppose f is not a polynomial. Let C = {x: there is no neighbourhood of x on which f is a polynomial}. A_n = {x: f^(n)(x) = 0}. Clearly C is closed and nonempty, and A_n closed with union_n A_n = [0,1]. Applying the Baire Category Theorem to C, A_n intersect C has nonempty interior in C for some n, i.e. there exist x0 in C and delta > 0 such that C intersect (x0 - delta, x0 + delta) is contained in A_n. Suppose f^(k)(x0) <> 0 for some k > n. By Taylor's Theorem, we get f^(n)(x) <> 0 for 0 < |x - x0| < eta (for some eta > 0). Taking eta < delta, this implies C intersect (x0 - eta, x0 + eta) = { x0 }, but then f is a polynomial on (x0 - eta, x0) and on (x0, x0 + eta) and hence on (x0 - eta, x0 + eta), contradicting x0 in C. So we must conclude that f^(k)(x0) = 0 for all k >= n. The same is true for all points in C intersect (x0 - delta, x0 + delta). Now suppose c in (x0 - delta, x0 + delta) \ C. Then f is a polynomial on some neighbourhood of c. Let [a,b] be the maximal interval containing c on which f is a polynomial. Then a or b (say b) is in (x0 - delta, x0 + delta) intersect C, so that f^(k)(b) = 0 for all k >= n. Now if f has degree p on [a,b], f^(p) <> 0 on [a,b] so p < n, and then f^(k)(c) = 0 for all k >= n. We conclude that f^(k)(x) = 0 for all k >= n and all x in (x0 - delta, x0 + delta). But then f is a polynomial on (x0 - delta, x0 + delta), contradiction. Robert Israel israel@math.ubc.ca Department of Mathematics (604) 822-3629 University of British Columbia fax 822-6074 Vancouver, BC, Canada V6T 1Z2