HMMT 二月 2015 · 团队赛 · 第 7 题
HMMT February 2015 — Team Round — Problem 7
题目详情
- [ 35 ] Let f : [0 , 1] → C be a nonconstant complex-valued function on the real interval [0 , 1]. Prove that there exists ǫ > 0 (possibly depending on f ) such that for any polynomial P with complex coefficients, there exists a complex number z with | z | ≤ 1 such that | f ( | z | ) − P ( z ) | ≥ ǫ .
解析
- [ 35 ] Let f : [0 , 1] → C be a nonconstant complex-valued function on the real interval [0 , 1]. Prove that there exists ǫ > 0 (possibly depending on f ) such that for any polynomial P with complex coefficients, there exists a complex number z with | z | ≤ 1 such that | f ( | z | ) − P ( z ) | ≥ ǫ . Answer: N/A Here’s a brief summary of how we’ll proceed: Use a suitably-centered high-degree roots of unity filter, or alternatively integrate over suitably-centered circles (“continuous roots of unity filter”). We claim we can choose ǫ ( f ) = max( | f ( x ) − f ( x ) | ) / 2. Fix P and suppose for the sake of contradiction 1 2 iθ that for all z with | z | ≤ 1 it is the case that | f ( z ) − P ( z ) | < ǫ . We can write z = re so that iθ | f ( r ) − P ( re ) | < ǫ. Let p be a prime larger than deg( P ) and set θ = 0 , 2 π/p, 4 π/p, . . . in the above. Averaging, and using triangle inequality, p − 1 ∑ 1 2 πik/p | f ( r ) − P ( re ) | < ǫ. p k =0 The sum inside the absolute value is a roots of unity filter; since p > deg( P ) it is simply the constant term of P - denote this value by p . Thus for all r , 0 | f ( r ) − p | < ǫ. 0 Setting r = x , x and applying triangle inequality gives the desired contradiction. 1 2 Remark. This is more or less a concrete instance of Runge’s approximation theorem from complex analysis. However, do not be intimidated by the fancy names here: in fact, the key ingredient behind 3 this more general theorem (namely, or similar ) uses the same core idea of the “continuous roots of unity filter”.