HMMT 二月 2015 · 团队赛 · 第 7 题

英文解析

7. [ 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-degreeroots of unity filter, or alternatively integrate over suitably-centered circles (“continuous roots of unityfilter”).
   We claim we can choose ǫ ( f ) = max( | f ( x ) − f ( x ) | ) / 2. Fix P and suppose for the sake of contradiction
   1 2
   that for all z with | z | ≤ 1 it is the case that | f ( z ) − P ( z ) | < ǫ . We can write z = re so thatiθiθ
   | 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 usingtriangle inequality,
   p − 1
   ∑
   2 πik/p 1
   | f ( r ) − P ( re ) | < ǫ.
   k =0 p
   The sum inside the absolute value is a roots of unity filter; since p > deg( P ) it is simply the constantterm of P - denote this value by p . Thus for all r ,
   | f ( r ) − p | < ǫ.0
   Setting r = x , x and applying triangle inequality gives the desired contradiction.0
   1 2
   Remark. This is more or less a concrete instance of Runge’s approximation theorem from complexanalysis. However, do not be intimidated by the fancy names here: in fact, the key ingredient behindthis more general theorem (namely, or similar ) uses the same core idea of the “continuous roots of 3
   unity filter”.