1

Let $\gamma_1,\dotsc,\gamma_n$ be nonnegative real numbers. Let $\eta_1,\dotsc,\eta_n$ be real numbers. Consider the function $f:\mathbb{R}\rightarrow\mathbb{R}$ defined by:

$$f(x)=\sum_{i=1}^n \gamma_i e^{2\pi i \cdot \eta_i x}\!.$$

Is there a function $G:\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0} $ such that

  1. $G(t)\rightarrow 0$ as $t\rightarrow 0^+$
  2. $\sum_{i=1}^n \gamma_i \leq G(\|f\|_\infty)$

I want $G$ not to depend on $n$.

In other words, I want to show that $\sum \gamma_i$ tends $0$ as $\|f\|_\infty$ tends of zero at a rate which depends only on the rate at which $\|f\|_\infty$ tends to zero, and not on $n$.

Michael Hardy
  • 1
Terry
  • 271
  • I doubt it, especially if you aren't choosing the $\eta_i$ according to any orthogonality condition. – Ian Apr 29 '16 at 16:40
  • @Ian: What do you mean by orthogonality condition? If I understand, I can think if it applies in my case. – Terry Apr 29 '16 at 16:49
  • @Ian: Actually, I can just say what "my case" is: $f$ comes from an arbitrary finite dimensional (unitary) complex representation of $\mathbb{R}$ by means of a diagonal matrix coefficient with respect to a unit vector. – Terry Apr 29 '16 at 16:52
  • Unless $e^{2 \pi i \cdot \eta_i x}$ are orthogonal, I see no good reason to be able to control $\sum_i \gamma_i$ by any norm of $f$. Even then I'm not so sure about controlling it by $L^\infty$. – Ian Apr 29 '16 at 17:12

1 Answers1

1

$$ f(0)=\sum_{i=1}^n\gamma_i\implies\sum_{i=1}^n\gamma_i\le\|f\|_\infty $$ Thus $G(t)=t$ works. Am I missing something?

Julián Aguirre
  • 76,354
  • Thank you. You're not missing anything :) I posted a variant here: http://math.stackexchange.com/questions/1765499/variant-bounding-fourier-coefficients-in-terms-of-supremum-norm – Terry Apr 30 '16 at 14:50