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Let $\{ \lambda_k \}_{k=-\infty}^{\infty}$ satisfy $\inf_{k} \{ \lambda_{k+1} - \lambda_{k} \} > 0.$ For any $ f \in L^{2}([-\pi, \pi])$ show that $\{ f_k \}= \{ e^{-i \lambda_k} \}_{k=-\infty}^{\infty}$ satisfies $$\sum_{k \in \mathbb{Z}}|\langle f , f_k \rangle|^2 \leq B \| f \|^2 ,$$ for some $B > 0.$

When an Orthonormal basis is known, say, $e_k$ we can write $$f=\sum_{k \in \mathbb{Z}} \langle f , e_k \rangle e_k \Rightarrow \| f \|^2 =\sum_{k \in \mathbb{Z}}|\langle f , e_k \rangle|^2.$$ I don't know how to use this and to show the given condition. Any help is much appreiciated.

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If $\mathscr E$ is an orthonormal set of a Hilbert space $\mathscr H$, then $$ \sum_{e\in \mathscr E}\lvert \langle h,e\rangle\rvert^2\leq \lVert h\rVert^2.$$

The condition $\lambda_k-\lambda_{k-1}>0$ implies that $\{\frac{1}{\sqrt{2\pi}}f_k\}$ is an orthonormal set of $L^2[-\pi,\pi]$. Hence the required inequality holds with $B=2\pi$.

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  • $\begingroup$ thank you. Could you explain how does it become orthonormal ? $\endgroup$ – user394036 Dec 6 '16 at 4:18
  • $\begingroup$ you can check directly. $\endgroup$ – C.Ding Dec 6 '16 at 4:23
  • $\begingroup$ I don't understand how does the given condition on $\lambda$s is applicable. $\endgroup$ – user394036 Dec 6 '16 at 5:43
  • $\begingroup$ It's ok. I see. I think I got it now. You can delete the comments if they can't be edited. Thanks again. $\endgroup$ – user394036 Dec 6 '16 at 5:56
  • $\begingroup$ The given condition implies $\lambda_m\neq \lambda_n$ for all $m\neq n$.So $\int_{-\pi}^{\pi}f_m\bar{f}_n=\int_{-\pi}^{\pi} e^{(\lambda_n-\lambda_m)i}=0(m\neq n)$. $\endgroup$ – C.Ding Dec 6 '16 at 10:13

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