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It is well-known that $\{e^{i n t}\}_{n\in\mathbb Z}$ is an orthonormal basis for $L^2(-\pi,\pi)$. A theorem by Kadec (Kadec $1/4$ theorem) studies the perturbed exponential system:

If $\{\lambda_n\}$ is a sequence of real numbers for which $$|\lambda_n-n|\leqq L<\frac{1}{4}, \ \ n=0, \pm 1, \pm 2, \dots$$ then $\{e^{i \lambda_n t}\}_{n\in\mathbb Z}$ satisfies the Paley-Wiener criterion and so forms a Riesz basis for $L^2(-\pi,\pi)$.

Any Hilbert space $\overline{\operatorname{span}}\left(\phi_k\right)$ can be endowed with different equivalent norms, i.e. if $\|\cdot\|$ is a norm of $\overline{\operatorname{span}}\left(\phi_k\right)$, then $\|\cdot\|'$ obeying $$c_1\|\Psi\|\leq \|\Psi\|'\leq c_2\|\Psi\|, \ \ \forall \Psi\in \overline{\operatorname{span}}\left(\phi_k\right); \ \ c_1,c_2>0$$ is also a norm of $\overline{\operatorname{span}}\left(\phi_k\right)$. If a basis set is a Riesz system with respect to $\|\cdot\|$, one can always choose a second equivalent norm $\|\cdot\|'$ such that the basis set becomes orthonormal with respect to the appertaining scalar product $(\cdot,\cdot)'$. If A is the Schmidt matrix, which orthonormalizes a Riesz system, then $$\|\Psi\|'=\|A\Psi\|, \ \ \forall \Psi\in \overline{\operatorname{span}}\left(\phi_k\right).$$ For an application that I'm studying, I'd like to calculate $A$, which orthonormalizes Riesz system $\{e^{i \lambda_n t}\}_{n\in\mathbb Z}$ (under Kadec assumption). But I do not know where to start. Any suggestions please? Bibliography references and answers are welcome.

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  • $\begingroup$ Is the existence of $(\cdot, \cdot)'$ automatic—that is, does a norm equivalent to one coming from an inner product always come from an inner product—or is the existence of such a product in this case part of the statement? $\endgroup$
    – LSpice
    Sep 13, 2015 at 20:00
  • $\begingroup$ Making use of Maple package OrthogonalExpansions, I tried a math experiment and obtained huge expressions (too long to be stated here). $\endgroup$
    – user64494
    Sep 13, 2015 at 20:10
  • $\begingroup$ @LSpice, I think that the existence of $(\cdot, \cdot)'$ is automatic, but I could not prove it, because I'm not an expert in this topic. $\endgroup$
    – Mark
    Sep 14, 2015 at 19:28
  • $\begingroup$ @user64494, ok. Maybe my question is less trivial than I thought at first. $\endgroup$
    – Mark
    Sep 14, 2015 at 19:29
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    $\begingroup$ If $T$ is a topological isomorphism that takes a Riesz basis $\{x_n\}$ to a orthonormal system $\{e_n\}$, ( $ T(x_n)=e_n$ ), then $(x,y)'=(Tx,Ty)$ is an equivalent inner product and $\{x_n\}$ is orthonormal with respect to this product. $\endgroup$
    – user75485
    Sep 15, 2015 at 9:47

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In the conditions of Kadec's result , you can write $A=(I-S)^{-1}=\sum_{m=0}^\infty S^m,$ where $S(f)(x)=\sum_{n=-\infty}^\infty \hat f(n) (e^{inx}-e^{i\lambda_n x})$ verifies $\Vert S\Vert<1.$ Here $\{ \hat f (n)\} $ are the Fourier coefficients of $f$.

Then $A(e^{i\lambda_n x})= e^{in x}$ and so $A$ orthogonalizes the Riesz basis. The fact that $\Vert S\Vert<1$ can be found for example in Benzinger, Nonharmonic Fourier series ans Spectral Theory, 1987, TAMS.

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