# Transform Riesz basis $\{e^{i \lambda_n t}\}_{n\in\mathbb Z}$ to an orthogonal basis

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.

• 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? – LSpice Sep 13 '15 at 20:00
• Making use of Maple package OrthogonalExpansions, I tried a math experiment and obtained huge expressions (too long to be stated here). – user64494 Sep 13 '15 at 20:10
• @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. – Mark Sep 14 '15 at 19:28
• @user64494, ok. Maybe my question is less trivial than I thought at first. – Mark Sep 14 '15 at 19:29
• 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. – user75485 Sep 15 '15 at 9:47

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.