Skip to main content
MathOverflow

Return to Answer

added 2 characters in body
Source Link
user75485
user75485

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^{\lambda_n x})= e^{n x}$$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.

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 $A(e^{\lambda_n x})= e^{n x}$ 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.

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.

Source Link
user75485
user75485

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 $A(e^{\lambda_n x})= e^{n x}$ 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.