# Derive an orthonormal system by Riesz basis $\{g(\cdot-\lambda_k),\ \lambda_k\in\mathbb R, \ k\in\mathbb Z\}$

Let $\{g(\cdot-k),k\in\mathbb Z\}$ be a Riesz basis, and let $\varphi\in L^2(\mathbb R)$ be a function defined by its Fourier transform $$\hat{\varphi}(\xi)=\frac{\hat{g}(\xi)}{\Gamma(\xi)},$$ where $$\Gamma(\xi)=\left(\sum_k |\hat{g}(\xi+2k\pi)|^2\right)^{1/2}$$ Then $\{\varphi(\cdot-k),\ k\in\mathbb Z\}$ is an orthonormal system.

The proof of this result uses Parseval's identity and the fact that the Fourier transform of $\varphi(x-k)$ is $e^{-ik\xi}\hat{\varphi}(\xi)$. Hence, after some steps, one arrives to: $$\frac{1}{2\pi}\int_0^{2\pi}e^{-i(k-l)\xi}d\xi=\delta_{kl}$$ A complete proof is located - for instance - in:

Härdle, Wolfgang, et al. Wavelets, approximation, and statistical applications. Vol. 129. Springer Science & Business Media, 2012.

Let us consider a "perturbed system" $\{g(\cdot-\lambda_k),\ \lambda_k\in\mathbb R, \ k\in\mathbb Z\}$ that is a Riesz basis for $|\lambda_k-k|\leq L<1$. Is it possible to derive an orthonormal system by this Riesz basis?

• Mark, What do you mean by "derive"? Do you want a function $f$ such that $f(\cdot-\lambda_k)$ form an orthonormal basis with the same span as $(g(\cdot-\lambda_k))$? – user71040 Oct 1 '15 at 11:30
• @user71040, Yes. Or a function f such that $f(\cdot−k)$ form an orthonormal basis with the same span as $(g(\cdot−\lambda_k))$. – Mark Oct 1 '15 at 19:58

This is an answer to the second version: a function $f$ such that $f(⋅−k)$ form an orthonormal basis with the same span as $(g(⋅−λ_k))$ exists only in very special cases. Assume it exists, then on the Fourier transform side span $(g-\lambda_k)$ is the space of functions of the form $\phi\cdot \hat f$ where $\phi$ is $2\pi$-periodic, so in particular we have $e^{i\lambda_k t}\hat g(t)=\phi_k(t)\hat f(t)$. Let $S$ be the support of $\hat g$. Clearly $\hat f$ is non zero on $S$, so on $S$ we have $e^{i\lambda_k t}=\phi_k(t)[\hat f(t)/\hat g(t)]$. Writing this for integers $k\neq s$ and dividing we see that $e^{i(\lambda_k-\lambda_s)t}$ is $2\pi$ periodic on $S$. If the measure of $S$ is $>2\pi$ there are two sets of positive measure $A$ and $B$ such that $A=2\pi \mu +B$ for some $\mu\in Z$. Thus for $t\in B$ we have $e^{i(\lambda_k-\lambda_s)(t+2\pi \mu)}=e^{i(\lambda_k-\lambda_s)t}$ which gives $e^{i(\lambda_k-\lambda_s)2\pi \mu}=1$ so $\lambda_k-\lambda_s$ is an integer for all $k,s\in Z$ which implies $\lambda_k=k+\delta$. If measure of $S$ is $\leq2\pi$ then such $f$ can exist only if $\sum_{s\in Z}I(t+2\pi s t)=1$ for $I$ an indicator function of $S$.
For the first version here is a counterexample. Let $g$ be the indicator function of the interval $[-1/2,1/2]$. Then $(g(x-k))$ is an orthonormal basis so a Riesz basis. Let $\lambda_k=k+\delta_k$ with $0<\delta_k<1/10$ and and linearly independent over rational numbers for $k$ odd and $\delta_{k}=0$ for $k$ even. It is a Riesz basis. We have $<g_j,g_s>$ equals $1$ if $j=s$ and equals $0$ if $|j-s|>1$. Otherwise it is small, so the Gram matrix is diagonally dominated, so invertible. Let $G$ denote span $(g(x-\lambda_j))$. If $f(x-\lambda_k))$ is orthogonal in $G$ all those translations are in the span so in particular $f\in G$. If $h\in G$ then $h=\sum_{j\in Z} x_j g_j$ with $(x_j)$ square summable. We calculate that $h$ is a piecewise constant function with jumps at points $k-\frac12+\delta_k$ or $k+\frac12+\delta_k$ for $k\in Z$. Thus if $x_0=s\pm\frac12+\delta_s$ with $s\neq 0$ is a jump of $f$ we see that $x_0+\lambda_\mu$ is in this set of jumps for each $\mu$. We check that this is impossible.
Thus there is no non-zero function $f\neq g$ in $G$ such that the whole sequence $(f(x-\lambda_j))$ is $G$.
• In the first answer, can you discard the case where for exemple $S=(0,2\pi)\cup(8\pi,10\pi)$ and $\lambda_k=k,$ for k even, $\lambda_k=k+1/2$ for k odd? – user75485 Oct 5 '15 at 16:36
• @Joseph. Thanks for the question. It points out that I was to optimistic. Clearly $e^{i(\lambda_k-\lambda_s)2\pi \mu}=1$ gives $\lambda_k-\lambda_s$ is an integer divided by $\mu$ which still gives a lot of space for counterexamples. – user71040 Oct 5 '15 at 19:29