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?


References and answers are welcome.



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


