Injectivity of Weighted Translation Operator on $L^2(\mathbb{R})$

Question

I'm studying the properties of an operator $L$ on $L^2(\mathbb{R})$ defined by:

$$(Lf)(x) = \sum_{n\in\mathbb{N} \neq 0} \phi(n) \left( f(x-n) + f(x+n) \right)$$

where $\phi(n) = \exp(-n^2)$. I understand that for $\phi = 1$, injectivity can fail (e.g., $f(x) = g(x-a)$ for integer $a$ will lead to $Lf = Lg$). I'm curious about the impact of the Gaussian $\phi$ on injectivity and whether a general method exists for analyzing such operators with varying $\phi$ or translation spacings.

Questions:

With $\phi(n) = \exp(-n^2)$, is $L$ injective? That is, does $Lf = Lg$ imply $f = g$ for all $f, g \in L^2(\mathbb{R})$?

Is there a general approach to determine injectivity and uniqueness of the image for different $\phi$ or translation spacings?

I'm looking for insights or methods to analyze the operator's behavior with various weighting functions or translation patterns. I'm not really sure where to start. I think there may be a proof using FT, but i am uncertain.

Thanks

Answer 1

To check if $L$ is injective, you need to study its kernel, a.k.a. its null space.

Suppose that $f\in L^2(\mathbb R)$ such that $Lf=0$. Then taking the Fourier transform, in the $L^2(\mathbb R)$ sense (and thus, almost everywhere): $$\sum_{n\neq 0} \phi(n) \left( e^{2i\pi n\xi}\widehat f(\xi)+e^{-2i\pi n\xi}\widehat f(\xi)\right)=0\tag{1}$$ Let $h$ be the $1$-periodic function defined as $$h(\xi) = \sum_{n\neq 0}\left(\phi(n)+\phi(-n)\right)e^{2i\pi n\xi}$$ Then $(1)$ can be rewritten as $$\widehat f(\xi)h(\xi)=0\tag{2}$$

• Thus if $h$ vanishes on an open set $I\subset(0, 1)$, then you can build $f$ such that $\widehat f$ is non zero on that open set, and zero elsewhere (for instance the indicator function of $I$). Then $(2)$ will be satisfied, and $Lf=0$. Therefore $L$ is not injective.
• If $h\neq 0$ almost everywhere in $(0, 1)$, then no function $f\neq 0$ can satisfy $(2)$, and thus $L$ is injective.

So a necessary and sufficient condition for $L$ to be injective is that $h\neq 0$ almost everywhere.

Let's look at a few examples:

• If $\phi(n)=1$ for all $n\neq 0$, then using the Poisson summation formula, in the sense of distributions $$h(\xi)=-2 + 2\sum_{n\in\mathbb Z} \delta(\xi-n)$$ Note that this choice of $\phi$, which doesn't decay, makes $L$ not preserve $L^2(\mathbb R)$. And also, $h\neq 0$ almost everywhere, thus $L$ is injective. This is in contradiction with your claim that it wouldn't be. For your counterexample to work (with $g$ being a shifted version of $f$), we would have to include $\phi(0)$ in the sum in the definition of $Lf$.

• If $\phi(n)=e^{-n^2}$, then $h$ is known as Jacobi's function. It doesn't vanish on $(0,1)$, thus in that case $L$ is injective.