Is there a dilation/scaling "equivalent" of Wiener's Tauberian theorems?

Question

Given an even function $g\in L^2(\mathbb{R})$, if it is orthogonal to $\exp(-\alpha^2 x^2)$ for all $\alpha$ (really, this is only needed on an open neighborhood of $1$), then it is identically zero. This is easy to see since differentiating with respect to $\alpha$ brings down powers of $x^2$. Evaluating at $\alpha = 1$ then yields that $g$ is orthogonal to all $x^{2m}\exp(-x^2)$, however the even Hermite-Gauss functions are a basis for the even functions in $L^2(\mathbb{R})$ which the $x^{2m}\exp(-x^2)$ are a linear combination thereof.

Based on some of my own published work, this result extends to functions of the form $\exp(-x^{2n})$.

This has some marked similarity to Wiener's Tauberian theorems but with translations supplanted by dilations. Likely there must be a condition on the function that is getting dilated (perhaps it must be even and nonzero everywhere?). Here would be a possible theorem statement:

Given sufficiently nice $f\in L^2(\mathbb{R}^+)$, if $g\in L^2(\mathbb{R}^+)$ is orthogonal to $f_{\alpha}$ for all $\alpha > 0$, then $g$ is identically 0, where $f_{\alpha}(x) = f(\alpha x)$.

(I replaced even functions in $L^2(\mathbb{R})$ for functions on $L^2(\mathbb{R}^+)$ for ease of discussion and generality.)

Is there any literature on dilation-based Tauberian theorems in this vein? The typical proof does not quite work in this setting since the Fourier inversion theorem pops out quite naturally with translations whereas no Fourier kernel appears in this case and the dilation turns into the inverse dilation further complicating matters.

Answer 1

This is maybe not the most satisfactory solution, but there is a way to reduce your setting to that of the original Wiener theorem.

To see how this is done, given $f \in L^2(\Bbb{R}_+)$, define $$ \widetilde{f} : \Bbb{R} \to \Bbb{C}, x \mapsto e^{x/2} \cdot f(e^x) . $$ By a standard change of variables, we see that $$ \int_{\Bbb{R}} |\widetilde{f}(x)|^2 \, d x = \int_{\Bbb{R}} e^x \cdot |f(e^x)|^2 \, d x = \int_0^\infty |f(y)|^2 \, d y , $$ so that the map $L^2(\Bbb{R}_+) \to L^2(\Bbb{R}), f \mapsto \widetilde{f}$ is well-defined and isometric (and in fact also surjective, i.e., unitary); in particular, this map preserves inner products.

Next, note that $$ \widetilde{f_\alpha} (x) = e^{x/2} \cdot f_\alpha (e^x) = e^{x/2} \cdot f(\alpha \, e^x) = \alpha^{-1/2} e^{ (x + \ln \alpha) / 2} \cdot f(e^{x + \ln \alpha}) = \alpha^{-1/2} (T_{\ln \alpha} \widetilde{f})(x), $$ where I use the usual notation $T_x f (y) = f(y - x)$ for the translation operator.

Finally, note that the condition $0 = \int_0^\infty f_\alpha(x) \, g(x) \, d x$ for all $\alpha > 0$ is equivalent to the condition $$ 0 = \int_{\Bbb{R}} \widetilde{f_\alpha} \, \widetilde{g} \, d x = \alpha^{-1/2} \int_{\Bbb{R}} (T_{\ln \alpha} \widetilde{f}) \cdot \widetilde{g} \, d x $$ for all $\alpha > 0$, and hence to the condition $$ \widetilde{g} \perp T_x \widetilde{f} \qquad \forall \, x \in \Bbb{R} . $$

From this, it is not hard to see that your property for $f$ holds if and only if $\{ T_x \widetilde{f} : x \in \Bbb{R} \}$ is a dense subspace of $L^2(\Bbb{R})$. By Wiener's theorem, this holds if and only if $\mathcal{F} \widetilde{f} \neq 0$ almost everywhere, with $\mathcal{F}$ denoting the Fourier transform. I don't think, however, that this last condition is straightforward to verify/recast in terms of $f$.