PhyslibAlpha.AlgebraicFramework.HilbertSpace.Unbounded.Existence.IteratedKernelGrowth

For every natural number $n$, there exists a non-negative constant $C$ such that for all real numbers $y$, the $n$-th Hermite polynomial $H_n(y)$ satisfies the bound $$|H_n(y)| \leq C(1 + |y|)^n.$$

For any natural number $n$ and any positive real number $\varepsilon$, there exists a non-negative constant $D$ such that for all real numbers $x$ satisfying $x^2 \leq 1$ and for all real numbers $u$, the $(n+1)$-th derivative of the Gaussian kernel $g_\varepsilon(t) = (\pi \varepsilon)^{-1/2} \exp(-t^2/\varepsilon)$ satisfies the pointwise bound: $$\left| g_\varepsilon^{(n+1)}(u - x) \right| \leq D (1 + |u|)^{n+1} \exp \left( - \frac{u^2}{2\varepsilon} \right).$$

For any natural number $n \in \mathbb{N}$ and any positive real number $c > 0$, the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(u) = |u|^n e^{-cu^2}$ is integrable.

For any natural number $n$ and any positive real number $c > 0$, the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(u) = (1 + |u|)^n e^{-cu^2}$ is integrable.

For any positive real number $\varepsilon > 0$, the Gaussian kernel function $t \mapsto (\pi \varepsilon)^{-1/2} \exp(-t^2/\varepsilon)$ is smooth (infinitely differentiable, or $C^\infty$) on $\mathbb{R}$.

For any natural number $n$ and any positive real number $\varepsilon > 0$, the $n$-th iterated derivative of the Gaussian kernel $t \mapsto \frac{1}{\sqrt{\pi \varepsilon}} \exp\left(-\frac{t^2}{\varepsilon}\right)$ is a continuous function.

Let $K_\varepsilon(t) = (\pi \varepsilon)^{-1/2} e^{-t^2/\varepsilon}$ be the Gaussian kernel. For any natural number $n$, any $\varepsilon > 0$, and any $t \in \mathbb{R}$, the $n$-th iterated derivative of the Gaussian kernel, $K_\varepsilon^{(n)}$, is differentiable at $t$ with derivative $K_\varepsilon^{(n+1)}(t)$.

Let $H$ be a complex Hilbert space and $U: \mathbb{R} \to \mathcal{L}(H)$ be a map such that for every $t \in \mathbb{R}$, $U(t)$ is a unitary operator, and for every $\xi \in H$, the map $t \mapsto U(t)\xi$ is continuous. Let $g_\varepsilon(t) = \frac{1}{\sqrt{\pi \varepsilon}} \exp\left(-\frac{t^2}{\varepsilon}\right)$ be the Gaussian kernel for $\varepsilon > 0$. Then for any $n \in \mathbb{N}$ and any vector $\psi \in H$, the function $$t \mapsto g_\varepsilon^{(n)}(t) U(t)\psi$$ is integrable on $\mathbb{R}$, where $g_\varepsilon^{(n)}$ denotes the $n$-th iterated derivative of $g_\varepsilon$ with respect to $t$.

Let $H$ be a complex Hilbert space and $U: \mathbb{R} \to \mathcal{L}(H)$ be a strongly continuous one-parameter group of unitary operators, satisfying $U(s + t) = U(s)U(t)$ and $U(t) \in \mathcal{U}(H)$ for all $s, t \in \mathbb{R}$. For a given $\varepsilon > 0$, let $g_\varepsilon(t) = (\pi \varepsilon)^{-1/2} e^{-t^2/\varepsilon}$ denote the Gaussian kernel, and let $g_\varepsilon^{(n)}$ denote its $n$-th derivative. For any vector $\psi \in H$, let $\phi_n$ be the Gårding vector defined by $\phi_n = \int_{-\infty}^{\infty} g_\varepsilon^{(n)}(t) U(t)\psi \, dt$. Then the map $s \mapsto U(s)\phi_n$ is differentiable at $s=0$ with derivative $$\left. \frac{d}{ds} (U(s)\phi_n) \right|_{s=0} = - \int_{-\infty}^{\infty} g_\varepsilon^{(n+1)}(u) U(u)\psi \, du.$$