PhyslibAlpha.ProbabilisticTheory.HilbertSpace.Unbounded.Existence.GardingVectors

Given real numbers $\epsilon$ and $t$, the Gaussian kernel (or heat kernel) is defined as $$g_\epsilon(t) = \frac{1}{\sqrt{\pi \epsilon}} \exp\left(-\frac{t^2}{\epsilon}\right)$$ This function represents a Gaussian distribution with variance $\sigma^2 = \frac{\epsilon}{2}$ and a total integral of $1$ over $\mathbb{R}$.

For any real number $\epsilon > 0$ and any real number $t$, the Gaussian kernel defined by $$g_\epsilon(t) = \frac{1}{\sqrt{\pi \epsilon}} \exp\left(-\frac{t^2}{\epsilon}\right)$$ is strictly positive, i.e., $g_\epsilon(t) > 0$.

For any positive real number $\epsilon > 0$, the integral of the Gaussian kernel $g_\epsilon(t) = \frac{1}{\sqrt{\pi \epsilon}} \exp\left(-\frac{t^2}{\epsilon}\right)$ over the real line is equal to $1$: $$\int_{-\infty}^{\infty} g_\epsilon(t) \, dt = 1$$

Let $H$ be a complex Hilbert space. Let $U: \mathbb{R} \to \mathcal{L}(H)$ be a map such that for all $t \in \mathbb{R}$, $U_t$ is a unitary operator, and for each vector $\psi \in H$, the orbit $t \mapsto U_t \psi$ is continuous. For any $\epsilon > 0$ and any $\psi \in H$, the function $t \mapsto g_\epsilon(t) U_t \psi$ is integrable on $\mathbb{R}$, where $g_\epsilon(t) = \frac{1}{\sqrt{\pi \epsilon}} \exp\left(-\frac{t^2}{\epsilon}\right)$ is the Gaussian kernel. This integrability arises because the Gaussian kernel is integrable and the function $t \mapsto U_t \psi$ is bounded (with $\|U_t \psi\| = \|\psi\|$) and continuous.

Given a Hilbert space $H$ and a one-parameter group of unitary operators $U(t)$, the analytic Gårding vector of a vector $\psi \in H$ with respect to a parameter $\epsilon \in \mathbb{R}$ is the vector obtained by smoothing $\psi$ along the orbit of the unitary group using the Gaussian heat kernel $g_\epsilon(t)$. It is defined by the Bochner integral: $$\int_{-\infty}^{\infty} g_\epsilon(t) U(t) \psi \, dt$$ where $g_\epsilon(t) = \frac{1}{\sqrt{\pi \epsilon}} \exp\left(-\frac{t^2}{\epsilon}\right)$ is the Gaussian kernel of width $\epsilon$.

The standard Gaussian kernel is the function from $\mathbb{R}$ to $\mathbb{R}$ defined by $x \mapsto \frac{1}{\sqrt{\pi}} e^{-x^2}$. This corresponds to the heat kernel $g_\epsilon(x) = \frac{1}{\sqrt{\pi \epsilon}} e^{-x^2 / \epsilon}$ evaluated at the parameter $\epsilon = 1$.

For any real number $x$, the standard Gaussian kernel, defined as $g_1(x) = \frac{1}{\sqrt{\pi}} e^{-x^2}$, is strictly positive, i.e., $g_1(x) > 0$.

The standard Gaussian kernel is equal to the Gaussian kernel $g_\varepsilon(t) = (\pi \varepsilon)^{-1/2} \exp(-t^2/\varepsilon)$ evaluated at the parameter $\varepsilon = 1$.

The integral of the standard Gaussian kernel $g_1(x) = \frac{1}{\sqrt{\pi}} e^{-x^2}$ over the real line $\mathbb{R}$ is equal to $1$.

The standard Gaussian kernel $g_1: \mathbb{R} \to \mathbb{R}$, defined as $g_1(x) = \frac{1}{\sqrt{\pi}} e^{-x^2}$, is integrable on $\mathbb{R}$.

For any positive real number $\varepsilon > 0$ and any real number $x$, the Gaussian kernel $g_\varepsilon(t) = \frac{1}{\sqrt{\pi \varepsilon}} e^{-t^2/\varepsilon}$ satisfies the scaling identity: $$\sqrt{\varepsilon} g_\varepsilon(\sqrt{\varepsilon} x) = g_1(x),$$ where $g_1(x) = \frac{1}{\sqrt{\pi}} e^{-x^2}$ is the standard Gaussian kernel. This identity shows that the $\varepsilon$-dependence in the kernel is cancelled by the Jacobian factor $\sqrt{\varepsilon}$ arising from the coordinate transformation $t = \sqrt{\varepsilon} x$.

Let $H$ be a Hilbert space and $U$ be a one-parameter group of unitary operators. For any positive real number $\varepsilon > 0$ and any vector $\psi \in H$, the analytic Gårding vector can be expressed by performing the substitution $t = \sqrt{\varepsilon}x$ in its defining integral, resulting in: $$\text{analyticGardingVector}(U, \varepsilon, \psi) = \int_{-\infty}^{\infty} g_1(x) U(\sqrt{\varepsilon}x) \psi \, dx$$ where $g_1(x) = \frac{1}{\sqrt{\pi}} e^{-x^2}$ is the standard Gaussian kernel.

Let $H$ be a Hilbert space and $U: \mathbb{R} \to \text{ops}(H)$ be a strongly continuous one-parameter group of operators such that $U(0) = I$. For any vector $\psi \in H$, the analytic Gårding vector is defined as $$\psi_\epsilon = \int_{-\infty}^{\infty} g_\epsilon(t) U(t) \psi \, dt$$ where $g_\epsilon(t) = \frac{1}{\sqrt{\pi \epsilon}} \exp\left(-\frac{t^2}{\epsilon}\right)$ is the Gaussian kernel. As $\epsilon$ approaches $0$ from the right ($\epsilon \to 0^+$), the analytic Gårding vectors converge to $\psi$ in the Hilbert space $H$: $$\lim_{\epsilon \to 0^+} \psi_\epsilon = \psi.$$

Let $H$ be a Hilbert space and $U(t)$ be a strongly continuous one-parameter group of unitary operators on $H$. For any $\epsilon > 0$, vector $\psi \in H$, and $s \in \mathbb{R}$, applying the operator $U(s)$ to the analytic Gårding vector results in a translation of the Gaussian kernel $g_\epsilon$: $$U(s) \left( \int_{-\infty}^{\infty} g_\epsilon(t) U(t) \psi \, dt \right) = \int_{-\infty}^{\infty} g_\epsilon(u - s) U(u) \psi \, du$$ where $g_\epsilon(t) = \frac{1}{\sqrt{\pi \epsilon}} \exp\left(-\frac{t^2}{\epsilon}\right)$.

Let $\varepsilon$ be a positive real number. The Gaussian kernel, defined as $g_\varepsilon(t) = (\pi \varepsilon)^{-1/2} \exp(-t^2/\varepsilon)$, is differentiable at any point $t \in \mathbb{R}$ with derivative $$\frac{d}{dt} g_\varepsilon(t) = -\frac{2t}{\varepsilon} g_\varepsilon(t).$$

For any $\varepsilon > 0$ and $t \in \mathbb{R}$, the derivative of the Gaussian kernel $g_\varepsilon(t) = (\pi \varepsilon)^{-1/2} \exp(-t^2/\varepsilon)$ with respect to $t$ is given by $$\frac{d}{dt} g_\varepsilon(t) = g_\varepsilon(t) \cdot \left(-\frac{2t}{\varepsilon}\right).$$

For any positive real number $\varepsilon > 0$, the derivative of the Gaussian kernel $g_\varepsilon(t) = \frac{1}{\sqrt{\pi \varepsilon}} e^{-t^2/\varepsilon}$ with respect to $t$ is a continuous function.

Let $\varepsilon$ be a positive real number. For any $x \in \mathbb{R}$ satisfying $x^2 \leq 1$ and for any $u \in \mathbb{R}$, the absolute value of the derivative of the Gaussian kernel $g_\varepsilon(t) = (\pi \varepsilon)^{-1/2} \exp(-t^2/\varepsilon)$ evaluated at $u - x$ satisfies the bound: $$|g_\varepsilon'(u - x)| \leq \left( \frac{2 e^{1/\varepsilon}}{\varepsilon \sqrt{\pi \varepsilon}} \right) (|u| + 1) \exp\left( -\frac{u^2}{2\varepsilon} \right).$$

For any positive real number $\epsilon > 0$ and any real constant $c$, the function $f: \mathbb{R} \to \mathbb{R}$ defined by $$f(u) = c(|u| + 1) \exp\left(-\frac{u^2}{2\epsilon}\right)$$ is integrable over $\mathbb{R}$.

Let $H$ be a complex Hilbert space and let $t \mapsto U(t)$ be a strongly continuous one-parameter group of unitary operators on $H$. For any $\varepsilon > 0$ and $\psi \in H$, let $\psi_\varepsilon$ be the analytic Gårding vector defined by $$\psi_\varepsilon = \int_{-\infty}^{\infty} g_\varepsilon(t) U(t)\psi \, dt$$ where $g_\varepsilon(t) = (\pi \varepsilon)^{-1/2} \exp(-t^2/\varepsilon)$ is the Gaussian kernel. Then the map $s \mapsto U(s)\psi_\varepsilon$ is differentiable at $s = 0$, and its derivative is given by $$\left. \frac{d}{ds} (U(s)\psi_\varepsilon) \right|_{s=0} = \int_{-\infty}^{\infty} -g_\varepsilon'(u) U(u)\psi \, du$$ where $g_\varepsilon'(u)$ denotes the derivative of the Gaussian kernel with respect to $u$.

Let $H$ be a complex Hilbert space and let $t \mapsto U(t)$ be a strongly continuous one-parameter group of unitary operators on $H$. For any $\epsilon > 0$ and any vector $\psi \in H$, the analytic Gårding vector defined by $$\psi_\epsilon = \int_{-\infty}^{\infty} g_\epsilon(t) U(t) \psi \, dt$$ where $g_\epsilon(t) = \frac{1}{\sqrt{\pi \epsilon}} \exp\left(-\frac{t^2}{\epsilon}\right)$ is the Gaussian kernel, belongs to the domain of the infinitesimal generator (the stone candidate generator) of the group $U$.

Let $H$ be a complex Hilbert space and let $U(t)$ be a strongly continuous one-parameter group of unitary operators on $H$. For any $\epsilon > 0$ and any vector $\psi \in H$, let $\psi_\epsilon$ be the analytic Gårding vector defined by the Bochner integral: $$\psi_\epsilon = \int_{-\infty}^{\infty} g_\epsilon(t) U(t) \psi \, dt$$ where $g_\epsilon(t) = \frac{1}{\sqrt{\pi \epsilon}} \exp\left(-\frac{t^2}{\epsilon}\right)$ is the Gaussian kernel. Then the infinitesimal generator $A$ (the stone candidate generator) of the group $U$ acts on $\psi_\epsilon$ as: $$A \psi_\epsilon = i \int_{-\infty}^{\infty} g_\epsilon'(u) U(u) \psi \, du$$ where $g_\epsilon'(u)$ denotes the derivative of the Gaussian kernel with respect to $u$.