Physlib.Mathematics.VariationalCalculus.IsTestFunction

Given a function $f: X \to U$ that satisfies the property of being a test function (meaning $f$ is smooth and has compact support), this definition constructs the corresponding compactly supported continuous map. The resulting object uses $f$ as its underlying function, with its continuity and compact support derived from the properties of a test function.

Let $f: X \to U$ be a continuous map with compact support. If $f$ is infinitely differentiable ($C^\infty$) over the real numbers $\mathbb{R}$, then $f$ is a test function.

Let $X$ be a measurable space where every open set is measurable, and let $\mu$ be a measure on $X$ that assigns finite measure to all compact sets. If $f: X \to U$ is a test function (meaning $f$ is smooth and has compact support), then $f$ is integrable with respect to $\mu$.

Let $f: X \to U$ be a function. If $f$ is a test function, then $f$ is differentiable over the real numbers $\mathbb{R}$.

Let $f: X \to U$ be a function. If $f$ is a test function, then $f$ is infinitely differentiable (of class $C^\infty$) over the real numbers.

The constant function $f: X \to U$ defined by $f(x) = 0$ for all $x \in X$ is a test function.

Let $f : X \to V$ be a test function. Let $g : V \to U$ be an infinitely differentiable function ($C^\infty$) such that $g(0) = 0$. Then the composite function $g \circ f$ is also a test function.

Let $\iota$ be a finite index set. Let $\varphi : X \to (\iota \to U)$ be a function mapping $x \in X$ to a collection of values indexed by $i \in \iota$. If for every index $i \in \iota$, the component function $x \mapsto \varphi(x, i)$ is a test function (i.e., a smooth function with compact support), then the function $x \mapsto \varphi(x, \cdot)$ is also a test function.

Let $\varphi : X \to \text{Space } d$ be a test function (a smooth function with compact support). Then for any index $i$, the component function $x \mapsto \varphi(x)_i$ is also a test function.

Let $f : X \to U$ and $g : X \to V$ be two functions. If $f$ and $g$ are test functions (smooth functions with compact support), then the function $x \mapsto (f(x), g(x))$ from $X$ to $U \times V$ is also a test function.

Let $f : X \to U \times V$ be a test function (a smooth function with compact support). Then the function mapping $x$ to the first component of $f(x)$, denoted by $x \mapsto (f(x))_1$, is also a test function.

Let $f : X \to U \times V$ be a test function (a smooth function with compact support). Then the function mapping $x$ to the second component of $f(x)$, denoted by $x \mapsto (f(x))_2$, is also a test function.

If $f : X \to U$ is a test function (an infinitely differentiable function with compact support), then its negation $-f$, defined by $x \mapsto -f(x)$, is also a test function.

Let $f, g : X \to U$ be two test functions (smooth functions with compact support). Then their pointwise sum, defined by the mapping $x \mapsto f(x) + g(x)$, is also a test function.

Let $f, g: X \to U$ be two test functions (smooth functions with compact support). Then their pointwise difference, defined by $x \mapsto f(x) - g(x)$, is also a test function.

Let $f: X \to \mathbb{R}$ and $g: X \to \mathbb{R}$ be two test functions (smooth functions with compact support). Then their pointwise product, defined by $x \mapsto f(x)g(x)$, is also a test function.

Let $X$ be a space. If $f: X \to \mathbb{R}$ is a smooth function ($C^\infty$) and $g: X \to \mathbb{R}$ is a test function (a smooth function with compact support), then their pointwise product $x \mapsto f(x)g(x)$ is also a test function.

Let $X$ be a space, $f: X \to \mathbb{R}$ be a test function (a smooth function with compact support), and $g: X \to \mathbb{R}$ be a smooth function ($C^\infty$). Then their pointwise product $x \mapsto f(x) \cdot g(x)$ is also a test function.

Let $X$ be a space and $V$ be a real inner product space. If $f: X \to V$ and $g: X \to V$ are test functions (smooth functions with compact support), then the scalar-valued function defined by their pointwise inner product, $x \mapsto \langle f(x), g(x) \rangle$, is also a test function.

Let $X$ be a space and $V$ be a generalized real inner product space. If $f: X \to V$ is a smooth function ($C^\infty$) and $g: X \to V$ is a test function (a smooth function with compact support), then the scalar-valued function mapping $x$ to the inner product $\langle f(x), g(x) \rangle$ is also a test function.

Let $V$ be a generalized inner product space over $\mathbb{R}$. If $f: X \to V$ is a test function (a smooth function with compact support) and $g: X \to V$ is a smooth ($C^\infty$) function, then the scalar-valued function defined by the pointwise inner product $x \mapsto \langle f(x), g(x) \rangle_{\mathbb{R}}$ is also a test function.

Let $f: X \to \mathbb{R}$ and $g: X \to U$ be two test functions (smooth functions with compact support). Then their pointwise scalar multiplication, defined by $(f \cdot g)(x) = f(x) \cdot g(x)$, is also a test function.

Let $f: X \to \mathbb{R}$ be a smooth ($C^\infty$) function and $g: X \to U$ be a test function (a smooth function with compact support). Then the pointwise scalar multiplication $(f \cdot g)(x) = f(x) \cdot g(x)$ is also a test function.

Let $f: X \to \mathbb{R}$ be a test function (a smooth function with compact support) and $g: X \to U$ be a smooth ($C^\infty$) function. Then the pointwise scalar multiplication $(f \cdot g)(x) = f(x) \cdot g(x)$ is also a test function.

Let $\iota$ be a finite index set. Let $\varphi : X \to (\iota \to U)$ be a function mapping $x \in X$ to a collection of values indexed by $i \in \iota$. If for every index $i \in \iota$, the component function $x \mapsto \varphi(x, i)$ is a test function (i.e., an infinitely differentiable function with compact support), then the sum of these functions $x \mapsto \sum_{i \in \iota} \varphi(x, i)$ is also a test function.

Let $\phi : X \to \text{Space } d$ be a test function (a smooth function with compact support). For any coordinate index $i \in \{0, \dots, d-1\}$, the component function $x \mapsto (\phi(x))_i$ is also a test function.

Let $f : X \to V$ be a test function (an infinitely differentiable function with compact support). Let $g : V \to U$ be an $\mathbb{R}$-linear map that is infinitely differentiable ($C^\infty$). Then the composite function $g \circ f$ is also a test function.

Let $X$ be a space and $V, U$ be real vector spaces. If $f: X \to V$ is a test function (a smooth function with compact support) and $g: X \to \mathcal{L}(V, U)$ is an infinitely differentiable ($C^\infty$) map from $X$ into the space of continuous linear maps from $V$ to $U$, then the function $x \mapsto g(x)(f(x))$ is also a test function.

Let $f: \mathbb{R} \to U$ be a function. If $f$ is a test function (meaning $f$ is smooth and has compact support), then its derivative $f' : \mathbb{R} \to U$ is also a test function.

Let $f: X \to U$ be a test function, which is an infinitely differentiable function with compact support. Then its Fréchet derivative $Df: X \to \mathcal{L}(X, U)$, defined as the map $x \mapsto Df(x)$ where $Df(x)$ is the continuous linear map representing the derivative of $f$ at $x$, is also a test function.

Let $f: X \to U$ be a test function (i.e., an infinitely differentiable function with compact support). For any vector $\delta x \in X$, the function mapping $x \in X$ to the Fréchet derivative of $f$ at $x$ applied to $\delta x$, denoted as $Df(x)(\delta x)$, is also a test function.

Let $X$ and $U$ be complete real generalized inner product spaces. Let $f: X \to U$ be a test function (an infinitely differentiable function with compact support). For any fixed vector $\delta y \in U$, the function $g: X \to X$ defined by $g(x) = (Df(x))^\dagger(\delta y)$, where $(Df(x))^\dagger$ is the adjoint of the Fréchet derivative of $f$ at $x$, is also a test function.

Let $X$ be a finite-dimensional real vector space. If $f : X \to X$ is a test function (that is, an infinitely differentiable function with compact support), then its divergence $\text{div } f$ is also a test function.

Let $d$ be a natural number. If $\phi: \mathbb{R}^d \to \mathbb{R}$ is a test function (that is, an infinitely differentiable function with compact support), then its gradient $\nabla \phi$ is also a test function.

For any natural number $d$, let $\phi: \text{Space } d \to \mathbb{R}^d$ be a vector-valued test function (i.e., an infinitely differentiable function with compact support). Then the divergence of $\phi$, denoted by $\nabla \cdot \phi$ (or `Space.div φ`), is also a test function.