Physlib.Mathematics.VariationalCalculus.HasVarAdjoint

Let $F: (X \to U) \to (Y \to V)$ and $F': (Y \to V) \to (X \to U)$ be transformations between function spaces. If $F'$ is the variational adjoint of $F$, and $F$ is a localized function transform, then $F$ is the variational adjoint of $F'$. A function transformation $F$ is said to be **localized** if for every compact set $K \subseteq Y$, there exists a compact set $L \subseteq X$ such that for any two functions $\phi, \phi' : X \to U$, if $\phi(x) = \phi'(x)$ for all $x \in L$, then $(F\phi)(y) = (F\phi')(y)$ for all $y \in K$.

The identity transformation $\text{id} : (X \to U) \to (X \to U)$, defined by $\text{id}(\phi) = \phi$ for any function $\phi$, is its own variational adjoint.

The zero transformation $F: (X \to U) \to (Y \to V)$, defined by $F(\phi) = 0$ for all functions $\phi: X \to U$, has a variational adjoint given by the zero transformation $F': (Y \to V) \to (X \to U)$, defined by $F'(\psi) = 0$ for all functions $\psi: Y \to V$.

Let $F: (Y \to V) \to (Z \to W)$ and $G: (X \to U) \to (Y \to V)$ be transformations between function spaces. Suppose $F$ has a variational adjoint $F'$, and $G$ has a variational adjoint $G'$. Then the composition of these transformations $F \circ G$ (where $(F \circ G)(\phi) = F(G(\phi))$) has a variational adjoint given by the composition of their adjoints in reverse order, $G' \circ F'$ (where $(G' \circ F')(\psi) = G'(F'(\psi))$).

Let $F, G: (X \to U) \to (Y \to V)$ and $F': (Y \to V) \to (X \to U)$ be operators between function spaces. If $F'$ is the variational adjoint of $G$, and for every test function $\phi$, the equality $F(\phi) = G(\phi)$ holds, then $F'$ is also the variational adjoint of $F$.

Let $F: (X \to U) \to (Y \to V)$ and $F', G': (Y \to V) \to (X \to U)$ be transformations between function spaces. Suppose that $F'$ is a variational adjoint of $F$, and that for every test function $\phi: Y \to V$, the condition $F'(\phi) = G'(\phi)$ holds. If $G'$ is a localized function transform, then $G'$ is also a variational adjoint of $F$. A function transformation $G': (Y \to V) \to (X \to U)$ is said to be **localized** if for every compact set $K \subseteq X$, there exists a compact set $L \subseteq Y$ such that for any two functions $\phi, \phi': Y \to V$, if $\phi$ and $\phi'$ agree on $L$ (i.e., $\phi(y) = \phi'(y)$ for all $y \in L$), then $G'\phi$ and $G'\phi'$ agree on $K$ (i.e., $(G'\phi)(x) = (G'\phi')(x)$ for all $x \in K$).

Let $X$ be a space equipped with a measure $\mu$ (the volume measure) such that open sets are measurable, $\mu$ is finite on compact sets, and $\mu$ is strictly positive on non-empty open sets. Let $F$ be an operator mapping functions from $X$ to $U$ to functions from $Y$ to $V$. If $F'$ and $G'$ are both variational adjoints of $F$, then for every test function $\phi: Y \to V$ (a smooth function with compact support), the equality $F'(\phi) = G'(\phi)$ holds.

Let $X$ and $Y$ be real inner product spaces equipped with measures, and let $Y$ be finite-dimensional. Assume that the volume measure on $X$ is finite on compact sets and strictly positive on non-empty open sets. Let $F$ be an operator mapping functions from $X$ to $U$ to functions from $Y$ to $V$. If $F'$ and $G'$ are both variational adjoints of $F$, then for any smooth ($C^\infty$) function $f: Y \to V$, it holds that $F'(f) = G'(f)$.

Let $F$ be a function transformation from functions on $X$ to functions on $V$, and let $F'$ be its variational adjoint. Then the transformation defined by $(-F)(\phi) = -F(\phi)$ has a variational adjoint defined by $(-F')(\psi) = -F'(\psi)$.

Let $F$ be a function transformation mapping functions from $X$ to $U$ into functions from $X$ to $V$, and let $F'$ be a function transformation mapping functions from $X$ to $V$ into functions from $X$ to $U$. If the transformation defined by $(-F)(\phi) = -F(\phi)$ has the variational adjoint $(-F')(\psi) = -F'(\psi)$, then $F$ has the variational adjoint $F'$.

Let $F$ and $G$ be function transformations mapping functions from $X$ to $U$ into functions from $X$ to $V$. If $F$ has a variational adjoint $F'$ and $G$ has a variational adjoint $G'$, then the pointwise sum transformation $(F + G)$, defined by $((F + G)\phi)(x) = F\phi(x) + G\phi(x)$, has the variational adjoint $(F' + G')$, defined by $((F' + G')\psi)(x) = F'\psi(x) + G'\psi(x)$.

Let $\iota$ be a finite index set. For each $i \in \iota$, let $F_i$ be an operator mapping functions from $X$ to $U$ to functions from $X$ to $V$, and let $F'_i$ be its corresponding variational adjoint mapping functions from $X$ to $V$ to functions from $X$ to $U$. If for every $i \in \iota$, $F_i$ has the variational adjoint $F'_i$, then the sum of these operators $\sum_{i \in \iota} F_i$ has a variational adjoint given by the sum of their individual adjoints $\sum_{i \in \iota} F'_i$. Specifically, the operator mapping $\phi$ to the pointwise sum $\sum_{i \in \iota} F_i(\phi)(x)$ has the variational adjoint mapping $\psi$ to $\sum_{i \in \iota} F'_i(\psi)(x)$.

Let $F$ and $G$ be transformations mapping functions from $X$ to $U$ into functions from $X$ to $V$. If $F$ has a variational adjoint $F'$ and $G$ has a variational adjoint $G'$, then the pointwise difference transformation $(F - G)$, defined by $((F - G)\phi)(x) = F\phi(x) - G\phi(x)$, has the variational adjoint $(F' - G')$, defined by $((F' - G')\psi)(x) = F'\psi(x) - G'\psi(x)$.

Suppose $F$ is an operator that maps functions $\phi: X \to U$ to functions $F(\phi): X \to \mathbb{R}$, and $F'$ is its variational adjoint. If $\psi: X \to \mathbb{R}$ is a smooth ($C^\infty$) function, then the operator mapping $\phi$ to the pointwise product $\psi F(\phi)$ has a variational adjoint which maps a function $\varphi$ to $F'(\psi \varphi)$, where $\psi \varphi$ is the pointwise product of $\psi$ and $\varphi$.

Let $X$ and $U$ be spaces. Let $F: (X \to U) \to (X \to \mathbb{R})$ be an operator that possesses a variational adjoint $F': (X \to \mathbb{R}) \to (X \to U)$. If $\psi: X \to \mathbb{R}$ is a smooth function (i.e., $\psi$ is $C^\infty$), then the operator defined by the pointwise product $\phi \mapsto F(\phi) \cdot \psi$ also has a variational adjoint. This adjoint is given by the mapping $\varphi \mapsto F'(\varphi \cdot \psi)$, where $(\varphi \cdot \psi)(x) = \varphi(x) \psi(x)$.

Let $F$ be an operator mapping functions $\phi: X \to U$ to functions $F(\phi): X \to V$, and let $F'$ be its variational adjoint. If $\psi: X \to \mathbb{R}$ is a smooth ($C^\infty$) function, then the operator that maps $\phi$ to the pointwise scalar product $\psi \cdot F(\phi)$ has a variational adjoint. This adjoint maps a function $\varphi: X \to V$ to $F'(\psi \cdot \varphi)$, where $(\psi \cdot \varphi)(x) = \psi(x) \cdot \varphi(x)$.

Let $F: (X \to U) \to (X \to V)$ be an operator that possesses a variational adjoint $F': (X \to V) \to (X \to U)$. If $\psi: X \to \mathbb{R}$ is a smooth ($C^\infty$) function, then the operator defined by the pointwise scalar multiplication $\phi \mapsto \psi \cdot F(\phi)$ also has a variational adjoint. This adjoint is given by the mapping $\varphi \mapsto F'(\psi \cdot \varphi)$, where $(\psi \cdot \varphi)(x) = \psi(x) \cdot \varphi(x)$.

Let $U$ and $V$ be complete generalized inner product spaces over $\mathbb{R}$. Let $f: X \to \mathcal{L}(U, V)$ be an infinitely differentiable ($C^\infty$) function that maps each point $x \in X$ to a continuous linear map $f(x) \in \mathcal{L}(U, V)$. Then the operator that maps a function $\phi: X \to U$ to the function $(f\phi)(x) = f(x)(\phi(x))$ has a variational adjoint. This variational adjoint maps a function $\psi: X \to V$ to the function defined by $x \mapsto f(x)^\dagger(\psi(x))$, where $f(x)^\dagger: V \to U$ is the adjoint of the continuous linear map $f(x)$.

Let $U$ be a real generalized inner product space. The derivative operator, which maps a function $\phi: \mathbb{R} \to U$ to its derivative $\phi'(x) = \frac{d\phi}{dx}(x)$, possesses a variational adjoint. This adjoint operator maps a function $\psi: \mathbb{R} \to U$ to its negative derivative $-\psi'(x) = -\frac{d\psi}{dx}(x)$. This relationship corresponds to the identity $\int_{-\infty}^{\infty} \langle \phi'(x), \psi(x) \rangle \, dx = \int_{-\infty}^{\infty} \langle \phi(x), -\psi'(x) \rangle \, dx$ for all test functions $\phi$ and $\psi$.

Let $X$ be a finite-dimensional real vector space equipped with an additive Haar measure (volume), and let $U$ be a real generalized inner product space. For any fixed vector $v \in X$, the linear operator $L$ that maps a test function $\phi: X \to U$ to its directional Fréchet derivative $(L\phi)(x) = \text{D}\phi(x)(v)$ has a variational adjoint $L^\dagger$. This adjoint operator is given by $(L^\dagger\psi)(x) = -\text{D}\psi(x)(v)$, where $\text{D}\psi(x)$ denotes the Fréchet derivative of the test function $\psi$ at $x$.

Let $X$ and $Y$ be real generalized inner product spaces, where $X$ is finite-dimensional, proper, and equipped with an additive Haar measure. For a fixed vector $\delta y \in Y$, let $L$ be the linear operator that maps a function $\phi: X \to Y$ to the function $x \mapsto (D\phi(x))^\dagger(\delta y)$, where $(D\phi(x))^\dagger: Y \to X$ is the adjoint of the Fréchet derivative of $\phi$ at $x$. The variational adjoint of $L$ is the operator $L^*$ that maps a function $\psi: X \to X$ to the function $x \mapsto -(\text{div } \psi(x)) \delta y$. This relationship is characterized by the identity: $$\int_X \langle (D\phi(x))^\dagger(\delta y), \psi(x) \rangle_X \, dx = \int_X \langle \phi(x), -(\text{div } \psi(x)) \delta y \rangle_Y \, dx$$ for all test functions $\phi$ and $\psi$.

Let $X$ be the $d$-dimensional real inner product space denoted by `Space d`. Let $L$ be the operator that maps a scalar function $\phi: X \to \mathbb{R}$ to its gradient field $\nabla \phi: X \to X$. The variational adjoint of $L$ is the operator $L^*$ that maps a vector field $\psi: X \to X$ to the scalar function $x \mapsto -\text{div}(\text{repr}(\psi(x)))$, where $\text{repr}$ denotes the representation of a vector in the canonical basis of $X$. This relationship is defined by the integral identity: $$\int_X \langle \nabla \phi(x), \psi(x) \rangle_X \, dx = \int_X \phi(x) \left( -\text{div}(\text{repr} \circ \psi)(x) \right) \, dx$$ for all test functions $\phi$ and $\psi$.

Let $X$ be a $d$-dimensional real inner product space (denoted as `Space d`). The operator that maps a scalar function $\phi: X \to \mathbb{R}$ to its gradient field $\nabla \phi: X \to X$ has a variational adjoint. This adjoint maps a vector field $\psi: X \to X$ to the scalar function defined by $x \mapsto -\text{div } \psi(x)$, where $\text{div}$ is the divergence operator. This relationship is characterized by the integral identity: $$\int_X \langle \nabla \phi(x), \psi(x) \rangle \, dx = \int_X \phi(x) (-\text{div } \psi(x)) \, dx$$ for all appropriate test functions $\phi$ and $\psi$.

Let $X$ be a $d$-dimensional real inner product space (denoted as `Space d`). The divergence operator, which maps a vector field $\phi: X \to \mathbb{R}^d$ to a scalar function $\text{div } \phi: X \to \mathbb{R}$, has a variational adjoint. This adjoint maps a scalar function $\psi: X \to \mathbb{R}$ to the vector field defined by $x \mapsto -\nabla \psi(x)$, where $\nabla$ is the gradient operator. This relationship is characterized by the integral identity: $$\int_X (\text{div } \phi(x)) \psi(x) \, dx = \int_X \langle \phi(x), -\nabla \psi(x) \rangle \, dx$$ for all appropriate test functions $\phi$ and $\psi$.

Let $X$ be a measurable space where open sets are measurable, and let the volume measure on $X$ be finite on compact sets. Let $F: (X \to U) \to (X \to V)$ and $G: (X \to U) \to (X \to W)$ be operators with variational adjoints $F'$ and $G'$, respectively. Then the product operator $H$, which maps a function $\phi: X \to U$ to the pair of functions $(F\phi, G\phi)$, has a variational adjoint $H'$. For any function $\psi: X \to V \times W$, this adjoint is given by $$H'(\psi) = F'(\psi_1) + G'(\psi_2)$$ where $\psi_1(x)$ and $\psi_2(x)$ are the first and second components of $\psi(x)$ in $V \times W$, respectively.

Let $X$ be a space and $U, W, V$ be target spaces. Let $F$ be an operator that maps functions $\phi: X \to U$ to functions $F(\phi): X \to W \times V$. If $F$ has a variational adjoint $F'$, then the operator $G: (X \to U) \to (X \to W)$ defined by the first component of $F$, $G(\phi)(x) = (F(\phi)(x))_1$, also has a variational adjoint $G'$. For any function $\phi: X \to W$, this adjoint is given by $G'(\phi) = F'(\psi)$, where $\psi: X \to W \times V$ is the function defined by $\psi(x') = (\phi(x'), 0)$ for all $x' \in X$.

Let $F$ be an operator that maps functions $\phi: X \to U$ to functions $F\phi: X \to W \times V$. Suppose $F$ has a variational adjoint $F'$. Then the operator $F_2$ defined by $F_2(\phi)(x) = (F\phi(x))_2$ (the projection onto the second component of the output) also has a variational adjoint. This adjoint maps a test function $\psi: X \to V$ to $F'(\tilde{\psi})$, where $\tilde{\psi}: X \to W \times V$ is the function defined by $\tilde{\psi}(x) = (0, \psi(x))$.