Physlib.Mathematics.Calculus.AdjFDeriv

Given a function $f: E \to F$ between two inner product spaces $E$ and $F$ over a field $\mathbb{k}$, let $Df(x): E \to F$ denote the Fréchet derivative of $f$ at the point $x \in E$. The adjoint Fréchet derivative at $x$ is the adjoint of this linear map, denoted $(Df(x))^*: F \to E$. For a vector $dy \in F$, the function returns the value $(Df(x))^*(dy) \in E$, which is the unique vector satisfying $\langle Df(x) \cdot v, dy \rangle_F = \langle v, (Df(x))^* \cdot dy \rangle_E$ for all $v \in E$.

Let $E$ and $F$ be inner product spaces over a field $\mathbb{k}$. For a function $f: E \to F$ and a point $x \in E$, if $f$ has an adjoint Fréchet derivative $f'$ at $x$ (expressed by the proposition `HasAdjFDerivAt 𝕜 f f' x`), then the defined adjoint Fréchet derivative $\text{adjFDeriv } \mathbb{k} \ f \ x$ is equal to $f'$.

Let $E$ and $F$ be complete generalized inner product spaces over a field $\mathbb{k}$. If a function $f: E \to F$ is Fréchet differentiable at a point $x \in E$, then $f$ possesses an adjoint Fréchet derivative at $x$, which is given by the adjoint of its Fréchet derivative, denoted as $\text{adjFDeriv } \mathbb{k} \ f \ x = (Df(x))^*$.

Let $X$ and $Y$ be real inner product spaces. The map that sends a pair $(L, y) \in \mathcal{L}(X, Y) \times Y$ to $L^* y$, where $L^*: Y \to X$ is the adjoint of the continuous linear map $L$, is a bounded bilinear map over $\mathbb{R}$.

Let $E$ be a real normed space, and let $F$ and $G$ be complete real generalized inner product spaces. Suppose $f: E \times F \to G$ is a function such that for every $x \in E$ and $y \in F$, the partial map $f_x: y \mapsto f(x, y)$ has an adjoint Fréchet derivative $f'(x, y): G \to F$ at $y$. If $f$ is $C^{n+1}$ as a function of $(x, y)$, then the map $(x, y, dz) \mapsto f'(x, y) \cdot dz$ from $E \times F \times G$ to $F$ is $C^n$. Here, the adjoint Fréchet derivative $f'(x, y)$ is the adjoint of the linear map $D_y f(x, y): F \to G$, satisfying $\langle D_y f(x, y) \cdot v, dz \rangle_G = \langle v, f'(x, y) \cdot dz \rangle_F$ for all $v \in F$ and $dz \in G$.

Let $U$ be an inner product space over a field $\mathbb{k}$ (where $\mathbb{k}$ is typically $\mathbb{R}$ or $\mathbb{C}$). For any function $f: U \to \mathbb{k}$ that is Fréchet differentiable at a point $x \in U$, the gradient $\nabla f(x)$ is equal to the adjoint Fréchet derivative of $f$ at $x$ applied to the unit $1 \in \mathbb{k}$. That is, $\nabla f(x) = (Df(x))^*(1)$, where $(Df(x))^*$ denotes the adjoint of the Fréchet derivative $Df(x): U \to \mathbb{k}$.

Let $E$ be an inner product space over a field $\mathbb{k}$. For any $x \in E$, the identity function $f : E \to E$ defined by $f(x) = x$ has an adjoint Fréchet derivative at $x$ which is the identity function $f'(dx) = dx$.

Let $E$ be an inner product space over a field $\mathbb{k}$. The adjoint Fréchet derivative of the identity function $f: E \to E$, defined by $f(x) = x$, is the identity map at every point. Specifically, for any point $x \in E$ and any vector $dx \in E$, the adjoint Fréchet derivative satisfies $(\text{adjFDeriv } \mathbb{k} \ f \ x) \ dx = dx$.

Let $E$ be an inner product space over a field $\mathbb{k}$. The adjoint Fréchet derivative of the identity function $\text{id} : E \to E$ is the identity map at every point. Specifically, for any point $x \in E$ and any vector $dx \in E$, the adjoint Fréchet derivative satisfies $(\text{adjFDeriv } \mathbb{k} \ \text{id} \ x) \ dx = dx$.

Let $E$ and $F$ be inner product spaces over a field $\mathbb{k}$. For any point $x \in E$ and any value $y \in F$, the constant function $f: E \to F$ defined by $f(z) = y$ has an adjoint Fréchet derivative at $x$ given by the zero linear map $dz \mapsto 0$.

Let $E$ and $F$ be inner product spaces over a field $\mathbb{k}$. For any constant $y \in F$, let $f: E \to F$ be the constant function $f(x) = y$. The adjoint Fréchet derivative of $f$ is zero at every point $x \in E$, meaning that for all $x \in E$ and all vectors $dy \in F$, $(\text{adjFDeriv } \mathbb{k} \ f \ x)(dy) = 0$.

Let $E, F$, and $G$ be inner product spaces over a field $\mathbb{k}$. Let $g: E \to F$ and $f: F \to G$ be functions. Suppose $g$ has an adjoint Fréchet derivative $g'$ at a point $x \in E$, and $f$ has an adjoint Fréchet derivative $f'$ at the point $g(x)$. Then the composition $f \circ g$ has an adjoint Fréchet derivative at $x$ given by the composition of the adjoint derivatives $g' \circ f'$.

Let $E, F$, and $G$ be complete inner product spaces over a field $\mathbb{k}$. Suppose $g: E \to F$ is a function differentiable at a point $x \in E$ and $f: F \to G$ is a function differentiable at $g(x)$. The adjoint Fréchet derivative of the composition $f \circ g$ at $x$ is the composition of the adjoint Fréchet derivatives of $g$ at $x$ and $f$ at $g(x)$. That is, for any $dy \in G$: \[ (\text{adjFDeriv } \mathbb{k} \ (f \circ g) \ x)(dy) = (\text{adjFDeriv } \mathbb{k} \ g \ x)((\text{adjFDeriv } \mathbb{k} \ f \ (g(x)))(dy)) \] In terms of the adjoint of the Fréchet derivative operator $D$, this is expressed as: \[ (D(f \circ g)(x))^* = (Dg(x))^* \circ (Df(g(x)))^* \]

Let $E, F$, and $G$ be inner product spaces over a field $\mathbb{k}$. Suppose $f: E \to F$ and $g: E \to G$ are functions such that at a point $x \in E$, $f$ has an adjoint Fréchet derivative $f': F \to E$ and $g$ has an adjoint Fréchet derivative $g': G \to E$. Then the map $h: E \to F \times G$ defined by $h(x) = (f(x), g(x))$ has an adjoint Fréchet derivative at $x$ given by the map $(dy, dz) \mapsto f'(dy) + g'(dz)$. Here, the product space $F \times G$ is equipped with the standard product inner product $\langle (y_1, z_1), (y_2, z_2) \rangle = \langle y_1, y_2 \rangle_F + \langle z_1, z_2 \rangle_G$.

Let $E, F$, and $G$ be generalized inner product spaces over a field $\mathbb{k}$. If a function $f: E \to F \times G$ has an adjoint Fréchet derivative $f' : F \times G \to E$ at a point $x \in E$, then the map $x \mapsto (f(x))_1$ (the projection onto the first component) has an adjoint Fréchet derivative at $x$ given by the function $dy \mapsto f'(dy, 0)$.

Let $E, F$, and $G$ be complete generalized inner product spaces over a field $\mathbb{k}$. For a function $f: E \to F \times G$ that is Fréchet differentiable at a point $x \in E$, the adjoint Fréchet derivative of its first component $f_1: E \to F$ (where $f_1(x) = (f(x))_1$) at $x$ is given by the linear map $dy \mapsto (Df(x))^*(dy, 0)$. Here, $(Df(x))^*: F \times G \to E$ denotes the adjoint Fréchet derivative of $f$ at $x$. The product space $F \times G$ is equipped with the standard product inner product $\langle (y_1, z_1), (y_2, z_2) \rangle = \langle y_1, y_2 \rangle_F + \langle z_1, z_2 \rangle_G$.

Let $E$ and $F$ be complete generalized inner product spaces over a field $\mathbb{k}$. For the projection map $\pi_1: F \times E \to F$ defined by $\pi_1(y, z) = y$, its adjoint Fréchet derivative at any point $x \in F \times E$ is the linear map that sends $a \in F$ to the pair $(a, 0) \in F \times E$.

Let $E, F$, and $G$ be inner product spaces over a field $\mathbb{k}$. Let $f: E \to F \times G$ be a function and $x \in E$. If $f$ has an adjoint Fréchet derivative $f': F \times G \to E$ at $x$, then the map $x \mapsto \pi_2(f(x))$ (the projection onto the second component) also has an adjoint Fréchet derivative at $x$, which is given by the linear map $dz \mapsto f'(0, dz)$.

Let $E, F$, and $G$ be complete generalized inner product spaces over a field $\mathbb{k}$. Let $f: E \to F \times G$ be a function that is differentiable at a point $x \in E$. Then the adjoint Fréchet derivative of the map $x \mapsto (f(x))_2$ (the projection of $f$ onto its second component) at $x$ is given by the linear map that sends $dy \in G$ to $(Df(x))^*(0, dy)$, where $(Df(x))^*: F \times G \to E$ denotes the adjoint Fréchet derivative of $f$ at $x$.

Let $E$ and $F$ be complete inner product spaces over a field $\mathbb{k}$. The adjoint Fréchet derivative of the second projection map $\text{proj}_2 : F \times E \to E$, defined by $\text{proj}_2(y, z) = z$, at any point $x \in F \times E$ is the linear map from $E$ to $F \times E$ that sends $a \in E$ to $(0, a)$.

Let $E, F$, and $G$ be inner product spaces over a field $\mathbb{k}$. Let $f: E \to F \to G$ be a function, and let $(x, y) \in E \times F$. Suppose that: 1. The uncurried function $(x, y) \mapsto f(x, y)$ is differentiable at $(x, y)$. 2. The partial map $x' \mapsto f(x', y)$ has an adjoint Fréchet derivative $fx': G \to E$ at $x$. 3. The partial map $y' \mapsto f(x, y')$ has an adjoint Fréchet derivative $fy': G \to F$ at $y$. Then the uncurried function $(x, y) \mapsto f(x, y)$ has an adjoint Fréchet derivative at $(x, y)$ given by the map $dz \mapsto (fx'(dz), fy'(dz))$ from $G$ into $E \times F$.

Let $E, F$, and $G$ be complete inner product spaces over a field $\mathbb{k}$. Let $f: E \to F \to G$ be a function such that the uncurried function $\hat{f}(x, y) = f(x, y)$ is differentiable at a point $(x, y) \in E \times F$. The adjoint Fréchet derivative of $\hat{f}$ at $(x, y)$, denoted $(D\hat{f}(x, y))^*: G \to E \times F$, is given by: $$(D\hat{f}(x, y))^*(dz) = \left( (D_1 f(x, y))^*(dz), (D_2 f(x, y))^*(dz) \right)$$ for all $dz \in G$, where $(D_1 f(x, y))^*: G \to E$ is the adjoint Fréchet derivative of the partial map $x' \mapsto f(x', y)$ at $x$, and $(D_2 f(x, y))^*: G \to F$ is the adjoint Fréchet derivative of the partial map $y' \mapsto f(x, y')$ at $y$.

Let $E$ and $F$ be inner product spaces over a field $\mathbb{k}$. If a function $f: E \to F$ has an adjoint Fréchet derivative $f'$ at a point $x \in E$, then the function $x \mapsto -f(x)$ also has an adjoint Fréchet derivative at $x$, given by the map $dy \mapsto -f'(dy)$.

Let $E$ and $F$ be complete inner product spaces over a field $\mathbb{k}$. If a function $f: E \to F$ is Fréchet differentiable at a point $x \in E$, then the adjoint Fréchet derivative of the negated function $x \mapsto -f(x)$ at $x$ is equal to the negation of the adjoint Fréchet derivative of $f$ at $x$. In terms of the adjoint operator $(Df(x))^*$, this is expressed as: $$(D(-f)(x))^* = -(Df(x))^*$$ Specifically, for any vector $dy \in F$, the value is $-(Df(x))^*(dy)$.

Let $E$ and $F$ be inner product spaces over a field $\mathbb{k}$. If $f, g: E \to F$ are functions such that $f$ has an adjoint Fréchet derivative $f'$ at $x \in E$ and $g$ has an adjoint Fréchet derivative $g'$ at $x$, then the sum function $f + g$ (defined by $x \mapsto f(x) + g(x)$) has an adjoint Fréchet derivative at $x$ given by $f' + g'$ (defined by $dy \mapsto f'(dy) + g'(dy)$).

Let $E$ and $F$ be complete inner product spaces over a field $\mathbb{k}$. If $f, g: E \to F$ are functions differentiable at $x \in E$, then the adjoint Fréchet derivative of their sum $f + g$ at $x$ is the sum of their individual adjoint Fréchet derivatives at $x$. That is, \[ (D(f+g)(x))^* = (Df(x))^* + (Dg(x))^*, \] where $(Df(x))^*$ denotes the adjoint of the Fréchet derivative of $f$ at $x$.

Let $E$ and $F$ be inner product spaces over a field $\mathbb{k}$. Suppose $f, g: E \to F$ are functions that have adjoint Fréchet derivatives $f', g': F \to E$ at a point $x \in E$, respectively. Then the function $f - g$ (defined by $x \mapsto f(x) - g(x)$) has an adjoint Fréchet derivative at $x$ given by $f' - g'$ (defined by $dy \mapsto f'(dy) - g'(dy)$).

Let $E$ and $F$ be complete inner product spaces over a field $\mathbb{k}$. Suppose $f, g: E \to F$ are functions that are Fréchet differentiable at a point $x \in E$. Then the adjoint Fréchet derivative of the difference $f - g$ at $x$ is equal to the difference of the adjoint Fréchet derivatives of $f$ and $g$ at $x$. That is, for any vector $dy \in F$: $$\text{adjFDeriv } \mathbb{k} \ (f - g) \ x \ (dy) = \text{adjFDeriv } \mathbb{k} \ f \ x \ (dy) - \text{adjFDeriv } \mathbb{k} \ g \ x \ (dy)$$ Using the notation $(Df(x))^*$ for the adjoint of the Fréchet derivative, this can be written as: $$(D(f - g)(x))^* = (Df(x))^* - (Dg(x))^*$$

Let $E$ and $F$ be generalized inner product spaces over a field $\mathbb{k}$ (where $\mathbb{k}$ is $\mathbb{R}$ or $\mathbb{C}$). Suppose $f: E \to F$ is a map with an adjoint Fréchet derivative $f'$ at $x \in E$, and $g: E \to \mathbb{k}$ is a map with an adjoint Fréchet derivative $g'$ at $x$. Then the adjoint Fréchet derivative of the scalar-vector product $x \mapsto g(x)f(x)$ at the point $x$ is the linear map $F \to E$ given by: $$dy \mapsto \overline{g(x)} f'(dy) + g'(\overline{\langle dy, f(x) \rangle})$$ where $\overline{z}$ denotes the conjugate of $z \in \mathbb{k}$ and $\langle \cdot, \cdot \rangle$ denotes the inner product on $F$.

Let $E$ and $F$ be complete inner product spaces over a field $\mathbb{k}$ (where $\mathbb{k}$ is $\mathbb{R}$ or $\mathbb{C}$). If $f: E \to F$ and $g: E \to \mathbb{k}$ are differentiable at a point $x \in E$, then the adjoint Fréchet derivative of the scalar-vector product $x \mapsto g(x)f(x)$ at $x$ is the linear map $F \to E$ given by: $$dy \mapsto \overline{g(x)} (Df(x))^*(dy) + (Dg(x))^*(\overline{\langle dy, f(x) \rangle})$$ where $(Df(x))^*$ and $(Dg(x))^*$ denote the adjoint Fréchet derivatives of $f$ and $g$ at $x$ respectively, and $\langle \cdot, \cdot \rangle$ denotes the inner product on $F$.

Let $E$ be a real generalized inner product space. For the inner product function $f: E \times E \to \mathbb{R}$ defined by $f(x_1, x_2) = \langle x_1, x_2 \rangle$, the adjoint Fréchet derivative at any point $x = (x_1, x_2) \in E \times E$ is the linear map from $\mathbb{R}$ to $E \times E$ defined by: $$y \mapsto y \cdot (x_2, x_1)$$ for any $y \in \mathbb{R}$.

Let $E$ be a real generalized inner product space. For the inner product function $f: E \times E \to \mathbb{R}$ defined by $f(x_1, x_2) = \langle x_1, x_2 \rangle$, the adjoint Fréchet derivative at any point $x = (x_1, x_2) \in E \times E$ is the linear map from $\mathbb{R}$ to $E \times E$ defined by: $$y \mapsto y \cdot (x_2, x_1)$$ for any $y \in \mathbb{R}$.