Physlib.Mathematics.FDerivCurry

Let $\mathbf{k}$ be a nontrivially normed field, and let $X, Y, Z$ be normed spaces over $\mathbf{k}$. Let $f: X \to Y \to Z$ be a curried function and $F: X \times Y \to Z$ be its uncurried version, defined by $F(x, y) = f(x)(y)$. If $F$ is differentiable at $(x, y)$, then for any vector $(dx, dy) \in X \times Y$, the Fréchet derivative $DF(x, y)$ applied to $(dx, dy)$ is given by the sum of the partial derivatives: $$DF(x, y)(dx, dy) = D(x' \mapsto f(x', y))(x)(dx) + D(y' \mapsto f(x, y'))(y)(dy)$$ where $D(x' \mapsto f(x', y))(x)$ denotes the derivative of $f$ with respect to the first argument at $x$, and $D(y' \mapsto f(x, y'))(y)$ denotes the derivative of $f$ with respect to the second argument at $y$.

Let $\mathbf{k}$ be a nontrivially normed field, and let $X, Y, Z$ be normed spaces over $\mathbf{k}$. Let $f: X \times Y \to Z$ be a function that is differentiable at $(x, y) \in X \times Y$. Then for any vector $dx \in X$, the Fréchet derivative of the function $x' \mapsto f(x', y)$ at $x$ applied to $dx$ is equal to the Fréchet derivative of $f$ at $(x, y)$ applied to the vector $(dx, 0)$: $$D(x' \mapsto f(x', y))(x)(dx) = Df(x, y)(dx, 0)$$

Let $\mathbf{k}$ be a nontrivially normed field, and let $X, Y, Z$ be normed spaces over $\mathbf{k}$. Let $f: X \times Y \to Z$ be a function. If $f$ is differentiable at $(x, y) \in X \times Y$, then for any vector $dy \in Y$, the Fréchet derivative of the partial map $y' \mapsto f(x, y')$ evaluated at $y$ and applied to $dy$ is equal to the total Fréchet derivative of $f$ at $(x, y)$ applied to the vector $(0, dy)$: $$D(y' \mapsto f(x, y'))(y)(dy) = Df(x, y)(0, dy)$$

Let $\mathbf{k}$ be a nontrivially normed field, and let $X, Y, Z$ be normed spaces over $\mathbf{k}$. Let $f: X \to Y \to Z$ be a function and let $F: X \times Y \to Z$ be its uncurried version, defined by $F(x, y) = f(x)(y)$. If $F$ is differentiable, then its Fréchet derivative $DF$ is given by the map: $$DF = (x, y) \mapsto D(x' \mapsto f(x', y))(x) \circ \pi_1 + D(y' \mapsto f(x, y'))(y) \circ \pi_2$$ where $\pi_1: X \times Y \to X$ and $\pi_2: X \times Y \to Y$ are the canonical projections from the product space, and $D(x' \mapsto f(x', y))(x)$ and $D(y' \mapsto f(x, y'))(y)$ denote the partial derivatives with respect to the first and second arguments, respectively.

Let $\mathbf{k}$ be a nontrivially normed field, and let $X, Y, Z$ be normed spaces over $\mathbf{k}$. Let $f: X \times Y \to Z$ be a function that is differentiable at a point $(x, y) \in X \times Y$. Then the Fréchet derivative of $f$ at $(x, y)$, denoted by $Df(x, y)$, is equal to the sum of its partial derivatives composed with the coordinate projections: $$Df(x, y) = D(x' \mapsto f(x', y))(x) \circ \pi_1 + D(y' \mapsto f(x, y'))(y) \circ \pi_2$$ where $\pi_1: X \times Y \to X$ and $\pi_2: X \times Y \to Y$ are the canonical continuous linear projections onto the first and second factors, respectively.

Let $\mathbf{k}$ be a nontrivially normed field, and let $X, Y, Z$ be normed spaces over $\mathbf{k}$. Let $f: X \times Y \to Z$ be a differentiable function. The Fréchet derivative of $f$, denoted by $Df$, is given by the map: $$Df(x, y) = D(x' \mapsto f(x', y))(x) \circ \pi_1 + D(y' \mapsto f(x, y'))(y) \circ \pi_2$$ where $\pi_1: X \times Y \to X$ and $\pi_2: X \times Y \to Y$ are the canonical continuous linear projections onto the first and second factors, respectively. Here $D(x' \mapsto f(x', y))(x)$ and $D(y' \mapsto f(x, y'))(y)$ represent the partial derivatives with respect to the first and second variables at the point $(x, y)$.

Let $X$, $Y$, and $Z$ be normed spaces over a field $\mathbb{k}$. Let $f: X \to \mathcal{L}_{\mathbb{k}}(Y, Z)$ be a function that is differentiable at $x \in X$, where $\mathcal{L}_{\mathbb{k}}(Y, Z)$ denotes the space of continuous linear maps from $Y$ to $Z$. For any $y \in Y$ and any vector $dx \in X$, the Fréchet derivative of $f$ at $x$ applied to $dx$ and then evaluated at $y$ is equal to the Fréchet derivative of the map $x \mapsto f(x)y$ at $x$ applied to $dx$: $$Df(x)(dx)(y) = D(x \mapsto f(x)y)(x)(dx)$$

Let $X, Y, Z$ be normed spaces over a field $\mathbb{k}$, and let $f: X \to Y \to Z$ be a function. Let $\hat{f} : X \times Y \to Z$ denote the uncurried version of $f$ (where $\hat{f}(x, y) = f(x)(y)$). If $\hat{f}$ is differentiable, then for any fixed $y \in Y$, the Fréchet derivative of the partial function $x \mapsto \hat{f}(x, y)$ is equal to the composition of the Fréchet derivative of $\hat{f}$ and the Fréchet derivative of the inclusion map $x \mapsto (x, y)$. That is: $$D(x \mapsto \hat{f}(x, y)) = x \mapsto D\hat{f}(x, y) \circ D(x \mapsto (x, y))$$ where $D$ denotes the Fréchet derivative `fderiv 𝕜`.

Let $X, Y,$ and $Z$ be normed spaces over a field $\mathbb{k}$, and let $f: X \to Y \to Z$ be a function. Let $\hat{f} : X \times Y \to Z$ denote the uncurried version of $f$ (where $\hat{f}(x, y) = f(x)(y)$). If $\hat{f}$ is differentiable, then for any fixed $x \in X$, the Fréchet derivative of the partial function $y \mapsto \hat{f}(x, y)$ is equal to the composition of the Fréchet derivative of $\hat{f}$ and the Fréchet derivative of the inclusion map $y \mapsto (x, y)$. That is: $$D(y \mapsto \hat{f}(x, y)) = y \mapsto D\hat{f}(x, y) \circ D(y \mapsto (x, y))$$ where $D$ denotes the Fréchet derivative `fderiv 𝕜`.

Let $X, Y, Z$ be normed spaces over a field $\mathbb{k}$ and $f: X \to Y \to Z$ be a function. Let $\hat{f}: X \times Y \to Z$ denote the uncurried version of $f$, defined by $\hat{f}(x, y) = f(x)(y)$. If $\hat{f}$ is differentiable, then for any $x, dx \in X$ and $y \in Y$, the value of the Fréchet derivative of the partial map $x' \mapsto f(x', y)$ at the point $x$ applied to the vector $dx$ is given by: $$(D(x' \mapsto f(x', y))(x))(dx) = (D\hat{f}(x, y)) ((D(x' \mapsto (x', y))(x))(dx))$$ where $D$ denotes the Fréchet derivative `fderiv 𝕜`.

Let $\mathbb{k}$ be a normed field and $X, Y, Z$ be normed spaces over $\mathbb{k}$. Let $f: X \to Y \to Z$ be a function such that its uncurried form $\tilde{f}: X \times Y \to Z$ (defined by $(x, y) \mapsto f(x)(y)$) is differentiable. For any fixed $x \in X$ and points $y, dy \in Y$, the Fréchet derivative of the partial map $y' \mapsto f(x)(y')$ at $y$ applied to the increment $dy$ is given by: $$\left( D(y' \mapsto f(x, y')) \big|_y \right) (dy) = D\tilde{f}(x, y) \left( \left( D(y' \mapsto (x, y')) \big|_y \right) (dy) \right)$$ where $D$ denotes the Fréchet derivative.

For any $x \in X$ and $y \in Y$, the Fréchet derivative of the map $y' \mapsto (x, y')$ at the point $y$ is the canonical continuous linear injection of the second factor into the product, $\text{inr} : Y \to X \times Y$, which is defined by $dy \mapsto (0, dy)$.

Let $\mathbb{k}$ be a normed field and $X, Y$ be normed spaces over $\mathbb{k}$. For any fixed $y \in Y$, the Fréchet derivative of the map $x \mapsto (x, y)$ at any point $x \in X$ is equal to the canonical continuous linear inclusion $\text{inl} : X \to X \times Y$, which is defined by $v \mapsto (v, 0)$.

Let $\mathbb{k}$ be a normed field and $X, Y, Z$ be normed spaces over $\mathbb{k}$. If a function $f: X \to Y \to Z$ is such that its uncurried version $(x, y) \mapsto f(x, y)$ is differentiable, then for any fixed $y \in Y$ and any point $x \in X$, the partial map $x' \mapsto f(x', y)$ is differentiable at $x$.

Let $\mathbb{k}$ be a normed field and $X, Y, Z$ be normed spaces over $\mathbb{k}$. Let $f : X \to (Y \to Z)$ be a function. If the uncurried function $(x, y) \mapsto f(x, y)$ is differentiable on $X \times Y$, then for any fixed $x \in X$, the map $y' \mapsto f(x, y')$ is differentiable at the point $y \in Y$.

Let $\mathbb{k}$ be a scalar field, and let $X, Y$, and $Z$ be normed spaces over $\mathbb{k}$. Let $f: X \times Y \to Z$ be a function (represented in the formal statement as the uncurrying of $f: X \to Y \to Z$). If $f$ is $C^2$ (twice continuously differentiable), then for any fixed $y \in Y$, the Fréchet derivative of the partial map $x \mapsto f(x, y)$ is differentiable.

Let $f: X \to Y \to Z$ be a function between normed vector spaces over a field $\mathbb{k}$. If the uncurried function $(x, y) \mapsto f(x, y)$ is twice continuously differentiable ($C^2$), then for any fixed $x \in X$, the map $y \mapsto D_y f(x, y)$, which assigns to each $y$ the Fréchet derivative of the partial function $y' \mapsto f(x, y')$, is differentiable.

Let $\mathbb{k}$ be a normed field, and $X, Y, Z$ be normed spaces over $\mathbb{k}$. Let $f: X \times Y \to Z$ be a function whose uncurried form $(x, y) \mapsto f(x, y)$ is twice continuously Fréchet differentiable ($C^2$). For any fixed $x \in X$, the mapping $y \mapsto D_1 f(x, y)$ is differentiable, where $D_1 f(x, y)$ denotes the Fréchet derivative of the partial map $x' \mapsto f(x', y)$ evaluated at $x$.

Let $\mathbb{k}$ be a normed field, and $X, Y, Z$ be normed spaces over $\mathbb{k}$. Let $f: X \times Y \to Z$ be a function such that its uncurried form $(x, y) \mapsto f(x, y)$ is twice continuously Fréchet differentiable ($C^2$). For any fixed $x \in X$ and direction vector $\delta x \in X$, the mapping $y \mapsto (D_1 f(x, y))(\delta x)$ is differentiable with respect to $y$, where $D_1 f(x, y)$ denotes the Fréchet derivative of the partial map $x' \mapsto f(x', y)$ evaluated at $x$.

Let $X, Y$, and $Z$ be normed spaces over a field $\mathbb{k}$. Let $f: X \times Y \to Z$ be a function such that its uncurried form $(x, y) \mapsto f(x, y)$ is twice continuously differentiable ($C^2$). For any fixed $y \in Y$, the function that maps $x$ to the Fréchet derivative of the partial map $y' \mapsto f(x, y')$ evaluated at $y$ is differentiable with respect to $x$.

Let $X, Y$, and $Z$ be normed spaces over a field $\mathbb{k}$. Let $f: X \times Y \to Z$ be a function such that its uncurried form $(x, y) \mapsto f(x, y)$ is twice continuously differentiable ($C^2$). For any fixed $y \in Y$ and $\delta y \in Y$, the function that maps $x$ to the Fréchet derivative of the partial map $y' \mapsto f(x, y')$ evaluated at $y$ and then applied to the vector $\delta y$ is differentiable with respect to $x$.

Let $\mathbb{k}$ be a normed field, and $X, Y, Z$ be normed spaces over $\mathbb{k}$. Let $f: X \times Y \to Z$ be a function such that its uncurried form $(x, y) \mapsto f(x, y)$ is twice continuously Fréchet differentiable ($C^2$). For any fixed $x \in X$ and direction vector $dx \in X$, the mapping $y' \mapsto (D_1 f(x, y'))(dx)$ is differentiable at $y \in Y$, where $D_1 f(x, y')$ denotes the Fréchet derivative of the partial map $x' \mapsto f(x', y')$ evaluated at $x$.

Let $X, Y$, and $Z$ be normed spaces over a field $\mathbb{k}$. Let $f: X \times Y \to Z$ be a function such that its uncurried form $(x, y) \mapsto f(x, y)$ is twice continuously differentiable ($C^2$). For any fixed $y \in Y$ and vector $\delta y \in Y$, the function that maps $x'$ to the Fréchet derivative of the partial map $y' \mapsto f(x', y')$ evaluated at $y$ and applied to the vector $\delta y$ is differentiable at $x$.

Let $\mathbb{k}$ be a normed field that is either the real numbers $\mathbb{R}$ or the complex numbers $\mathbb{C}$. Let $X, Y,$ and $Z$ be normed spaces over $\mathbb{k}$. Suppose $f: X \times Y \to Z$ is a twice continuously Fréchet differentiable ($C^2$) function. For any points $x \in X, y \in Y$ and vectors $dx \in X, dy \in Y$, the mixed partial derivatives commute: $$D_1 (x' \mapsto D_2 f(x', y)(dy))(x)(dx) = D_2 (y' \mapsto D_1 f(x, y')(dx))(y)(dy)$$ where $D_1$ denotes the partial Fréchet derivative with respect to the first variable and $D_2$ denotes the partial Fréchet derivative with respect to the second variable.