# Abstract Nonsense

## Further Properties of the Total Derivative (Pt. I)

Point of Post: In this post we finally prove the majority of the basic theorems regarding the total derivative (differentiable functions form a vector space, etc.)

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Motivation

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So we now finish proving the ‘obvious’ facts one would hope that the total derivative would share with the normal derivative, such as the sum of differentiable functions is differentiable, the derivative of a sum is the sum of the derivatives, the product of two real valued differentiable functions is differentiable, the derivative of such a product is the product ‘rule’, the derivative of a vector valued function is differentiable if and only if each of its coordinate functions is, etc. A lot of the work is already done because of the corollaries of the total derivative of a multilinear function

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Properties of the Total Derivative

We now show the claimed properties actually hold:

Theorem: Let $f:U\to\mathbb{R}^m$, $U\subseteq\mathbb{R}^n$, with coordinate functions $f_1,\cdots,f_m$. Then, $f$ is differentiable at $a\in\mathbb{R}^n$ if and only if $f_j$ is differentiable at $a$ for $j=1,\cdots,n$. Moreover,

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$\displaystyle \text{Jac}_f(a)=\left(\begin{array}{ccc} & \text{Jac}{f_1}(a) & \\\hline & \vdots & \\\hline & \text{Jac}_{f_m}(a) & \end{array}\right)$

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where the above makes sense since $\text{Jac}_{f_j}(a)$ is a $1\times n$ matrix or just a row vector.

Proof: Suppose first that $f$ is differentiable at $a$. Then, each coordinate function $f_j$  is differentiable since the canonical projection map $\pi_j:\mathbb{R}^m\to\mathbb{R}$ is differentiable everywhere and $f_j=\pi_j\circ f$ from where the differentiability follows from the chain rule.

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Conversely, suppose that each $f_j$ is invertible and let $T$ be the linear operator whose matrix with respect to the canonical basis is

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$\left(\begin{array}{ccc} & \text{Jac}{f_1}(a) & \\\hline & \vdots & \\\hline & \text{Jac}_{f_m}(a) & \end{array}\right)$

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Note then that

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\displaystyle \begin{aligned}\frac{\|f(a+h)-f(a)-T(h)\|}{\|h\|} &=\frac{\left\|\left(f_1(a+h)-f_1(a)-\text{Jac}_{f_1}(a)(h),\cdots,f_m(a+h)-f_m(a)-\text{Jac}_{f_m}(a)(h)\right)\right\|}{\|h\|}\\ &\leqslant \sum_{j=1}^{m}\frac{\|f_j(a+h)-f_j(a)-\text{Jac}_{f_j}(a)(h)\|}{\|h\|}\end{aligned}

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and so taking the limit as $h\to\bold{0}$ on both sides (noting that each summand goes to zero) we may conclude that

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$\displaystyle \lim_{h\to\bold{0}}\frac{\left\|f(a+h)-f(a)-T(h)\right\|}{\|h\|}=0$

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and thus $f$ is differentiable at $a$ and $D_f(a)=T$. $\blacksquare$

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Theorem: Let $f_1,\cdots,f_k:U\to\mathbb{R}^m$ (where $U\subseteq\mathbb{R}^n$) be differentiable at $a\in U$. Then, $f_1+\cdots+f_k$ is differentiable at $D_{f_1+\cdots+f_k}(a)=D_{f_1}(a)+\cdots+D_{f_k}(a)$.

Proof: Consider the functions $K:\mathbb{R}^n\to \mathbb{R}^{mk}$ given by $x\mapsto (f_{1,1}(x),\cdots,f_{1,m}(x),\cdots,f_{k,1}(x),\cdots,f_{k,m}(x))$ (where $f_{i,j}$ is the $j^{\text{th}}$ coordinate function of $f_i$) and $J:\mathbb{R}^{mk}\to\mathbb{R}^m$ given by

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$\displaystyle (x_{1,1},\cdots,x_{1,m},\cdots,x_{k,1},\cdots,x_{k,m})\mapsto \sum_{j=1}^{k}(x_{j,1},\cdots,x_{j,m})$

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Clearly then $K$ is differentiable at $a$ since by the last theorem each $f_{i,j}$ is differentiable at $a$ and so each coordinate function of $K$ is differentiable. Moreover, it’s clear that $J$ is differentiable since it’s linear. Noting then that $f_1+\cdots+f_k=J\circ K$ we may conclude from the chain rule that $f_1+\cdots+f_k$ is differentiable at $a$ and $D_{f_1+\cdots+f_k}(a)=D_J(K(a))\circ D_K(a)$. That said, we note from the previous theorem that

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$\text{Jac}_K(a)=\left(\begin{array}{ccc}& \text{Jac}_{f_{1,1}}(a) &\\ \hline & \vdots & \\ \hline & \text{Jac}_{f_{1,m}}(a) & \\ \hline& \vdots & \\ \hline & \text{Jac}_{f_{k,1}}(a) & \\ \hline & \vdots & \\ \hline & \text{Jac}_{f_{k,m}}(a) & \end{array}\right)$

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and since $J$ is linear we know that $D_J(f(a))=J$. Thus,

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\displaystyle \begin{aligned}D_{f_1+\cdots+f_k}(a) &= D_{J}(f(a))\left(D_K(a)\right)(x)\\ & =J\left(\text{Jac}_{f_{1,1}}(a)(x),\cdots,\text{Jac}_{f_{1,m}}(a)(x),\cdots,\text{Jac}_{f_{k,1}}(a)(x),\cdots,\text{Jac}_{f_{k,m}}(a)(x)\right)\\ &= \sum_{j=1}^{k}\left(\text{Jac}_{f_{j,1}}(a)(x),\cdots,\text{Jac}_{f_{j,m}}(a)(x)\right)\\ &= \sum_{j=1}^{k}D_{f_j}(a)(x)\end{aligned}

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Since $x$ was arbitrary the conclusion follows. $\blacksquare$

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References:

1. Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.

May 25, 2011 -

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