Abstract Nonsense

Crushing one theorem at a time

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.

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May 25, 2011 - Posted by | Analysis | , , , , , ,

4 Comments »

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  3. […] evidently suffices to prove this for mappings by our prior theorem regarding the relationship between the total differentiability of a function and the total […]

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  4. […] first and last one we have proven before (here and here respectively) and the second can be proved thinking of multiplication as a bilinear map […]

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