Abstract Nonsense

Crushing one theorem at a time

Relationship Between the Notions of Directional and Total Derivatives (Pt.II)

Point of Post: This is a continuation of this post.

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What we shall now show that there is an almost converse to the above. Namely, if we impose the not-so-crazy condition that the first partial order derivatives of all types exist on some neighborhood of some point a and they are all continuous there then the function is, in fact,totally  differentiable there too.

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Theorem: Let f:U\to\mathbb{R}^m, where U\subseteq\mathbb{R}^n is open, be such that D_jf exists everywhere on U for j\in[n] and be continuous at a\in U. Then, f is totally differentiable at U.

Proof: It evidently suffices to prove this for mappings f:U\to\mathbb{R} by our prior theorem regarding the relationship between the total differentiability of a function and the total differentiability of its coordinate functions. So, let f:U\to\mathbb{R} where U\subseteq\mathbb{R}^n is open and a=(a_1,\cdots,a_n)\in U. Note then that for sufficiently small h=(h_1,\cdots,h_n) (so that a+h\in U) we have that f(a+h)-f(a) is equal to

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\displaystyle \sum_{j=1}^{n}\bigg(f(a_1+h_1,\cdots,a_j+h_j,a_{j+1},\cdots,a_n)-f(a_1+h_1,\cdots,a_{j-1}+h_{j-1},a_j,\cdots,a_n)\bigg)

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(everything just telescopes). But, using logic similar to that in the proof of Clairaut’s theorem we may apply the mean value theorem to find \xi_j\in\mathbb{R} with \|\xi_j-a_j\|\leqslant \|(a_j+h_j)-a_j\|=\|h_j\| such that  f(a_1+h_1,\cdots,a_j+h_j,a_{j+1},\cdots,a_n)-f(a_1+h_1,\cdots,a_{j-1}+h_{j-1},a_j,\cdots,a_n) is equal to h_jD_jf(a_1+h_1,\cdots,\xi_j,a_{j+1},\cdots,a_n). Thus, in particular we have that

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\displaystyle f(a+h)-f(a)=\sum_{j=1}^{n}h_j D_jf(a_1+h_1,\cdots,\xi_j,a_{j+1},\cdots,a_n)

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Thus, we may conclude that

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\displaystyle \begin{aligned}\frac{\left\|f(a+h)-f(a)-\nabla f(a)\cdot h\right\|}{\|h\|} &= \frac{\displaystyle \left\|\sum_{j=1}^{n}h_j \bigg(D_j(a_1+h_1,\cdots,\xi_j,a_{j+1},\cdots,a_n)-D_jf(a)\bigg)\right\|}{\|h\|}\\ &\leqslant \sum_{j=1}^{n}\frac{|h_j|}{\|h\|}\left\|D_jf(a_1+h_1,\cdots,\xi_j,a_{j+1},\cdots,a_n)-D_f(a)\right\|\\ &\leqslant \sum_{j=1}^{n}\left\|D_jf(a_1+h_1,\cdots,\xi_j,a_{j+1},\cdots,a_n)-D_jf(a)\right\|\end{aligned}

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Note though that as h\to\bold{0} we have that (a_1+h_1,\cdots,\xi_j,a_{j+1},\cdots,a_n)\to a and thus by continuity of D_jf at a for j=1,\cdots,n we have that D_jf(a_1+h_1,\cdots,\xi_j,a_{j+1},\cdots,a_n)\to D_jf(a). It thus follows from the above that

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

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Thus, f is differentiable at a as desired. \blacksquare

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Remark: This gives an alternate proof (once one knows that total differentiability implies partial differetiability in all directions) that \text{Jac}_f is the matrix of partials–once you combine it with our previous half-uncovered form of the derivative. Also, there is a stronger form of this theorem where one only needs all but one of the partial derivatives to be continuous–but this seems excessive in most cases.

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

2. Apostol, Tom M. Mathematical Analysis. Reading, MA: Addison-Wesley Pub., 1974. Print.


June 2, 2011 - Posted by | Analysis | , , , , , ,

1 Comment »

  1. […] This follows immediately from our previous theorem pertaining to […]

    Pingback by Functions of Class C^k « Abstract Nonsense | June 4, 2011 | Reply

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