# Abstract Nonsense

## 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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References:

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.