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I have seen two definitions of differentiability of a real valued function and I wonder why they are equivalent.

The first definition: For a function $\mathbf{F}:\mathbb R^n\to \mathbb R$ is differentiable at $\mathbf x$, if there exist a matrix $\mathbf{DF}(\mathbf x)$ such that $$\lim_{\Delta\mathbf x\to\mathbf0}\frac{\mathbf F(\mathbf x+\Delta\mathbf x)-\mathbf F(\mathbf x)-\mathbf{DF}(\mathbf x)\Delta\mathbf x}{||\Delta\mathbf x||}=0$$

It turns out that $\mathbf{DF}=\nabla \mathbf F$

So if $\mathbf F$ is differentiable, $$\lim_{\Delta\mathbf x\to\mathbf0}\frac{\Delta\mathbf F-\nabla \mathbf F\cdot\Delta\mathbf x}{\Delta \mathbf x}=0$$

The second definition: If $\mathbf F$ is differentiable, $$\Delta\mathbf F=\nabla\mathbf F\cdot\Delta\mathbf x+\epsilon\cdot\Delta\mathbf x$$ where $\epsilon=(\epsilon_1,\epsilon_2,...,\epsilon_n)$ and $\epsilon\to\mathbf0$ as $\Delta \mathbf x\to \mathbf0$

I know how to prove 2 implies 1 by Cauchy Schwarz inequality ($\epsilon\cdot\Delta\mathbf x\le||\epsilon||||\Delta\mathbf x||$) but I don't know how to prove 1 implies 2

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Y.H. Chan
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  • There seems to be something wrong in the second definition. On the right hand side you sum the vector $\nabla\mathbf F\cdot\Delta\mathbf x$ and what I infer from your comments to be a scalar $\varepsilon\cdot\Delta\mathbf x$. That makes no sense ... – String Aug 05 '14 at 10:48
  • @String $\epsilon$ is a vector as well (I can't bold it) and bot right and left hand sides are scaler – Y.H. Chan Aug 05 '14 at 12:31
  • Ah, right you are! I just got pre-occupied with the statement about $\mathbf{DF}(\mathbf x)$ being a matrix. You are talking about a multivariate real valued function $\mathbf{F}$ and not a vector valued one, so should not $\mathbf{DF}(\mathbf x)$ be a vector rather than a matrix? Then it starts to make sense, I think. – String Aug 05 '14 at 12:38
  • @String ya, actually I got this question when I was reading https://proofwiki.org/wiki/Characterization_of_Differentiability – Y.H. Chan Aug 05 '14 at 13:39
  • Sorry about the confusion! I gave an answer to the question know, I hope it is satisfactory ... – String Aug 05 '14 at 19:44

1 Answers1

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For the other direction, define a function $k:\mathbb R^n\setminus\{\mathbf0\}\rightarrow\mathbb R$ by $$ k=k(\Delta\mathbf x)=\frac{\Delta\mathbf F-\nabla \mathbf F\cdot\Delta\mathbf x}{||\Delta \mathbf x||} $$ then $\Delta\mathbf F-\nabla \mathbf F\cdot\Delta\mathbf x=k\cdot||\Delta\mathbf x||$. Since $\Delta\mathbf x\neq\mathbf 0$ we have $||\Delta\mathbf x||\neq 0$. Thus we may define $$ \mathbf\epsilon=k\cdot\frac{\Delta\mathbf x}{||\Delta\mathbf x||} $$ in order to have $\mathbf\epsilon\cdot\Delta\mathbf x=k\cdot||\Delta\mathbf x||=\Delta\mathbf F-\nabla \mathbf F\cdot\Delta\mathbf x$. Finally, since we have assumed 1 to hold we know that $k\longrightarrow 0$ for $\Delta\mathbf x\longrightarrow \mathbf0$ and since the normed vector $\dfrac{\Delta\mathbf x}{||\Delta\mathbf x||}$ is bound to be on the unit sphere having length $1$ it follows that $\mathbf\epsilon=k\cdot\dfrac{\Delta\mathbf x}{||\Delta\mathbf x||}\longrightarrow\mathbf 0$. This proves that $\mathbf{(1)}\implies\mathbf{(2)}$.

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