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So, here's what I'm working with. I'm trying to prove that, given some function $f: \mathbb{R}^n \to \mathbb{R}, a \in \mathbb{R}^n$ such that f is differentiable at a, and some open ball $B = Ball(a, \delta)$ where $f(a) \geq f(x)$ for all $x \in B$, $Df(a) = 0$

I've been trying to work off the definition of differentiability- that is, $\lim_{h\to 0}\frac{f(a+h) - f(a) - Df(a)h}{\|h\|} = 0$

I've been trying to prove it by contradiction. After chipping at it some more, I get to this $\lim_{h\to 0} f(a+h)-f(a) - Df (a) = 0$.

After applying the epsilon-delta definition, I manipulate it into the following: $\|h\| < \delta_1 \rightarrow |f(a+h) - f(a) - Df(a)h| < \epsilon$

Unpacking the absolute value and adding $Df(a)h$ to both sides, I get $Df(a)h - \epsilon < f(a + h) - f(a) < Df(a) + \epsilon$

From here i'm pretty sure that I want to establish that the lower bound is (or at least can be) positive, but I am unsure of how to do that. My knowledge of epsilon-delta limits is a bit shaky, and I'm not exactly sure what I'm allowed to "choose" in this situation.

Could I have some help wrapping this proof up?

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    Show all partial derivatives are $0$ at $a;$ this follows from the one variable theory. Since $Df(a)$ is given by the dot product with $ \nabla f (a), Df(a) = 0.$ – 2017-01-29
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    @zfw the dot product of what with what? – 2017-01-29
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    $Df(a)(v) = \langle \nabla f (a), v\rangle.$ – 2017-01-29
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    @zfh Ok I understand that, but I don't see how it plays into the proof. Am I going in the wrong direction with my epsilon-delta approach? – 2017-01-29
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    You're trying to reinvent the wheel. If all partial derivatives are $0$ at $a,$ then $\nabla f(a)$ is the $0$ vector, hence $Df(a)=0$ – 2017-01-29

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