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This a proof from a book:

Theorem: let $f:\Bbb R^p\to\Bbb R$ with continuous partial derivatives at $a$. Then $f$ is differentiable at $a$.

Proof: without lose of generality suppose that $p=2$ and let $a:=(b,c)$ we have $$f(b+h,c+k)-f(b,c)=f(b+h,c+k)-f(b+h,c)+f(b+h,c)-f(b,c)\tag1$$ On one hand we have $$f(b+h,c)-f(b,c)=\partial_1 f(b,c)h+o(h)\tag2$$ On the other hand we can apply the mean value theorem to the function $t\mapsto f(b+h,t)$, hence $$f(b+h,c+k)-f(b+h,c)=\partial_2 f(b+h,c+\theta k)k,\quad\text{for some }\theta\in(0,1)\tag3$$ However, because of the continuity at $a$, we have that $$\partial_2 f(b+h,c+\theta k)=\partial_2 f(b,c)+o(h,k)\tag4$$ $\Box$

I cant see where $(4)$ comes and it seems wrong. If it would be true then we will had that

$$\lim_{(h,k)\to(0,0)}\frac{\partial_2 f(b+h,c+\theta k)-\partial_2 f(b,c)}{\|(h,k)\|}=0$$

what means that $\partial\partial_2 f(b,c)=0$, what doesnt make sense.

QUESTION: it is a typo in the proof or Im wrong in some place? In any case, can you enlighten the attempted proof?

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What the author means by the notation $o(h,k)$ (which is not the same thing as $o(\|(h,k)\|$) is

$$\lim_{(h,k) \to (0,0)}\partial_2 f(b+h,c+\theta k)=\partial_2 f(b,c).$$

This is true since the partial derivatives are continuous at $a = (b,c)$ by hypothesis, and facilitates proving the desired result that $f$ is differentiable at $a$:

$$f(b+h,c+k) - f(b,c) - \underbrace{[\partial_1f(b,c)h+ \partial_2f(b,c)k]}_{\text{linear operator acting on }(h,k)} = o(\|(h,k\|)$$

Had this been written as

$$\partial_2 f(b+h,c+\theta k)=\partial_2 f(b,c) +o(\|(h,k)\|)$$

it would mean

$$\lim_{(h,k) \to (0,0)}\frac{\partial_2 f(b+h,c+\theta k)-\partial_2 f(b,c)- \mathcal{O\cdot(h,k)}}{\sqrt{h^2+k^2}} = 0,$$

where $\mathcal{O}$ is the null (or zero) linear operator and you would be correct saying that

$$\partial\partial_2f(b,c) = 0.$$

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  • $\begingroup$ but the, supposed, meaning of $o(h,k)$ doesnt correspond with the definition of little o. The definition of little o is $$f(x)\in o(g(x))\iff |f(x)|\le\epsilon |g(x)|,\forall\epsilon>0$$ Then use $o(h,k)$ or $o(\|h,k\|)$ in this context is the same, in any case it means that $\partial\partial_2 f(b,c)=0$, what doesnt make sense. $\endgroup$ – Masacroso Feb 27 '18 at 19:05
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    $\begingroup$ Yes -- the author is abusing the little o notation. He should not use $o(h,k)$. I'm just explaining what he meant. The argument of little o needs to be clearly expressed in terms of a function. In proofs like this, I would prefer to see everything written more clearly in terms of limits or $\epsilon - \delta$ arguments. $\endgroup$ – RRL Feb 27 '18 at 19:09

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