# If $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ has coordinates $f^1 \ldots f^m$, and each $f^i$ is differentiable at $0$…

...then does it follow that $f$ is differentiable at 0?

My motivation for asking this is as follows: in Spivak's Calculus on Manifolds, in theorem 2.9, he uses this with the additional condition that each $f^i$ is continuously differentiable in a nbh of 0, to conclude that $f$ is differentiable and I don't think is necessary.

Namely, if each $f^i$ has derivative $Df^i$, I claim the matrix with $i^{th}$ row $Df^i$ will serve as $Df$. Indeed, let $v_j$ be a sequence tending to 0 in $\mathbb{R}^n$, we have (by the triangle inequality, if you wish)$$\frac{| f(v_j) - f(0) - \sum_i Df^i(v) |}{|v_j|} \leqslant \frac{\sum_i |f^i(v_j) - f^i(0) - Df^i(v_j)|}{|v_j|}$$Taking the limit as $j \rightarrow \infty$, each summand goes to 0 by the differentiability of $Df^i$ (and there's only $m$ of them), hence the limit is 0.

Is this wrong? Thanks in advance!

EDIT: btw, conditional on the above proof being right and/or the claim being right, does anyone know maybe what Spivak was going for?

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Er wait I'm not asking that why if the partials exist, the derivative exists...or maybe I'm misunderstanding your comment? – uncookedfalcon Sep 6 '12 at 1:49
Lol k no worries – uncookedfalcon Sep 6 '12 at 1:50
I think someone else has made the same mistake as I. At any rate, looking at the proof I think he's just stating something that is locally superfluous. Theorem 2–3 clearly confirms that differentiability can be checked for each component separately. – Dylan Moreland Sep 6 '12 at 1:54
I think so as well :p. It's a super reasonable mistake to make. – uncookedfalcon Sep 6 '12 at 1:56
Dear uncooked falcon, the idea of your calculation seems correct but your notation is not. You must write $Df^i(0)(v_j)$ on the right hand side . Similarly the sum $\sum_i Df^i(v)$ does not make sense: you want to replace it with a suitable vector in $\mathbb R^m$. – Georges Elencwajg Sep 6 '12 at 7:15

In case this is useful to anyone else, let me record the comments of Georges Elencwajg and Dylan Moreland-the answer is yes it's true, and the condition in Spivak is superfluous.

The proof is the one liner I wrote above, albeit with better notation: If $\lambda^i$ are the derivatives of the $f^i$ at 0, I claim the matrix with $i^{th}$ row $\lambda^i$ will serve as $Df(0)$. Indeed we have $$\frac{|f(v) - f(0) - \lambda^i(v)e_i|}{|v|} \leqslant \sum_i \frac{|f^i(v) - f^i(0) - \lambda^i(v)|}{|v|}$$Taking any $v_i \rightarrow 0$, applying the above inequality, and using the differentiability of each $f^i$ at 0 gives the result.

Thanks all for the assistance!

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The "continuously differentiable" condition for the partial derivatives is essential to showing that $f$ is differentiable $0$. This is the canonical counterexample $$f(x,y) = \left\{ \begin{array}{lr} \frac{x^2y}{x^2+y^2} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } x = (0,0) \end{array} \right.$$

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Hey Shankara, thanks for the answer, but I'm not sure your answering my question. I think you answered: if all the partials exist, then is $f$ differentiable? – uncookedfalcon Sep 6 '12 at 1:57
The answer is no. – Integral Sep 6 '12 at 2:04
lmao yeah I agree, but again: not my question :p – uncookedfalcon Sep 6 '12 at 2:07
This is indeed not an answer to the question. – Georges Elencwajg Sep 6 '12 at 6:19
-1: This is the answer to another question. – Giuseppe Negro Sep 6 '12 at 22:22