# 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?

• Er wait I'm not asking that why if the partials exist, the derivative exists...or maybe I'm misunderstanding your comment? Sep 6, 2012 at 1:49
• Lol k no worries Sep 6, 2012 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. Sep 6, 2012 at 1:54
• 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 $\mathbb R^m$. Sep 6, 2012 at 7:15
• Dear uncookedfalcon, I have checked Spivak's booklet and I think there is a misunderstanding on your part.The theorem you are asking about is Theorem 2-3 (3) on page 20. It is proved (without any continuity assumption) on the next page and the proof is essentially your one-liner, with the caveat I mentioned in my preceding comment. So Spivak is 100% correct, and optimal as to the hypotheses. His Theorem 2.9 that you mention is about the differentiability of the composition of maps and is an entirely different problem. Sep 6, 2012 at 7:35

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!

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

• 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? Sep 6, 2012 at 1:57
• The answer is no. Sep 6, 2012 at 2:04
• lmao yeah I agree, but again: not my question :p Sep 6, 2012 at 2:07
• This is indeed not an answer to the question. Sep 6, 2012 at 6:19
• -1: This is the answer to another question. Sep 6, 2012 at 22:22