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...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?

Michael Hardy
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uncookedfalcon
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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 06 '12 at 01:49
  • Lol k no worries – uncookedfalcon Sep 06 '12 at 01:50
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    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 06 '12 at 01:54
  • I think so as well :p. It's a super reasonable mistake to make. – uncookedfalcon Sep 06 '12 at 01:56
  • I remember thinking that this section was strangely written, and perhaps this is why. It is not apparent how the $g_i$ being continuously differentiable is being used, if it is being used at all. He only seems to want to apply the chain rule, which doesn't need continuous partials. – Dylan Moreland Sep 06 '12 at 01:59
  • Yeah man I totally agree, hence my question :p. My 1-liner is actually a proof yeah? – uncookedfalcon Sep 06 '12 at 02:03
  • Yes, if each $f^i$ is differentiable at $0$, then $Df(0)$ exists and is the matrix with $Df^i(0)$ as its lines . Continuity of the differential is irrelevant and actually does not even make sense because there is no reason that $Df^i(x)$ should be defined at points $x \neq0$. – Georges Elencwajg Sep 06 '12 at 06:31
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    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 06 '12 at 07:15
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    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. – Georges Elencwajg Sep 06 '12 at 07:35
  • Dear Georges thank you for the comments. You're absolutely right that I was careless with my notation, by $Df^i$ I mean what you call $Df^i(0)$, and by $\sum_i Df^i(v)$ I mean $\sum_i Df^i(v) e^i$, where $e^i$ are the standard basis for $\mathbb{R}^m$. – uncookedfalcon Sep 06 '12 at 19:42
  • (argh I mean $\sum_i Df^i(0)(v)e^i$) Also, as you say, Theorem 2-3(3) is exactly what I wanted, and answers my question. In theorem 2.9, he writes, ``since $g_i$ is continuously differentiable at $a$, it follows from Theorem 2.8 that $g$ is differentiable at $a$". This was the motivation for my question, and I guess the continuous condition is superfluous. If you care to write anything in the answers section I'd be more than happy to accept it. – uncookedfalcon Sep 06 '12 at 19:49
  • Dear uncookedfalcon, thank you for this kind offer but since I just gave a reference, I don't think it is appropriate to upgrade it to an answer. Anyway, I am happy that all is now perfectly clear to you: only that really matters. And good luck with the rest of the book! – Georges Elencwajg Sep 06 '12 at 21:39
  • Aww thanks man :D ! – uncookedfalcon Sep 06 '12 at 21:55

2 Answers2

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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!

uncookedfalcon
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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 $$f(x,y) = \left\{ \begin{array}{lr} \frac{x^2y}{x^2+y^2} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } x = (0,0) \end{array} \right.$$

Shankara Pailoor
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