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Given a subset $\Omega$ of $\mathbb{R}^n$ and a function $\sigma: \Omega \to \mathbb{R}^n$, we define its derivative at $x$ to be a linear operator $\sigma'(x)$ such that $$ \lim_{y \to 0} \frac{\|\sigma(x + y) - \sigma(x) - \sigma'(x)y\|}{\|y\|} = 0 $$

If this operator exists (i.e. if $\sigma$ is differentiable at $x$) then how do we prove that $\sigma'(x)$ is unique?

Sine of the Time
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Ayesha
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    Do you know the proof for $\mathbb{R}$? It's exactly the same. But note that for some $f$ defined in $U$ and $x$ not an interior point, then the derivative is NOT unique at $x$. – user40276 Dec 18 '13 at 02:42
  • Is $U$ a subset of $\mathbb{R}^n$? – Ayesha Dec 18 '13 at 02:58
  • Yes, it's a subset of the euclidean space. – user40276 Dec 18 '13 at 03:12
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    Not sure if its a good idea to put a comment on such an old question! Is the definition even correct? Shouldn't the middle term be $f(x)$, instead of $f(y)$? – Mike V.D.C. Mar 13 '19 at 17:17
  • Yes, it should be $f(x),$ and Disintegrating By Parts keeps reverting edits, so it perpetually stays incorrect in question and their solution. – sunspots Feb 24 '23 at 00:30

1 Answers1

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Suppose there are linear operators $\sigma_{1}'$ and $\sigma_{2}'$ such that $$ \lim_{\|y\|\rightarrow 0}\frac{\|\sigma(x+y)-\sigma(x)-\sigma_{j}'(x)y\|}{\|y\|} =0, \\\;\;\; j =1,2. $$ Because $\sigma_{2}'(x)$ and $\sigma_{1}'(x)$ are linear, $$ (\sigma_{2}'(x)-\sigma_{1}'(x))(y)= \\ = \{\sigma(x+y)-\sigma(x)-\sigma_{1}'(x)y\}\\ -\{\sigma(x+y)-\sigma(x)-\sigma_{2}'(x)y\} $$ Applying the triangle inequality gives $$ \lim_{\|y\|\rightarrow 0}\frac{\|(\sigma_{1}'-\sigma_{2}')(y)\|}{\|y\|}=0 $$ In particular, for any non-zero vector $y$, one has $\alpha y \rightarrow 0$ as the scalar $\alpha$ tends to 0. Therefore, by linearity of $\sigma_{1}'-\sigma_{2}'$, $$ 0= \lim_{\alpha\rightarrow 0}\frac{\|(\sigma_{1}'-\sigma_{2}')(\alpha y)\|}{\|\alpha y\|} \\= \lim_{\alpha\rightarrow 0}\frac{\|(\sigma_{1}'-\sigma_{2}')(y)\|}{\|y\|} \\= \frac{\|(\sigma_{1}'-\sigma_{2}')(y)\|}{\|y\|}. $$ So $\|(\sigma_{2}'-\sigma_{1}')(y)\|=0$ for all non-zero vectors $y$, which implies that $\sigma_{2}'=\sigma_{1}'$.

Disintegrating By Parts
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  • Isn't the numerator usually defined to be $f(x+y) - f(x) - Df(x)(x+y-x) = f(x+y) - f(x) - Df(x)(y)?$ Then, $f$ is the $\sigma$ and $Df(x)$ is the $\sigma^{'}.$ – sunspots Feb 22 '23 at 20:30
  • Also, you later have $\sigma^{'}$ take $y$ as input, so the edit reversion doesn't make sense. I only added the grammar part since I need 6 character changes to submit the edit. – sunspots Feb 22 '23 at 20:38
  • Not sure about why this is deleted? The proof seems to be correct. – Arctic Char Mar 01 '23 at 01:14
  • @ArcticChar : Sorry about that. Trying to deal with someone's edits on a question and answer that are ten years old ended up in a situation where I was lost in the paper trail of edits, and I didn't know how to roll it back. So I deleted the answer, hoping that would delete the answer acceptance and points as well. Unfortunately, it did not. If someone else wants to post their own, they're free to do so now. Lately I've been getting several people wanting to edit accepted answers. I don't quite understand this trend. – Disintegrating By Parts Mar 01 '23 at 01:37
  • Sometimes it's edits of minor typos that were somehow missed for a long time, sometimes minor grammar issues. If you like I'll be happy to check whether that's the case with the two recent edits at issue. – Lee Mosher Mar 01 '23 at 01:55
  • @LeeMosher : Thank you for your offer. I'd say that the answer should be deleted. Then the person who is suggesting edits can post a correct solution. – Disintegrating By Parts Mar 01 '23 at 03:12
  • Okay, I'm going to bring this one up on meta, but before I do that I want to run this by you one more time. The original post, and your original answer, had the same minor error, namely a $y$ which should have been an $x$. This was corrected in the post itself. But rather than correct that error in the answer, your desire is that the answer be deleted. Is that a good summary? – Lee Mosher Mar 01 '23 at 13:34
  • @LeeMosher : If that's all it takes to fix the answer, then that would be good. I should be able to manage the edits then. – Disintegrating By Parts Mar 01 '23 at 15:08
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    Yeah, that's what it looks like to me. From what I can tell, everything would be fine by first rolling back to your own Revision 5, followed by a single additional edit in the first displayed formula, replacing the expression $\sigma'_j(x)$ with $\sigma'_j(y)$. – Lee Mosher Mar 01 '23 at 16:09
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    @LeeMosher : Thank you for your help. I rolled things back, and made a few changes for formatting purposes. I think this should be okay. If you want to check, your help would be greatly appreciated. – Disintegrating By Parts Mar 02 '23 at 13:20