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I've found an answer of the question in Proof that the derivative is unique?, but I don't quite understand the answer, how the proof uses triangle inequality? And how it use $\alpha y \rightarrow 0$ and linearity of $\sigma_{1}'-\sigma_{2}'$ to get the equation below? Thanks very much!

  • It doesn't as far as I can see. It is just subtraction. Choose $y = \alpha (\sigma_1'-\sigma_2')^T$ to get the desired result. (Also, some conditions on $\Sigma$ are needed to make sure we have uniqueness.) – copper.hat Oct 07 '18 at 21:00

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$$\|\frac{(\sigma_2'-\sigma_1')(y)}{y}\| \leq \frac{\|(\sigma_2'-\sigma_1')(y)\|}{\|y\|} \leq \frac{\|\sigma(x+y)-\sigma(y)-\sigma_1'(y)\|}{\|y \|}+ \frac{\|\sigma(x+y)-\sigma(y)-\sigma_2'(y) \|}{\| y\|}$$

Since norms are non-negative, the limit when $\|y\|$ goes to $0$ is equal to $0$. As for the second question, linearity means we can pull out $\alpha$ from both numerator and denominator, so they cancel.

user667
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  • Thanks! It's maybe silly, why the limit when ‖y‖ goes to 0 is equal to 0? And why we need the step with $\alpha$? –  Oct 07 '18 at 21:13
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    By assumption, we have that both $ \frac{|\sigma(x+y)-\sigma(y)-\sigma_1'(y)|}{|y |}$ and $ \frac{|\sigma(x+y)-\sigma(y)-\sigma_2'(y) |}{| y|}$ go to zero as $|y|$ go to zero. So that the limit of the sum also goes to zero. – user667 Oct 07 '18 at 21:19
  • The step with $\alpha$ is more of a trick so that we can end up with the result we want- equality. Note that $\alpha y $ goes to zero satisfies the condition that $| y|$ goes to zero. – user667 Oct 07 '18 at 21:23
  • Thanks again! Just still a little confused about $\alpha$, isn't the inequalities already proved the limit equals to 0 which implies the two derivative are equal. Why need to proved with $\alpha y$ again? –  Oct 07 '18 at 21:29
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    No, the limit as $|y|$ goes to zero does not mean that the two are equal. This just means that the two are essentially equivalent close to the origin. We need to prove that the two are equal for any non-zero vector $y$. Hence is why we introduce the scalar $\alpha$ and use the linearity of $\sigma_1'$ and $\sigma_2'$. – user667 Oct 07 '18 at 21:34
  • Note that the in last display of equations the rightmost expression does not have a limit involved. – user667 Oct 07 '18 at 21:35
  • Just still a little confusion that why the rightmost expression does not have a limit involved? –  Oct 09 '18 at 05:50
  • By linearity of $\sigma_1'$ and $\sigma_2'$ we get $\lim_{\alpha\rightarrow 0}\frac{|(\sigma_{1}'-\sigma_{2}')(\alpha y)|}{|\alpha y|} = \lim_{\alpha\rightarrow 0}\frac{|\alpha||(\sigma_{1}'-\sigma_{2}')(y)|}{|\alpha||y|}$. $|\alpha|$ cancels out and we get the desired expression. – user667 Oct 09 '18 at 19:20