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I want to find the derivative of the function $f:\mathbb R^n\to \mathbb R^m$ at a point $x_0\in \mathbb R^n$, where $f(x)=c\in \mathbb R^m$, is a constant function. What I did is as follows:

If $f$ is differentiable at $x_0$, then there exists a linear function $L_{x_{0}}:\mathbb R^n\to \mathbb R^m$ such that $\lim\limits_{\parallel h\parallel \to 0}\frac{\parallel f(x_0+h)-f(x_0)-L_{x_0}(h)\parallel }{\parallel h\parallel }=0$.

This gives $\lim\limits_{\parallel h\parallel \to 0}\frac{\parallel L_{x_0}(h)\parallel }{\parallel h\parallel }=0$. Now how to show that $L_{x_0}(h)=0?$ Please help!

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    $\begingroup$ Use the linearity of $L_{x_0}(h)=ah+b$ and show that $a=b=0$ must be true for that limit to hold. $\endgroup$
    – lemon
    Commented Jul 27, 2014 at 15:29
  • $\begingroup$ Can I use the fact that there exists $M>0$ such that $\parallel L_{x_0}(h)\parallel\leq M\parallel h\parallel $? $\endgroup$
    – Anupam
    Commented Jul 27, 2014 at 15:46
  • $\begingroup$ I personally don't see how that would help. Why not do $\lim \frac{\|ah+b\|}{\|h\|}\leq\lim \frac{|a|\|h\|}{\|h\|}+\frac{|b|}{\|h\|}$? $\endgroup$
    – lemon
    Commented Jul 27, 2014 at 15:52
  • $\begingroup$ $0\leq \vert a\vert+\lim \frac{\vert b\vert}{\parallel h\parallel}$...then? $\endgroup$
    – Anupam
    Commented Jul 27, 2014 at 15:59
  • $\begingroup$ How about using the linearity of $L$ to write $lim_{|h|->0} L_{x_{0}}({h \over |h|})=0 $.Since $h$ is arbitrary this shows that $L_{x_0} = 0$ on the unit sphere. I'm not sure if that's correct. Please do correct me if something is wrong. $\endgroup$
    – Srinivas K
    Commented Jul 27, 2014 at 16:30

1 Answer 1

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Just check that $L_{x_0} \equiv 0 $ verifies the limit. We have: $$\lim_{h \to 0} \frac{\|f(x_0 + h) - f(x_0)\|}{\|h\|} = \lim_{h \to 0} \frac{\|c - c\|}{\|h\|} = 0$$

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  • $\begingroup$ That means I have to guess the derivative? $\endgroup$
    – Anupam
    Commented Jul 27, 2014 at 16:00
  • $\begingroup$ You can use some intuition like in basic calculus. If $f$ is constant, its derivative is zero. If $f$ is linear, its derivative is itself. You have some rules like $\mathrm{d}(fg) = g~\mathrm{d}f + f~\mathrm{d}g$, etc. $\endgroup$
    – Ivo Terek
    Commented Jul 27, 2014 at 16:03
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    $\begingroup$ +1. @Anupam: all partial derivatives $\frac{\partial f_i}{\partial x_j}$ exist and are continuous functions at each $x\in \mathbb R^n$ (they are all equal 0, in particular). So the Jacobian matrix of $f$ at any $x$ is equal to the zero matrix. Use this fact to justify the choice in Ivo Terek's answer and you are done. $\endgroup$
    – Avitus
    Commented Jul 27, 2014 at 17:58
  • $\begingroup$ Remember a bit of linear algebra. The total derivative is a linear map. What is its matrix in the standard bases of $\Bbb R^n$ and $\Bbb R^m$? $\endgroup$
    – Ivo Terek
    Commented Jul 27, 2014 at 18:03

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