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I was reading a proof for the multi-variable chain rule and in the proof the mean-value theorem was used. The use of the theorem requires that a function is continuous between two points.

Hence the motivation for the question, if a function is differentiable at a point, is it continuous in some neighborhood? If not, then the multi-variable chain rule proof is fake?

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  • $\begingroup$ Actually the mean-value theorem requires also that the function is differentiable, which is shown to not always be true in a different question. So is the proof of the multi-variable chain-rule wrong? $\endgroup$ – Tony Jul 10 at 20:17
  • $\begingroup$ in the chain rule all the functions are differentiable at any point, and thus, they are also continuous at every point $\endgroup$ – Exodd Jul 10 at 20:25
  • $\begingroup$ @Exodd Not according to this (answer by Ben) math.stackexchange.com/questions/1566376/… $\endgroup$ – Tony Jul 10 at 20:28
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    $\begingroup$ yeah Stefan appears to be assuming differentiability in a neighbourhood of the point, so it's not a fake proof, but it makes stronger assumptions than necessary $\endgroup$ – shibai Jul 10 at 20:37
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    $\begingroup$ @Tony Ben's result is the "stronger" version of the classic chain rule, meaning that it is more difficult to prove. Stefan proof is not sufficient and needs to be changed a lot $\endgroup$ – Exodd Jul 10 at 20:45
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To answer the question in the title: a function that is differentiable at a point need not be continuous in a neighbourhood. For example, consider the function $$ f(x) = \begin{cases} x^2, & x\in\mathbb Q \\ 0, & x\notin\mathbb Q\end{cases} $$ then $\frac{\mathrm{d}f}{\mathrm{d}x}(0)=0$, but $\lim_{x\to a}f(x)$ cannot exist for $a\neq0$ because $\lim_{\substack{x\to a\\x\in\mathbb Q}}f(x)=\lim_{x\to a}x^2=a^2$ whereas $\lim_{\substack{x\to a \\ x\notin\mathbb Q}}f(x)=0$.

This doesn't mean the multivariable chain rule is fake, however, though perhaps the proof you were reading made stronger assumptions than just being differentiable at the point of interest?

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It usually isn't. Consider the function $f : [0,1[ \rightarrow \mathbb R$ defined by

$$ f(x) = \frac{1}{n} \quad \text{if} \quad x \in \left[\frac{1}{n+1},\frac{1}{n} \right[,n \in \mathbb N^* $$ $$ f(0) = 0 $$

It is surely continuous at $x = 0$ and you can easily check that it is differentiable at $0$ with derivative $1$. And yet there is a discontinuity in any neighborhood of $0$.

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No, differentiability in a point only implies continuity in that specific point. A counterexample can already be found in one dimension:

$$f:\mathbb R\to\mathbb R,\quad f(x)=\begin{cases}x^2&x\in\mathbb Q\\0&\textrm{otherwise}\end{cases}$$

This function is differentiable in $0$, but not continuous anywhere other than $0$.

There is also a proof of the multi-variable chain rule which does not depend on the mean value theorem. A function $f:U\to\mathbb R^m$ with $U\subseteq \mathbb R^n$ open is differentiable in $x_0\in U$ iff there exists a linear map $\mathrm Df(x_0):\mathbb R^n\to\mathbb R^m$ and a function $R_f:U\to\mathbb R^m$ with $\lim_{x\to x_0}\frac{R_f(x)}{\Vert x-x_0\Vert}=0$ such that

$$f(x)=f(x_0)+\mathrm Df(x_0)(x-x_0)+R_f(x).$$

The same goes for a function $g:V\to\mathbb R^k$ with $U\subseteq\mathbb R^m$ open and $y_0\in V$. So if $g$ is differentiable in $y_0:=f(x_0)$, then

$$g(y)=g(y_0)+\mathrm Dg(y_0)(y-y_0)+R_g(y).$$

Insert $y_0=f(x_0)$ and $y=f(x)=f(x_0)+\mathrm Df(x_0)(x-x_0)+R_f(x)$ to get

$$\begin{align*}g\circ f(x)&=g(f(x_0))+\mathrm Dg(f(x_0))[f(x_0)+\mathrm Df(x_0)(x-x_0)+R_f(x)-f(x_0)]+R_g(y)\\ &=g(f(x_0))+\mathrm Dg(f(x_0))[\mathrm Df(x_0)(x-x_0)+R_f(x_0)]+R_g(x_0)\\ &=g(f(x_0))+\underbrace{\mathrm Dg(f(x_0))\mathrm Df(x_0)}_{=\mathrm D(g\circ f)(x_0)}(x-x_0)~+~\underbrace{\mathrm Dg(f(x_0))R_f(x_0)+R_g(y_0)}_{=:R_{g\circ f}} \end{align*}$$

It is now straightforward to show that $R_{g\circ f}$ has the required property of

$$\lim_{x\to x_0}\frac{R_{g\circ f}(x)}{\Vert x-x_0\Vert}=0,$$

which makes

$$\mathrm D(g\circ f)(x_0)=\mathrm Dg(f(x_0))\mathrm Df(x_0).$$

And that's exactly the multi-variable chain rule.

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Define $f(1/n) =0, n=1,2,\dots $ and define $f(x)=x^2$ everywhere else. Then $f'(0)=0.$ But in any neighborhood of $0,$ $f$ has infinitely many points of discontinuity–namely at the points $1/n$ that fit into that neighborhood.

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