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Consider the following theorem:

Let $E$ be a subset of ${\bf R}^n$, $f:E\to {\bf R}^m$ be a function, $F$ be a subset of $E$, and $x_0$ be an interior point of $F$. If all the partial derivatives $\frac{\partial f}{\partial x_j}$ exist on $F$ and are continuous at $x_0$, then $f$ is differentiable at $x_0$.

It is trivial to show that the converse is NOT true when $m=1$. It seems no hope that it will be true when $m\geq 2$. Here is my question:

Is the converse true when $m\geq 2$? If is not true, how to construct the counterexample?

Edit: The title is corrected.

Edit: Since another question is not relevant to the first one here, I think, I put it into another post.

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$x\sin\frac{1}{x}$ at $0$ – Dactyl Jun 9 '11 at 16:14
@Dactyl that function is not differentiable at zero (no matter how you define the value at 0) – mt_ Jun 9 '11 at 16:23
@mt_ Hi. I know. But that function is so familiar, that its a good hint, and now the OP can construct what ever counter-example he wants and not just for $C^0$ or $C^1$. – Dactyl Jun 9 '11 at 16:32
Trivial. After seeing the solution maybe. – jspecter Jun 11 '11 at 2:39

2 Answers 2

up vote 5 down vote accepted

Re your first question, the converse is false and one most usually mentions counterexamples based on the oscillatory behaviour of $x\mapsto\sin(1/x)$ near $x=0$.

Re your second question, Chern's and Folland's definitions are identical since both require $D^kf$ to be continuous for $f$ to belong to $C^k$, as they should.

Edit Due to a radical modification of the question, the above only partly applies to the current version of the question. Great...

However, one can say this: the case $m\ge2$ cannot yield stronger results, simply consider a function like $(f,0,0,\ldots,0)$ where $f$ is a counterexample when $m=1$.

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No, there are differentiable functions which are not $C^1.$ For example consider $f:\mathbb{R} \rightarrow \mathbb{R}$ given by $f(x) = x^2 \sin(\frac{1}{x})$ for $x \neq 0$ and $f(x) = 0$ if $x = 0.$ Then $f$ is differentiable but not $C^1.$

To see that $f$ is differentiable away from zero is not hard and follows from the product and chain rules. On the other hand to show $f$ is diffentiable at $0$ we must appeal to the definition of the derivative.


$$\displaystyle\lim_{x\rightarrow 0} \frac{f(x) - f(0)}{x-0} = \displaystyle\lim_{x\rightarrow 0} \text{ } x \sin(1/x)$$

which converges to $0$ as $|x \sin(1/x)| < |x|$.

Hence, $f$ is everywhere diffentiable.

It follows $f'(x)$ is given by $2x\sin(\frac{1}{x}) - \cos(\frac{1}{x})$ if $x \neq 0$ and $0$ if $x = 0.$

We claim $f'$ is not continuous at $0.$ To see this, note that if $f'$ were continuous at $0,$ then so too would be the function $g(x)$ defined by $\cos(1/x)$ if $x \neq 0$ and $0$ if $x = 0.$ But this is false as $\langle\frac{1}{\pi + n2\pi}\rangle_{n\in\mathbb{Z}_+}$ is a sequence of real numbers converging to $0$ such that

$$\displaystyle\lim_{n\rightarrow \infty} g\left(\frac{1}{\pi + n 2\pi}\right) = 1 \neq 0 = g(0).$$

Hence, $f$ is differentiable but not $C^1.$

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Please use $\backslash\cos$ and $\backslash\sin$. – Did Jun 10 '11 at 6:26

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