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Let $U \subseteq \mathbb R^n$ and $F: U \to \mathbb R^m$ a function with coordinate functions $f_i$. My notes say that:

If $F$ is differentiable on $U$ the Jacobian of $F$ is defined at each point in $U$, its $nm$ entries are functions on $U$. These functions need not be continuous on $U$; they are continuous on $U$ if and only if $F$ is of class $C^1$.

My problem is with the only if part of last part of the statement: it should be $C^k$ with $k \ge 1$, right? (I'm merely asking because there is this nagging doubt I have that it might not be a typo after all.)

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If a function $f$ is $C^k$ for $k\geq 1$ then it automatically is $C^1$, too. So the statement is valid. –  Gregor Bruns Jul 29 '13 at 7:20
    
@user84127 Beware the term "differentiable". Some texts intend "differentiable" to mean smooth or $C^{\infty}$ to be precise. Other texts merely insist that the differential at each point exist. The criteria that these differentials continuously link together gives $C^1$. –  James S. Cook Jul 29 '13 at 7:23
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You should note that the affirmative: "if the entries of the Jacobian are continuous then $F$ is of class $\mathcal{C}^1$" is only true when $U$ is a convex open set. –  Marra Jul 29 '13 at 7:23
    
@GustavoMarra I did not know that. This would provide an answer to my question. Could you post it and maybe add a proof or a reference to where I can read a proof? –  Michael de Santa Jul 29 '13 at 7:28
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Dude, forget everything I said, it's almost 5AM and I've been studying all night. The result I said is correct if $f$ is convex (not $U$! Definition: en.wikipedia.org/wiki/Convex_function#Definition )and $m=1$ (that is, $f:U\subset \mathbb{R}^n\rightarrow\mathbb{R}$). I'm terribly sorry. –  Marra Jul 29 '13 at 7:44

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up vote 5 down vote accepted

No, it's not a typo. To say $F$ is $C^1$ on $U$ is to say that it is continuously differentiable on $U$. This means the mapping $p \mapsto dF_p$ is continuous as a mapping from $U$ to the space of operators on smooth functions at $p$. It is a very nice result that this continuity of operators reduces to the much easier to understand criteria of continuity of the component functions of the Jacobian.


Again, I think you are misinterpreting the term "differentiable" to mean that derivatives of all orders exist. Of course, if it was the case that $f \in C^k$ for $k>1$ then $f \in C^2$ implies $f' \in C^1$. But, I don't believe that is the intended use of the term "differentiable". Consider the example below: for $x \neq 0$ define $$ f(x) = x^2\sin(1/x) $$ and for $x=0$ define $f(0)=0$. Differentiate for $x \neq 0$ and obtain, [ f'(x) = 2x\sin(1/x)-\cos(1/x) ] thus clearly $f'$ is not continuous at $x=0$. However, $f'(0)$ does exist and it can be shown to be zero by careful arguments from the definition of the derivative. Therefore, $f$ is differentiable at $x=0$ however, $f \notin C^1(0)$. Here's a picture I show my calculus I class to show how this happens geometrically but it's for a slightly different example $f(x) = \frac{x}{2}+x^2\sin(1/x)$ to give slope $1/2$ instead of zero at $(0,0)$


Incidentally, your suffering this day is precisely the reason I think we should use the term smooth instead of "differentiable" to describe a function which supports arbitrarily many continuous derivatives. But, minds much greater than mine disagree, so the suffering must continue.

enter image description here

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But if $F$ is $C^2$ then isn't the mapping $p \to dF_p$ also continuous? –  Michael de Santa Jul 29 '13 at 7:25
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Yes, it is, and also it's derivative is continuous. –  Marra Jul 29 '13 at 7:26
    
@GustavoMarra But then it would be a typo since only the if part of the statement would be true. Wouldn't it? –  Michael de Santa Jul 29 '13 at 7:27
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@user84127: You seem to be under the impression that a function in $C^2$ is not in $C^1$. This is false. –  Chris Eagle Jul 29 '13 at 7:39
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@user84127 note that $f$ differentiable does NOT imply $f \in C^k$ for $k>1$. It does imply $f$ is in $C^0$ for the domain in question, but that is not the same as supposing $f' \in C^0$. In other words, existence of the derivatives alone does not imply the continuity of the derivatives. I'll add an example to emphasize the distinction. –  James S. Cook Jul 29 '13 at 16:07

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