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Let $f \in C^1([0,T]\times S(s))$ where $S(r) \subset \mathbb{R}^n$ is compact for each $r \in [0,T].$

Does it not automatically follow that $f \in C([0,T], C^1(S(s)))$?

In the paper I'm reading, apparently it is true only because $f$ and its first derivative wrt. $x$ are continuous and thus uniformly continuous on the compact set $[0,T] \times S(s)$. Can someone explain this to me please?

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What is $\Gamma$? –  user53153 Dec 18 '12 at 22:47
    
@PavelM Sorry, that was a typo. It's $S(s).$ –  soup Dec 18 '12 at 22:54
    
OK, I'm still trying to understand the notation. Apparently $s$ is a variable that takes values in $[0,T]$, and for each such value we have a compact set $S(s)$? If so, then $[0,T]\times S(s)$ is not really a product but can be understood as $\{(s,x)\colon 0\le x\le T, x\in S(s)\}$. (Correct me if this is wrong). Next question would be: how do we interpret $C^1$ on a compact set (which could be something like a Cantor set)? –  user53153 Dec 18 '12 at 23:02
    
@PavelM Sorry for late reply. In fact, just set $S(s) = S$. I should not have put in the extra detail as it is not required. So just say $[0,T]\times S$ for a compact surface $S$. As for your last question, why is that a problem for compact sets? I am unsure of why that's an issue? –  soup Dec 22 '12 at 21:44
    
Perhaps it's not. I don't really know what is the issue here, anyway. You ask if something follows from $f\in C^1(product)$. The paper says that something follows from $f$ with its derivatives being continuous on the product. Potato, potato. –  user53153 Dec 22 '12 at 21:51

1 Answer 1

up vote 2 down vote accepted

If $S$ is not a compact set, then $f\in C^1([0,T]\times S)$ does not imply $f\in C([0,T],C^1(S))$ in general. For example, let $S=(0,1)$ and $f(t,x)=x^2\sin \frac{t}{x}$. Then $f_t$ and $f_x$ are bounded and continuous on $[0,T]\times S$, hence $f\in C^1([0,T]\times S)$. However, $\|f(t_1,\cdot)-f(t_2,\cdot)\|_{C^1(S)}$ is (roughly) $\sup_{x\in S}\left|\cos \frac{t_1}{x}-\cos\frac{t_2}{x}\right|$ which is equal to $2$ whenever $t_1\ne t_2$. Therefore, the map $t\mapsto f(t,\cdot)$ is not continuous into $C^1(S)$.

In a nutshell: continuity of a map into $C^1(S)$ automatically imposes some uniformity, since $C^1(S)$ is equipped with uniform norm. When $S$ is compact, we gain such uniformity and therefore can conclude that $f\in C([0,T],C^1(S))$.

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