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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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    @PavelM I'm wondering why the paper stresses the *uniform* continuity. (Maybe I am being stupid but I can't see it.)2012-12-22

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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 $f\in C([0,T],C^1(S))$.