What is the meaning of a function $f : \mathbb{R}^N \to \mathbb{R}$ being $\mathcal{C^k}$ where $k \in \mathbb{N}$? I need an explanation for the case $N \geq 2$.
What is the meaning of a function $f : \mathbb{R}^N \to \mathbb{R}$ being $\mathcal{C^k}$?
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$\begingroup$
calculus
real-analysis
2 Answers
7
All partial derivatives up to order $k$ exist and are continuous.
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0Is there any restriction as to what happens to partial derivatives of order higher than $k$ ? – 2011-04-15
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0Also wonder what happens to the case of $f : \mathbb{R^N} \to \mathbb{R^M}$ ? – 2011-04-15
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3@Rajesh, no restriction. For the general case, you impose the conditions on all coordinate functions of $f$. – 2011-04-15
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0by no restriction do you mean one or more of them may not exist at some point. Please clarify whether i am getting it correct. – 2011-04-15
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0@Rajesh, being $\mathcal C^k$ has *no* implication on being $\mathcal C^r$ for $r>k$. In other words, $\mathcal C^r \subset \mathcal C^k$ and the inclusion is strict. – 2011-04-15
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The function is $k$ times differentiable everywhere on $\mathbb{R}^n$, and $D^kf$ is continuous on $\mathbb{R}^n$.
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1But defining $D^k f$ when $N>1$ makes this answer harder to understand than the lhf answer... – 2011-04-15