How do you prove the following:
In general, a $C^k$ function is contained in $C^{k-1}$ for any $k$.
Why is this true? Thanks for helping.
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asked Mar 3, 2018 at 22:18
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4$\begingroup$ Do you know the definition of $C^k$, and can you write it here? Once it's written out properly it should be virtually trivial to see why this is true.... $\endgroup$– user296602Commented Mar 3, 2018 at 22:19
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$\begingroup$ ...if a function has k continuous derivatives, it has k-1 continuous derivatives. $\endgroup$– WojowuCommented Mar 3, 2018 at 22:19
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1 Answer
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$f\in C^k$ means that $f$ is $k$ times differentiable and the $k$th derivative is continuous. In particular $f$ is $k-1$ times differentiable and the $(k-1)$th derivative is continuous (since $f^{(k)}$ exists)
answered Mar 3, 2018 at 22:25