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Given $(M^n, g)$ closed riemannian manifold, I am wondering about the definition of the $C^k$ norms with respect to the metric, and how these norms depend on $g$. For example, I would assume that if $x$ is a local coordinate on $M$, then for $f: M \to \mathbb{R}$, we can define

$$f'(x) = \lim_{y \to x} \frac{f(y) - f(x)}{\text{dist}_g(x,y)}$$

and

$$||f||_{C^1,g} = \text{sup}_{x \in M} |f(x)| + \text{sup}_{x \in M} |f'(x)|$$

Now a priori, if the first equation is the correct definition for $f'(x)$, then it may be that higher derivatives of $f$ depend on derivatives of $g$ a priori. This raises the question of how does $||f||_{C^k, g}$ depend on the regularity of $g$?

In addition, if we use exponential map coordinates to define $f'(x)$, does the regularity of $g$ affect the smoothness of these coordinates? And hence the $|| \cdot ||_{C^k, g}$ norm as well?

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  • $\begingroup$ That is not how I would define $f'$. By contrast, if $M = \Bbb R^n$ and I assume $f(0) = 0$, you're attempting to define $f'(0) = \lim_{x\to 0} f(x)/\|x\|$. This will usually not be defined. Better is the gradient, defined by $g(\nabla f, v) = (df)(v)$, and then $\|f\|_{C^1} = \|f\|_{C^0} + \|g(\nabla f, \nabla f)\|_{C^0}$. This norm varies continuously as $g$ varies continuously in $C^0$. $\endgroup$
    – mme
    Commented Sep 26 at 15:55
  • $\begingroup$ Sure, do you have reference for that definition? I've definitely seen holder norms as $$||f||_{C^{\alpha}, g} = \sup_{x \neq y} \frac{|f(x) - f(y)|}{dist_g(x,y)^{\alpha}}$$ So I'm extrapolating from that $\endgroup$
    – JMK
    Commented Sep 26 at 16:15

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