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A theorem by Ghomi (https://link.springer.com/article/10.1007/s10711-007-9151-y) says that any $C^2$ curve $\gamma:[a,b]\to\mathbb R^{n≥3}$ can be $C^1$-approximated by curves with prescribed constant curvature as long as the prescribed curvature is larger than the maximum of the curvature of $\gamma$. I wonder if one could relax the assumption on the prescribed curvature to be constant and weaken the $C^1$-approximability by $C^0$-approximability. More precisely I wonder if the following statement holds true:

Let $\gamma:[a,b]\to\mathbb R^{n≥3}$ be a curve of class $C^2$ with curvature $k$ and let $k':[a,b]\to\mathbb R$ be a smooth function that satisfies $k'>k$ point wise. There is a smooth curve $\delta:[a,b]\to\mathbb R^{n≥3}$ with curvature $k'$ that is $C^0$-close to $\gamma$.

I'd be thankful for any reference.

EDIT: I corrected the question, thanks.

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    $\begingroup$ If the prescribed curvature $k'$ is only continuous, then the curve $\delta$ cannot be smooth (by which I mean $C^\infty$) so the answer to the question the way it is stated is: no. Correct your question. $\endgroup$ Commented Oct 8, 2019 at 11:41

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This has been established in a recent paper by Micha Wasem:

h-principle for curves with prescribed curvature. Geom. Dedicata 184 (2016), 135–142.

In particular Wasem showed that one can prescribe curvature as you suggest, and it is not even necessary to give up $C^1$ closeness.

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  • $\begingroup$ Thank you for the reference! In this paper, the approximating curves are of class $C^2$ and I wonder if one could get $C^\infty$ ones (maybe at the expense of giving up $C^1$-closeness). $\endgroup$
    – Stadionok
    Commented Oct 10, 2019 at 6:14
  • $\begingroup$ @Stadionok: I think that it should be possible to get $C^\infty$ ones, without giving up $C^1$-closeness. $\endgroup$ Commented Oct 10, 2019 at 13:43
  • $\begingroup$ This is what I guess too but it is not clear to me, how to adapt your proof... $\endgroup$
    – Stadionok
    Commented Oct 11, 2019 at 9:40

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