How to make a $C^1$ knot into a $C^\infty$ knot Suppose I have a $C^1$ imbedding $f: S^1 \rightarrow S^3$.  From the point of view of knot theory, what's the "best" way to get a $C^\infty$ curve that "looks like" or is "equivalent to" $f$?  For instance, one of the stronger things I can think to try to prove is that there exists a $C^1$ isotopy, H, of $S^3$ which takes $f$ to a $C^\infty$ imbedding and such that $H_t  (f(S^1))= f(S^1)$ for all times $t$ (so the isotopy doesnt move the image of $f$ off of itself).  But I would be happy with weaker results.  Also at the risk of asking two questions I'll say that I'd be happy to learn of any particular books or notes that deal with very technical and foundational knot-theoretic questions such as this.  Thank you for reading my question,
 A: In general it's impossible to keep $H_t(f(S^1))=f(S^1)$. This is because the image of $S^1$ under a $C^1$ embedding is only a $C^1$ manifold in general, while the image under a $C^\infty$ embedding is a $C^\infty$ manifold. For a concrete example, the curve $y=x\sqrt{|x|}$ in the $xy$-plane is the image of $\mathbb R$ under the $C^1$-embedding $x\mapsto (x,x\sqrt{|x|})$. Since its curvature blows up at $(0,0)$, it cannot be the image of $\mathbb R^2$ under a $C^2$-embedding. 
But you can construct an isotopic $C^\infty$ knot arbitrarily close to the original one, by using convolution with a smooth compactly supported mollifier. Fix $\delta>0 $ such that the values of $f'/|f'|$ do not differ by more than $1/10$ within any interval of length $\delta$. For $\epsilon>0$  let $f_\epsilon=f*\varphi_\epsilon$ where $\varphi_\epsilon$ is a mollifier with support of size $\epsilon$. When $\epsilon<\delta$, the derivative of $f_\epsilon$, which is $f'*\varphi_\epsilon$, does not vanish: there is no substantial cancellation in the integral, because the derivative vector points essentially in the same direction. By a similar argument, $f_\epsilon(a)\ne f_\varepsilon(b)$ when $\epsilon<\delta/2$ and $|a-b|<\delta/2$. 
Finally, let $\sigma=\inf\{|f(a)-f(b)|:|a-b|\ge \delta/2\}$. For sufficiently small $\epsilon $ we have  $\max |f_\epsilon-f|<\sigma/2$, which implies that $f_\epsilon(a)\ne f_\epsilon(b)$ whenever $|a-b|\ge \delta/2$. This completes the proof.
