Can anyone reference/disprove the theorem in the case where the embedded submanifold is merely $C^1$ instead of smooth? I have a compact $C^1$ embedded submanifold of $\mathbb{R}^n$ without boundary that I want to show there exists a tubular neighborhood of (with the radius of the tube not necessarily constant). Actually, I want to be able to create one piecewise using a finite covering. Much obliged.



Tubular Neighborhood Theorem for $C^1$ Submanifold
The answer to your question depends on whether you are looking for a tubular neighborhood in the general differential topological sense or the more restrictive geometric sense. The answer in the topological sense is **Yes**, but in the geometric sense the answer in general is **No**. These two conceptions may coincide for $C^2$ submanifolds, but for $C^1$ submanifolds the construction is a bit more subtle, or non-canonical, as I will elaborate below. 


A $C^1$ submanifold $M\subset R^n$ may not have any neighborhoods fibrated by normal vectors. So $C^1$ submanifolds do not in general have canonical tubular neighborhoods generated by the exponential map, and "radius" of the neighborhood does not really make sense; however, they still have a tubular neighborhood in the topological sense, i.e., a $C^1$ embedding of the normal bundle which extends that of $M$, as the zero section of the bundle. These may be constructed as follows.


The fastest way to construct a topological tubular neighborhood for $M$ is by noting that there exists a $C^1$ diffeomorphism $\phi\colon R^n\to R^n$ such that $\phi(M)$ is $C^\infty$ (e.g., see Thm. 3.6 in Chap. 2 of [Hirsch's book](http://www.springer.com/us/book/9780387901480)). Then $\phi(M)$ is going to have a tubular neighborhood $U$ in the standard sense, obtained by applying the inverse function theorem to the exponential map (or taking the union of all sufficiently short normal vectors), and $\phi^{-1}(U)$ yields the desired tubular neighborhood of $M$ (which will be $C^1$-diffeomorphic to the normal bundle).


An example of a $C^1$ submanifold without a neighborhood fibrated by normals may be constructed as follows. Take an ellipse in $R^2$, which is not a circle, and let $M$ be the inner parallel curve of the ellipse which passes through the foci. Then $M$ will be $C^1$, but it will not have a tubular neighborhood in the geometric sense. Indeed, the nearest point projection map into $M$ will not be one-to-one in any neighborhood of $M$.


