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If $M \subset \mathbb R^n$ is a compact smooth manifold with boundary, and ${M_\varepsilon }$ is the closed $\varepsilon$-neighborhood of $M$ in $\mathbb R^n$, then whether for sufficiently small $\varepsilon$, ${M_\varepsilon }$ is a smooth manifold?

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Yes. This follows from the tubular neighborhood theorem, which you may find in many differential geometry/topology books. See e.g. http://www.google.com/search?q=tubular%20neighborhood%20theorem&um=1&ie=UTF-8&hl=en&tbo=u&tbm=bks

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    Maybe it is not so easy for this situation. The book I am reading says that when $M$ is boundaryless ${M_\varepsilon }$ is indeed a smooth manifold by tubular neighborhood for some $\varepsilon$. But when $M$ has boundary, we can only prove that ${M_\varepsilon }$ is $C^1$ without higher smoothness. That is where I cannot figure out how.2011-11-13