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When dealing with differentiable surfaces one defines a function $f:S\rightarrow \mathbb{R}$ as being differentiable if its expression in local coordinates is differentiable. But one could also define it to be differentiable if there exists differentiable function $F: V\subset \mathbb{R}^3 \rightarrow \mathbb{R}$ from an open set $V$ of $\mathbb{R}^3$ such that $S\subset V$ and $F|_{S} = f$, i.e. a differentiable extension of $f$.

Are these two definitions of differentiability equivalent? More precisely, when given a differentiable function on a surface can you always extend it to a differentiable function of an open set of $\mathbb{R}^3$ containing the surface?

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up vote 3 down vote accepted

Georges's answer generalizes to functions on submanifolds. That is, if $N$ is a submanifold of $M$ and you have a function $u : N \rightarrow \mathbb{R}$ then:

1) If dim $N$ < dim $M$, then you can always find an open neighborhood $U \subseteq M$ of $N$ in $M$ and an extension $\tilde{u} : U \rightarrow \mathbb{R}$ of $u$ to $U$. You do this by working in a slice coordinate system, extending locally, and patching with a partition of unity.

2) If dim $N$ < dim $M$ and $N$ is a closed subset of $M$, then you can find an extension $\tilde{u} : M \rightarrow \mathbb{R}$. You can do this by extending the local extension to the whole of $M$ with a smooth cutoff function.

3) If dim $N$ = dim $M$ (so $N$ is just an open subset of $M$), then $u$ is already defined on an open subset of $M$ and you can't necessarily find an extension of $u$ to a larger open subset (think uniform continuity).

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1) If $S\subset \mathbb R^3$ is a closed smooth surface every differentiable function on $S$ extends smoothly to $\mathbb R^3$.

2) If $S\subset \mathbb R^3$ is an arbitrary (=locally closed) smooth surface every smooth function on $S$ extends smoothly to an open subset $ V\subset \mathbb R^3 \: (S\subset V)$.
However it is not always true that you can take $V=\mathbb R^3$.
For example if $S$ is the hemisphere $S$ given by $x^2+y^2+z^2=1, z\gt 0$ you can't extend $\frac {1}{z}$ smoothly to $\mathbb R^3$

Proof via partitions of unity.

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