Let $M \subset \mathbb{R}^d$ be some $C^2$ submanifold and $f:M \rightarrow \mathbb{R}$ be some $C^2$ function.


Since $f$ is $C^2$, there is $U$ a neighborhood of $M$ and $F:U \rightarrow \mathbb{R}$ a local $C^2$ extension of $f$ such that $F\_{M}= f$.
The gradient of $f$ at $x$ can then be defined as $\nabla f(x) = P\_x \nabla F(x)$, where $P\_x$ is the projection onto the tangent space of $M$ at $x$. This definition doesn't depend on the local extension $F$.


**Question:** Can $\nabla f : M \rightarrow \mathbb{R}^d$ be extended to a $C^1$ function on a neighborhood of $M$? And if so, is there a simple way to prove it?


Intuitively, since $f$ is $C^2$, it seems to me that the answer should be positive.


If $M$ is $C^3$, then I managed to prove this result using the fact that, $\pi\_{M}$, the projection onto $M$ is $C^2$. In this case $F = f \circ \pi\_{M}$ is $C^2$ (on a neighborhood of $M$) and its gradient is indeed equal to $\nabla f$ (and is $C^1$)


However, i don't know how to approach the case where $M$ is only $C^2$. Any help would be greatly appreciated.



Does the gradient of a twice differentiable function on a submanifold can be extended to a differentiable vector field?
Suppose your submanifold is a graph of a function $\mathbb{R}^{k}\to\mathbb{R}^{d-k}$.
Then you can extend your vector field by moving it in the directions on $ \mathbb{R}^{d-k}$; the obtained vector field is as smooth as the original.


In general case you can cover $M$ by such graphs, extend the vector field separately for each graph and mix the results using a partition of unity.


