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Let $M\subset \mathbb{R}^n$ be a compact smooth manifold embedded in $\mathbb{R}^n$, we define $$\mathfrak{X}(M) := \{X: M \to \mathbb{R}^n;\ X\mbox{ is smooth and }\ X(p) \in T_p M \subset \mathbb{R}^n,\ \forall\ p \in M \}.$$

Choosing an atlas $\{(\varphi_i,U_i)\}_{i=1}^{n}$, and compacts $K_i \subset U_i$, such that $$\bigcup_{i=1}^n K_i = M,$$ we define the $\|\cdot \|_r$ norm as

\begin{align*}\|\cdot\|_r : \mathfrak{X}(M)&\to \mathbb{R}\\ X &\to \max_{\substack{i\in\{1,...,n\} \\ j\in \{0,...,r\}}}\left\{\sup_{x \in \varphi^{-1}_i(K_i)}\left\| \text{d}^{j}\left( X\circ\varphi_i \right) \right\|\right\}, \end{align*}

then we named $\mathfrak{X}^r(M)$ as the complete Banach space $(\mathfrak{X}(M),\|\cdot\|_r)$ (it is possible to prove that the topology of $\mathfrak{X}^r(M)$ does not depend on the selected atlas).

My Question: Let $X \in \mathfrak{X}(M)$ and $Y$ be a smooth vector field on $M$ defined just in a compact $K \subset M$ such that $$\max_{\substack{i\in\{1,...,n\} \\ j\in \{0,...,r\}}}\left\{\sup_{x \in \varphi^{-1}_i(K_i\cap K)}\left\| \text{d}^{j}\left( X\circ\varphi_i \right) - \text{d}^{j}\left( Y\circ\varphi_i \right) \right\|\right\}<\varepsilon,$$ is it possible extend $Y$ to a vector field $\tilde{Y}$ such that

1) $\left.\tilde{Y}\right|_{K} = Y$,

2) $\|X-\tilde Y\|_r < A\cdot\varepsilon$ , where $A $ is a constant that depends only on the manifold $K$ ?

The compact $K$ is a connected submanifold with boundary of $M$, such that $\dim K = \dim M$.

Edit: I changed $\|X-\tilde Y\|_r < \varepsilon$ to $\|X-\tilde Y\|_r < A\cdot\varepsilon$ after Moishe Kohan's comment.


My ideas

First, I extend $Y$ by a smooth vector field $Z$ $\in \mathfrak{X}(M)$, by the continuity of $Z$, so there exists a neighborhood $U$ of $K$, such that $$\max_{\substack{i\in\{1,...,n\} \\ j\in \{0,...,r\}}}\left\{\sup_{x \in \varphi^{-1}_i(K_i\cap U)}\left\| \text{d}^{j}\left( X\circ\varphi_i \right) - \text{d}^{j}\left( Z\circ\varphi_i \right) \right\|\right\}<\varepsilon,$$

and then choosing a partition of unity $ \{\phi_1, \phi_2\}$ subordinate to the cover $\{U,M\setminus K\}$ we can define $$\tilde{Y} = \phi_1 Z + \phi_2 X, $$ however I could not guarantee that $\|X - \tilde{Y}\|_r < \varepsilon$, because I can not control de derivatives of $\phi_1$ and $\phi_2$. Does anyone know how should I proceed?

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  • $\begingroup$ This is a form of Whitney extension theorem/problem. Take a look here: annals.math.princeton.edu/wp-content/uploads/… I suspect you cannot keep the same $\epsilon$: Fefferman's result (and some follow up papers) show that you can find an extension with an error $A\epsilon$ where $A$ is some constant depending only on dimension. $\endgroup$ – Moishe Kohan Mar 28 at 3:38
  • $\begingroup$ Thx for the reference. Once the same $A$ holds for every function. It would be enough to solve my problem. The only complication I am seeing right know it is the fact that this result is for $\mathbb{R}^n $ and not manifolds. Do you know how to generalize to manifolds this result? $\endgroup$ – Matheus Manzatto Mar 28 at 9:13

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