0
$\begingroup$

Let $I_1$ and $I_2$ be two closed bounded intervals. Suppose $W(x,y)$ is a smooth function whose support is contained inside $I_1 \times I_2$.

Suppose I have $\Phi= (\Phi_1(x,y), \Phi_2(x,y)) : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ which is a diffeomorphism restricted to $B$, where $B$ is an open bounded box such that $I_1 \times I_2 \subseteq B$.

I want to define $\widetilde{W} (u,v) = W \circ \Phi^{-1} (u,v)$ which is well defined on $\Phi(B)$. I would like to extend it to be a function on all of $\mathbb{R}^2$ by defining it to be $0$ outside $\Phi(B)$.

My question is, from this does it follow that $\widetilde{W} (u,v)$ differnetiable everywhere? or do I need extra assumptions? (I guess my concern is what happens near the boundaries, does everything work out nicely so that this is the case?) Thank you very much.

$\endgroup$
2
$\begingroup$

The function $\widetilde W$ is a smooth function on the open set $\Phi(B)$ and is supported in the compact set$\Phi(I_1\times I_2)$. As a result it can be extended by 0 as you wish and this is simply due to the continuous canonical injection $$ C_{comp}^\infty(\Omega)\subset C_{comp}^\infty(\mathbb R^n), $$ for $\Omega$ open subset of $\mathbb R^n$.

$\endgroup$
  • $\begingroup$ Are there no possible issues at the boundary or something? or am I just thinking too much? $\endgroup$ – Johnny T. Feb 19 '18 at 18:29
  • 1
    $\begingroup$ No, you just have to think about the continuous canonical injection mentioned above. $\endgroup$ – Bazin Feb 20 '18 at 11:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.