Let $\Omega \subseteq \mathbb{R}^n$ be a nonempty open set, and $C_{c}^{\infty}(\Omega)$ the set of all infinitely differentiable functions $f:\mathbb{R}^n \rightarrow \mathbb{R}$, whose support is contained in $\Omega$. For any $f \in C_{c}^{\infty}(\Omega)$ and any nonnegative integer $N$, define
\begin{equation}
\left| \left| f \right| \right|_{N} = \max \left \{
\left| D^{\alpha}f(x) \right| : x \in \Omega, \left| \alpha \right| = 0, 1, \dots, N \right \},
\end{equation}
where $\alpha=(\alpha_1,\dots,\alpha_n)$ is a multi-index and $| \alpha | = \alpha_1 + \dots + \alpha_n$.
Consider the following conjectures.
(C1) Let $N$ be a nonnegative integer, $M > 0, \epsilon > 0$, and $x_0 \in \Omega$. Does there exists $f \in C_{c}^{\infty}(\Omega)$ such that
(i) $||f||_{N} \leq \epsilon$,
(ii) $|D^{\alpha}f(x_0)| > M$ for some $\alpha$ such that $| \alpha| = N+1$.
A much more stronger statement is the following.
(C2) Let $N$ be a nonnegative integer, $\epsilon > 0$, and $x_0 \in \Omega$. Associate to each multi-index $\alpha$, with $| \alpha | > N$, a real number $c_{\alpha}$. Then there exists $f \in C_{c}^{\infty}(\Omega)$ such that
(i) $||f||_{N} \leq \epsilon$,
(ii) $D^{\alpha}f(x_0)=c_{\alpha}$ for each $\alpha$ such that $|\alpha|>N$.
I could not even prove (C1), which seems evidently true. As to (C2) it was suggested to me by the following result, which is an immediate corollary of Whitney Extension Theorem.
Theorem
Let $x_0 \in \Omega$, and associate to each multi-index $\alpha$ a real number $c_{\alpha}$. Then there exists $f \in C_{c}^{\infty}(\Omega)$ such that $D^{\alpha}f(x_0)=c_{\alpha}$ for every $\alpha$.
Proof. Let $B(x_0,r)$ an open ball of center $x_0$ and radius $r > 0$ contained in $\Omega$. Then take $E={x_0} \cup (\mathbb{R}^n \backslash B(x_0,r/2))$, $f_{\alpha}(x_0)=c_{\alpha}$, $f_{\alpha}(x)=0$ for $x \in \mathbb{R}^n \backslash B(x_0,r/2)$ in Whitney, Theorem I.
Any help in solving these two problems is welcome.
Best Answer
Let $\phi_r$ be a smooth function which is one on $B(x_0,r)$, zero outside of $B(x_0, 2r)$, and $\partial^\alpha \phi_r (x) \lesssim r^{-|\alpha|}$, $r$ to be chosen. Now fix $f$ so that $\partial^\alpha f(x_0) = c_\alpha$ for $|\alpha| > N$, and $\partial^\alpha f(x_0) = 0$ for $|\alpha| \leq N$. It follows from Taylor's theorem that, for $|\alpha| \leq N$, $$\partial^\alpha f(x) = \partial^\alpha f(x) - \sum\limits_{|\alpha + \beta| \leq N} {1 \over \beta!} \partial^{\alpha + \beta} f(x_0)(x-x_0)^\beta = o(|x-x_0|^{N - |\alpha|}).$$
We have $$\partial^\alpha (\phi_r f)(x) = \sum\limits_{\beta + \beta' = \alpha} {\alpha \choose \beta} \partial^\beta \phi_r (x) \partial^{\beta'}f(x).$$ Now, if $r \leq 2|x-x_0|$, $|\partial^\beta \phi_r (x)| \lesssim |x-x_0|^{-|\beta|}$. On the other hand, $$\partial^{\beta'} f(x) = o(|x-x_0|^{N-|\beta'|}) = o(|x-x_0|^{N-|\alpha| + |\beta|}),$$ so if $r$ is chosen suitably we can bound the sum above to be as small as we like. Since $\phi_r$ is identically one on a neighborhood of $x_0$ for any $r$, $\partial^\alpha (\phi_r f)(x_0) = c_\alpha$ for all $|\alpha| > N$.