2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ (C1) arose in answering the question in my post Topology of $\mathcal{D}(\Omega)$. At first glance, it seemed to me evidently true, but thinking of it I realized that I could not give a proof. $\endgroup$ – Maurizio Barbato Aug 12 '16 at 17:20
3
$\begingroup$

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 $|\alpha| > N$.

$\endgroup$
  • $\begingroup$ Thank you very very much, Justthisguy for your extraordinary help! $\endgroup$ – Maurizio Barbato Aug 13 '16 at 10:59
2
$\begingroup$

This is not a separate answer. I only write down here carefully the solution sketched by Justthisguy.

First of all, not that for any two smooth functions $f$ and $g$ we have for any multi-index $\alpha$

\begin{equation} D^{\alpha}(fg) = \sum_{\beta \leq \alpha} { \alpha \choose \beta} D^{\beta}f D^{\alpha - \beta} g, \end{equation} where $\beta \leq \alpha$ means that the j-th component $\beta_j$ of $\beta$ is less than or equal to the j-th component $\alpha_j$ of $\alpha$ for any $j=1,\dots,n$, and we have set \begin{equation} {\alpha \choose \beta} = \prod_{j=1}^{n} {\alpha_j \choose \beta_j}. \end{equation} This formula for the derivative of the product can be easily checked by induction. Of course, we can also write as Justthisguy did \begin{equation} D^{\alpha}(fg) = \sum_{\beta + \beta' = \alpha} { \alpha \choose \beta} D^{\beta}f D^{\beta'} g. \end{equation}

Now, let us come to our problem, and let us show that (C2), and so (C1), is true. 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 $c_{\alpha}$. Set $c_{\alpha}=0$ for all $\alpha$ such that $| \alpha | \leq N$. Then by the theorem stated in the post there exists a function $f \in C_{c}^{\infty}(\Omega)$ such that $D^{\alpha}f(x_0)=c_{\alpha}$ for all $\alpha$.

Denote with $B(x,r)$ the open ball of center $x$ and radius $r$. From Urysohn Lemma, in the version given e.g. in Lieb and Loss, Analysis, we know that there exists a function $\phi \in C_{c}^{\infty}(B(0,1))$ such that $0\leq \phi \leq 1$, and $\phi(x)=1$ for all $x$ in the closed ball of center $0$ and radius $1/2$. Define for any $r > 0$, $\phi_r(x)=\phi((x-x_0)/r)$ for all $x$. We have $\phi_r \in C_{c}^{\infty}(B(x_0,r))$, $0\leq \phi \leq 1$, $\phi_r(x)=1$ for all $x$ in the closed ball of center $x_0$ and radius $r/2$. Moreover, we have for any multi-index $\alpha$ \begin{equation} (D^{\alpha} \phi_r)(x) = r^{-|\alpha|} (D^{\alpha} \phi) \left( \frac{x- x_0}{r} \right). \end{equation} So, with the notation used in the post, if $M > 0$ is such that \begin{equation} \left| \left| \phi \right| \right|_{N} \leq M, \end{equation} we have for any $\alpha$ such that $| \alpha | \leq N$ \begin{equation} \max \{ |(D^{\alpha} \phi_r)(x)| : x \in \mathbb{R}^n \} \leq M r^{-|\alpha|}. \end{equation} For any multi-index $\beta=(\beta_1,\dots,\beta_n)$, let us set \begin{equation} \beta!=\beta_1!\dots\beta_n!. \end{equation} Now, since $D^{\alpha} f(x_0)=0$ for $| \alpha | \leq N$, from Taylor's Theorem we get that for every $\alpha$ such that $| \alpha | \leq N$ \begin{equation} D^{\alpha}f(x) = \sum_{|\beta | \leq N - | \alpha |} \frac{1}{\beta!}D^{\alpha + \beta} f(x_0) (x-x_0)^{\beta} + o( |x - x_0|^{N - | \alpha |}) = o( |x - x_0|^{N - | \alpha |}). \end{equation} So we can find $r \in (0,2)$ such that for all $\alpha$, with $ | \alpha | \leq N$, we have \begin{equation} \left| D^{\alpha}f(x) \right| \leq \frac{\epsilon |x - x_0|^{N - | \alpha |}}{M2^{N}} \end{equation} for all $x$ such that $|x - x_0| \leq r$. Then for each $x$ such that $|x - x_0| \leq r$ and for every $\alpha$ such that $| \alpha | \leq N$, we have \begin{multline} \left| D^{\alpha} (\phi_r f)(x) \right| = \left| \sum_{\beta \leq \alpha} { \alpha \choose \beta} (D^{\beta}\phi_r)(x) (D^{\alpha - \beta} f)(x) \right| \leq \frac{\epsilon}{2^N} \sum_{\beta \leq \alpha} { \alpha \choose \beta} r^{-|\beta|}|x - x_0|^{N-|\alpha|+|\beta|} \leq \\ \leq \frac{\epsilon}{2^N} \sum_{\beta \leq \alpha} { \alpha \choose \beta} |x-x_0|^{-|\beta|}|x - x_0|^{N-|\alpha|+|\beta|} \leq \frac{\epsilon}{2^N} |x - x_0|^{N - | \alpha|} \sum_{\beta \leq \alpha} { \alpha \choose \beta} = \epsilon \left( \frac{|x - x_0|}{2} \right)^{N - | \alpha|} \leq \epsilon. \end{multline} So we conclude that \begin{equation} ||\phi_r f ||_{N} \leq \epsilon. \end{equation} Moreover, since $\phi_r(x)=1$ for every $x$ in the closed ball of center $x_0$ and radius $r/2$, we also have $D^{\alpha}(\phi_rf)(x_0)=c_{\alpha}$ for all $\alpha$.

QED

$\endgroup$

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.