Proposition: Extendability of a trace and of its smooth multiple

Let $S_0\subset {\mathbb{R}}^d$ be open, $Z\subset S_0$ relatively closed, $c\in {\mathcal{D}}^{\prime }(S_0\setminus Z)$, and let $p_k(\psi )={\max }_{|\alpha |\le k}\Vert {\partial }^{\alpha }\psi {\Vert }_{\infty }$.

1. If $v\in C^{\infty }(S_0)$ and $c$ extends to ${\mathcal{D}}^{\prime }(S_0)$, then $vc$ extends to ${\mathcal{D}}^{\prime }(S_0)$. Equivalently: if $vc$ has no extension, then $c$ has no extension. No invertibility of $v$ is needed.

2. The converse of (a) fails: there are $v\in C^{\infty }(S_0)$ and $c$ without extension such that $vc$ extends.

3. If $v$ is only assumed smooth on $S_0\setminus Z$, (a) fails: there are such $v$ and an extendable $c$ with $vc$ not extendable.

4. (Corrected witness transfer.) Let $v\in C^{\infty }(S_0)$, let $K\subset S_0$ be compact, and let $({\varphi }_n)\subset \mathcal{D}(S_0\setminus Z)$, $\mathrm{supp}{\varphi }_n\subset K$, satisfy $p_n({\varphi }_n)\le 1$ and $|⟨vc,{\varphi }_n⟩|>n$. Put ${\psi }_n=v{\varphi }_n$. Then $\mathrm{supp}{\psi }_n\subset K$, $|⟨c,{\psi }_n⟩|>n$, and for each fixed $k$ there is $C_k=C_k(v,K)$ with $p_k({\psi }_n)\le C_k$ for all $n\ge k$. Any sequence with these properties excludes an extension of $c$. However, $p_n({\psi }_n)\le 1$ need not hold, so $({\psi }_n)$ is in general not a normalized failure sequence in the sense of XV-C; a normalized failure sequence for $c$ exists by XV-C's failure proposition but is in general a different sequence.

Proof. (a) If $\tilde{c}\in {\mathcal{D}}^{\prime }(S_0)$ extends $c$, then $v\tilde{c}\in {\mathcal{D}}^{\prime }(S_0)$ and $(v\tilde{c}){|}_{S_0\setminus Z}=v{|}_{S_0\setminus Z}c=vc$.

(b) Take $S_0=(-1,1)$, $Z=\{0\}$, $c(x)=e^{1/x^2}$ for $x>0$ and $c(x)=0$ for $x<0$, which has no extension by XV-B's Proposition prop:nonextendable XVB. Take $v(x)=e^{-2/x^2}$ for $x\ne 0$, $v(0)=0$, which is in $C^{\infty }(S_0)$ and flat at $0$. Then $vc(x)=e^{-1/x^2}$ for $x>0$ and $0$ for $x<0$, a bounded continuous function on $(-1,1)\setminus \{0\}$, which extends to ${\mathcal{D}}^{\prime }(S_0)$ as the corresponding $L^{\infty }$ function.

(c) Take $S_0$, $Z$ as in (b), $c=1$ (extendable), and $v(x)=e^{1/x^2}$ on $S_0\setminus Z$. Then $vc$ is XV-B's nonextendable function on $x>0$ (and $e^{1/x^2}$ on $x<0$, which does not restore extendability: XV-B's test functions are supported in $x>0$).

(d) $\mathrm{supp}{\psi }_n\subset \mathrm{supp}{\varphi }_n\subset K$ and $⟨c,{\psi }_n⟩=⟨vc,{\varphi }_n⟩$. By the Leibniz rule, $p_k(v{\varphi }_n)\le 2^k{\max }_{|\alpha |\le k}{\sup }_K|{\partial }^{\alpha }v|\cdot p_k({\varphi }_n)\le C_k$ for $n\ge k$, since $p_k({\varphi }_n)\le p_n({\varphi }_n)\le 1$. If $c$ had an extension $\tilde{c}$, then on tests supported in $K$, $|⟨\tilde{c},\psi ⟩|\le C{p}_k(\psi )$ for some $C,k$, so $n<|⟨c,{\psi }_n⟩|\le C{C}_k$ for all $n\ge k$, a contradiction. For the last assertion, $p_n(v{\varphi }_n)$ involves derivatives of $v$ of order up to $n$ on $K$, whose suprema are not bounded uniformly in $n$ in general; so the bound $p_n({\psi }_n)\le 1$ is not implied. Existence of a normalized sequence for $c$ is XV-C's Proposition prop:failure applied to $c$, which is nonextendable by (a) XVC.