# Extension of continuous and smooth functions

Let us consider any subset $U \subset \mathbb{R}^{n}$. By definition, a function $f: U \rightarrow \mathbb{R}^m$ is smooth if, for every $x \in U$, there exist an open neighbourhood $\Omega_{x}$ of $x$ in $\mathbb{R}^{n}$ and a smooth function $F_{x}: \Omega_{x} \rightarrow \mathbb{R}^m$ such that $F_{x}\vert_{\Omega_{x} \cap U} = f$.

If we consider a continuous function $f: U \rightarrow \mathbb{R}^m$, such a local extension does not necessarily exist. For example, we can consider $U = \mathbb{Q} \subset \mathbb{R}$ and the following function $f: \mathbb{Q} \rightarrow \mathbb{R}$: $f(q) = 0$ for $q \leq 0$, $f(q) = \frac{1}{n}$ for $\frac{\sqrt{2}}{n+1} < q < \frac{\sqrt{2}}{n}$ and $f(q) = 1$ for $q > \sqrt{2}$. The function $f$ is continuous, but it has no local extension in $0$.

When $U$ is locally closed, such a local extension always exists because of the Tietze extension theorem. Moreover, in this case, there exists an extension of $f$ to a whole neighbourhood of $U$. In fact, a locally closed set is the intersection between a closed set $A$ and an open set $B$, hence we extend $f$ from $U = A \cap B$ to $B$ by Tietze extension theorem. If $f$ is smooth, this extension can be chosen to be smooth, because $B$ is an $n$-manifold and $U$ is a closed subset.

Finally, if I am not wrong, if $U$ is an $F_{\sigma}$ set (even not locally closed), a continuous function, that admits a local extension in each point, admits also an extension to a neighbourhood of the whole $U$, because of paracompactness. If $f$ is smooth, such an extension can be chosen to be smooth, using a smooth partition of unity.

My questions are the following:

1) If $U$ is not locally closed and not an $F_{\sigma}$ set, is it possible to find a continuous function with a local extension in each point, but without any extension to a neighbourhood of the whole $U$?

2) Is it possible to find a smooth function (that, by definition, has a local smooth extension in each point) in such a way that there is no smooth extension to a neighbourhood of the whole $U$?

3) If the reply to question 2 is positive, is it possible to find an example of a smooth function having a continuous but not smooth extension to a neighbourhood of the whole $U$?

• Your statement about $F_\sigma$ sets seems wrong: take $U=\mathbb{R}\setminus\{0\}$, and $f(x)=1$ if $x>0$ and $0$ if $x<0$. Then $f$ is continuous on $U$, and $U$ is $F_\sigma$ (in fact, open), but $f$ has no local extension at $0$. – Noah Schweber Nov 4 '15 at 15:00
• This function is not defined in $0$, and $\mathbb{R} \setminus \{0\}$ is an open neighbourhood of itself, in which $f$ is smooth. – Fabio Nov 4 '15 at 15:11
• I think I found the answer. Every subspace of $\mathbb{R}^n$ is paracompact (Dugundji, ex 5 p. 164). More generally, every subspace of a metrizable space is paracompact (E. Michael, "A note on paracompact spaces", Proc. AMS, p. 834). Anyway, if I am not wrong, this is not necessary. For every $x \in U$, let us consider an open neighbhourhood $\Omega_{x}$ and an extension $F_{x}$ of $f$. The union of such neighbhourhoods is open, hence it is paracompact (it is an $F_{\sigma}$). Hence, we can choose a (smooth) partition of unity subodinate to the cover $\{\Omega_{x}\}$ and extend $f$, – Fabio Nov 4 '15 at 23:38
• P.S. Alternatively, an open subspace of $\mathbb{R}^{n}$ is a Hausdorff and second-countable topological manifold, hence it is paracompact. – Fabio Nov 4 '15 at 23:47

I think I found the answer. Every subspace of $\mathbb{R}^n$ is paracompact (Dugundji, ex 5 p. 164). More generally, every subspace of a metrizable space is paracompact (E. Michael, "A note on paracompact spaces", Proc. AMS, p. 834). Anyway, if I am not wrong, this is not necessary. For every $x\in U$, let us consider an open neighbhourhood $\Omega_x$ and an extension $F_x$ of $f$. The union of such neighbhourhoods is open, hence it is paracompact (it is an $F_\sigma$). Hence, we can choose a (smooth) partition of unity subodinate to the cover $\{\Omega_x\}$ and extend $f$,
P.S. Alternatively, an open subspace of $\mathbb{R}^n$ is a Hausdorff and second-countable topological manifold, hence it is paracompact