Smooth extension of functions at corners

Question

Let $\mathbb{B}_1(0)\subseteq\mathbb{R}^n$ be the ball of radius $1$ in the Euclidean space, $n>1$. Suppose we have a cylinder $C=[0,1]\times \mathbb{B}_1(0)$ and suppose we are given smooth functions

• $\rho_0\colon [0,\varepsilon)\times \mathbb{B}_1(0)\to\mathbb{R}$;
• $\rho_1\colon (1-\varepsilon,1]\times \mathbb{B}_1(0)\to\mathbb{R}$;
• $F_0\colon[0,1]\to\mathbb{R}$,

such that $F_0\equiv\rho_0(\cdot,0)$ on $[0,\varepsilon)$ and $F_0\equiv\rho_1(\cdot,0)$ on $(1-\varepsilon,1]$.

Can we find a smooth function $F(x,s)$ on $C$ such that

• $F(x,s)=\rho_0(x,s)$ for $(x,s)\in[0,\varepsilon)\times \mathbb{B}_1(0)$;
• $F(x,s)=\rho_1(x,s)$ for $(x,s)\in(1-\varepsilon,1]\times \mathbb{B}_1(0)$;
• $F(x,0)=F_0(x)$ for all $x\in[0,1]$?

If not always, which assumptions do we need on $F_0,\rho_0,\rho_1$?

When we can, do we have control on derivatives?

I know Tietze's extension yields the result with $F(x,s)$ continuous, but I can't find much in the smooth case. Whitney's extension doesn't work either as on $[0,1]\times\{0\}$ we have $F_0$ but we don't know whether it can be extended locally in the sense of smooth functions on closed subsets.

I think the main difficulty lies in the fact that the boundary of where the functions are defined is not a manifold; in particular the points $(\varepsilon,0)$ and $(1-\varepsilon,0)$ cause problems even locally.

Is this a known topic? Thank you for your replies!

Answer 1

$\newcommand\ep\varepsilon$Take any $\ep\in(0,1/2)$. Let $$A:=A_0\cup A_1\cup F,$$ where $A_0:=[0,\ep)\times B$, $A_1:=(1-\ep,1]\times B$, $F:=[0,1]\times\{0\}$, and $B$ is the unit ball (you did not say if your unit ball is closed or open; let us assume that it is closed).

You have a well-defined function $f$ on $A$ such that $f|_{A_0}=\rho_0$, $f|_{A_1}=\rho_1$, and $f(\cdot,0)=F_0$.

Your question is then when $f$ can be extended to a smooth function $F$ on $C=[0,1]\times B$.

For that it is clearly necessary that $\rho_0$ and $\rho_1$ be extendible to smooth functions $\bar\rho_0$ and $\bar\rho_1$ on the closures $\bar A_0$ and $\bar A_1$ of $A_0$ and $A_1$. Then $f$ can be accordingly extended to a uniquely determined function $\bar f$ on the closure $\bar A$ of $A$.

So, the question then becomes when $\bar f$ can be extended to a smooth function $F$ on $C=[0,1]\times B$.

By Whitney's Theorem I, a sufficient condition for this is that $\bar f$ be of class $C^\infty$ on $\bar A$ in the sense of this paper by Whitney (the set $A$ in the mentioned theorem can be any closed subset of a Euclidean space -- it does not have to be a manifold).

It is clear that this sufficient condition is also necessary.

Therefore and because the functions $\bar\rho_0$ and $\bar\rho_1$ are smooth, we conclude:

$f$ can be extended to a smooth function $F$ on $C=[0,1]\times B$ iff condition (3.2) in the mentioned paper by Whitney holds for the function $\bar f$ and all the points $x^0$ in the set $[\ep,1-\ep]\times\{0\}$ (with respect to the set $\bar A$).

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The case $\ep>1/2$ is trivial. In the case when $\ep=1/2$, the obvious necessary and sufficient condition is that the values of the function $\bar\rho_0$ and all its partial derivatives on the set $\bar A_0\cap\bar A_1=\{1/2\}\times B$ be the same as the corresponding values of the function $\bar\rho_1$ and all its partial derivatives.