3
$\begingroup$

Let $\mathbb{D}^n \subseteq \mathbb{R}^n$ be the closed $n$-dimensional unit ball. Suppose we are given an open connected* subset $U \subseteq \mathbb{D}^n$ of full measure in $\mathbb{D}^n$, and a smooth bounded map $f:U \to \mathbb{R}^{k}$, where $k>1$. Note that $f \in L^2(\mathbb{D}^n, \mathbb{R}^{k})$.

Now, let $h:\mathbb{D}^n \to \mathbb{R}^{k}$ be a smooth map satisfying $h(x) \neq 0$ for every $x \in U$ (in particular $h(x) \neq 0$ a.e.).

Assume that $ (df_x)^T\big(h(x)\big)=0$ for every $x \in U$.

Do there exist smooth maps $f_k:\mathbb{D}^n \to \mathbb{R}^{k}$ which converge to $f$ in $L^2$ and satisfy $ ((df_k)_x)^T\big(h(x)\big)=0$?

Note that the derivatives of the original $f$ may explode when we approach $\partial U=\mathbb{D}^n \setminus U$. I also don't assume that $f$ can be continuously extended to all of $\mathbb{D}^n$ or that it is a Sobolev map.

Edit: Why $U$ must be connected in general:

Take $h(x)=(1,0,...,0)$ to be constant. The condition $(df_x)^T\big(h(x)\big)=0$ is equivalent to $$ \langle\partial_j f(x),h(x)\rangle=\partial_jf^1(x)=0 \,\,\quad \forall j=1,...,n\quad\forall x \in U, $$

i.e. $\nabla f^1=0$ on $U$. If $U$ is not connected, we can make $f^1$ to obtain two different constant values on different connected components of $U$. Now, any smooth map $g:\mathbb{D}^n \to \mathbb{R}^{k}$ that satisfies $ (dg_x)^T\big(h(x)\big)=0$ on $\mathbb{D}^n$ must have constant first component, hence cannot approximate well our map $f$.

I guess a reasonable start would be to start with $h$ being constant:

Suppose that $h(x)=(h^1,h^2,...,h^k)$. Then our condition becomes $$ \langle\partial_j f(x),h(x)\rangle=\partial_j(\sum_{i=1}^k h^if^i)=0 \,\,\quad \forall j=1,...,n\quad\forall x \in U, $$

i.e. $\nabla (\sum_{i=1}^k h^if^i)=0$ on $U$. Since $U$ is connected, this implies that $\sum_{i=1}^k h^if^i$ is constant.

Now, the question is whether or not $f$ can be approximated by smooth maps $f_k$ with constant $\sum_{i=1}^k h^if_k^i$.

$\endgroup$

1 Answer 1

1
$\begingroup$

Not in general. As shown in the answer to this question, for specific $h$ it is possible that there are no non-constant smooth maps $f \colon \mathbb{D}^n \rightarrow \mathbb{R}^k$ which satisfy $(df|_{x})^T h(x) = 0$ for all $x \in \mathbb{D}^n$. This is the case for example if $n = 2$ and $h(x,y) = (x,y)^T$. The set $U = \mathbb{D}^2 \setminus \{ (0,0) \}$ is open, connected subset of full measure and $h(p) \neq 0$ for all $p \in U$. If we take

$$ f(x,y) = \left( \frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}} \right)^T $$

then $f$ is defined on $U$, smooth and bounded (of norm one) and satisfies $(df|_{x})^T h(x) = 0$ for all $x \in U$. However, the only smooth maps which satisfy the equality on the whole of $\mathbb{D}^2$ are the constant maps and they definitely can't be used to approximate $f$ in $L^2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .