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 $\sum_{i=1}^k h^if_k^i$.


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 $L^2$.


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.