Boundedness of the derivative of the trace of an H^1 function - MathOverflow

Boundedness of the derivative of the trace of an H^1 function

2013-01-08T12:27:45Z

As a research preface, this question is linked to a problem of increasing magnetism in Ginzburg-Landau equations that I have distilled for the purpose of getting to the bottom of this technical matter.

Suppose $u_n\in H^1(B_1)$ (actually, $u_n$ are smooth), where $B_1$ is the unit ball in $\mathbb{R}^N$, and that we know $||u_n||_{H^1} \leq C$, so that up to a subsequence, $u_n \to u$ in $H^1(B_1)$ weakly. What can we say about the boundedness of the quantity

$\int_{\partial B_1} \nabla u_n(x) \cdot n(x)\;d\mathcal{H}^{N-1}(x)$,

where $n(x)$ is the unit normal to $\partial B_1$?

In particular, is this quantity always finite under these hypotheses, or is there a counterexample that shows this blows up for a bounded subset of $H^1$?

Answer by Peter Michor for Boundedness of the derivative of the trace of an H^1 function

2013-01-08T13:25:10Z

If you restrict a $H^s$-function $f$ to a submanifold of codimension $k$ you get a $H^{s-k/2}$ function.

EDIT: But only if $s$ is large enough (Sobolev lemma threshold). Thus the rest of my answer does not work, and I deleted it.

Answer by Liviu Nicolaescu for Boundedness of the derivative of the trace of an H^1 function

2013-01-08T14:13:25Z

Clearly, what you call $\newcommand{\bn}{\boldsymbol{n}}$ $\nabla u\cdot \bn $ is the normal derivative $\frac{\partial u}{\partial \bn}$. The trace theorem (see e.g. Lions and Magenes, Non-Homogeneous Boundary Value Problems and Applications. I, Thm. 9.4, Chap 1) shows that for $s> \frac{3}{2}$ the restriction map $\newcommand{\pa}{\partial}$

$$R: C^\infty(B_1)\ni u\mapsto (u|_{\pa B_1}, \frac{\pa u}{\pa \bn})\in C^\infty(\pa B_1)\times C^\infty(\pa B_1) $$

extends to a continuous split surjective map

$$ H^s(B_1)\to H^{s-\frac{1}{2}}(\pa B_1) \times H^{s-\frac{3}{2}}(\pa B_1). $$

The result is optimal because Theorem 9.5, Chap. 1 op. cit. shows that if $s\leq \frac{3}{2}$, then for any $\phi\in C^\infty(\pa B_1)$ the linear functional

$$ C^\infty( B_1)\ni u\mapsto \int_{\pa B_1} \frac{\pa u (x)}{\pa \bn} \phi (x) dA(x)\in\mathbb{R} $$

is not continuous in the topology induced by $H^s(B_1)$. In particular, if $s\leq \frac{3}{2}$ there cannot exist a constant $C>0$ such that

$$ \left|\int_{\pa B_1} \frac{\pa u}{\pa \bn} dA\right| \leq C\Vert u\Vert_{H^s(B_1)},\;\;\forall u\in C^\infty(B_1), $$

so that there exists a sequence $u_k\in C^\infty(B_1)$ such that $\Vert u_k\Vert_{H^1(B_1)}\leq 1$ and

$$ \left|\int_{\pa B_1} \frac{\pa u_k}{\pa \bn} dA\right| \to \infty. $$