Boundedness of the derivative of the trace of an H^1 function - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:22:03Z http://mathoverflow.net/feeds/question/118352 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118352/boundedness-of-the-derivative-of-the-trace-of-an-h1-function Boundedness of the derivative of the trace of an H^1 function Daniel Spector 2013-01-08T12:27:45Z 2013-03-08T15:26:58Z <p>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.</p> <p>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</p> <p>$\int_{\partial B_1} \nabla u_n(x) \cdot n(x)\;d\mathcal{H}^{N-1}(x)$,</p> <p>where $n(x)$ is the unit normal to $\partial B_1$?</p> <p>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 $H^1$?</p> http://mathoverflow.net/questions/118352/boundedness-of-the-derivative-of-the-trace-of-an-h1-function/118358#118358 Answer by Peter Michor for Boundedness of the derivative of the trace of an H^1 function Peter Michor 2013-01-08T13:25:10Z 2013-01-10T00:56:19Z <p>If you restrict a $H^s$-function $f$ to a submanifold of codimension $k$ you get a $H^{s-k/2}$ function. </p> <p>EDIT: But only if $s$ is large enough (Sobolev lemma threshold). Thus the rest of my answer does not work, and I deleted it. </p> http://mathoverflow.net/questions/118352/boundedness-of-the-derivative-of-the-trace-of-an-h1-function/118360#118360 Answer by Liviu Nicolaescu for Boundedness of the derivative of the trace of an H^1 function Liviu Nicolaescu 2013-01-08T14:13:25Z 2013-01-08T20:24:57Z <p>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, <strong>Non-Homogeneous Boundary Value Problems and Applications. I</strong>, Thm. 9.4, Chap 1) shows that for $s> \frac{3}{2}$ the restriction map $\newcommand{\pa}{\partial}$</p> <p>$$R: C^\infty(B_1)\ni u\mapsto (u|_{\pa B_1}, \frac{\pa u}{\pa \bn})\in C^\infty(\pa B_1)\times C^\infty(\pa B_1)$$</p> <p>extends to a continuous split surjective map</p> <p>$$H^s(B_1)\to H^{s-\frac{1}{2}}(\pa B_1) \times H^{s-\frac{3}{2}}(\pa B_1).$$</p> <p>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</p> <p>$$C^\infty( B_1)\ni u\mapsto \int_{\pa B_1} \frac{\pa u (x)}{\pa \bn} \phi (x) dA(x)\in\mathbb{R}$$</p> <p>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</p> <p>$$\left|\int_{\pa B_1} \frac{\pa u}{\pa \bn} dA\right| \leq C\Vert u\Vert_{H^s(B_1)},\;\;\forall u\in C^\infty(B_1),$$</p> <p>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</p> <p>$$\left|\int_{\pa B_1} \frac{\pa u_k}{\pa \bn} dA\right| \to \infty.$$</p>