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

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 $H^1$?

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

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

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

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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.

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 This conflicts with the above answer, which is confusing to me. Can you shed some light on the difference? – Daniel Spector Jan 9 at 17:33