Math

Why is trace embedding $W^{1,2}(M) \hookrightarrow L^p(\partial M)$ compact

Step-by-step mathematics solution: Why is trace embedding $W^{1,2}(M) \hookrightarrow L^p(\partial M)$ compact

As an Amazon Associate, I earn from qualifying purchases. For more practice problems like this, see Schaum’s Outline of Calculus, 7th Edition.

1. Restatement in plain language

Let

  • (M) be a compact (n)-dimensional Riemannian manifold with (non‑empty) smooth boundary (\partial M);
  • (W^{1,2}(M)=H^{1}(M)) be the usual Sobolev space of functions whose first weak derivatives are square‑integrable;
  • (L^{p}(\partial M)) be the Lebesgue space on the boundary equipped with the induced surface measure.

The trace operator
[ \operatorname{Tr}:W^{1,2}(M)\longrightarrow L^{p}(\partial M),\qquad u\mapsto u|_{\partial M} ] is known to be continuous for every
[ 1\le p\le p^{*}:=\frac{2(n-1)}{n-2}\qquad (n\ge 3), ] (and for all (p<\infty) when (n=2)).

The question asks to prove that this trace map is compact when the exponent is strictly sub‑critical:

[ \boxed{\; 1\le p<p^{*}\ \Longrightarrow\ \operatorname{Tr}:W^{1,2}(M)\hookrightarrow L^{p}(\partial M)\ \text{is compact.}\;} ]

In addition we should comment on how much regularity of the boundary is really needed (Lipschitz, (C^{1}), …).

2. Full proof

The proof proceeds in three stages:

Step What we do Reason
1 Reduce the problem to a local statement on a half‑space (\mathbb{R}^{n}_{+}) using charts and a partition of unity. Manifolds are locally Euclidean; the trace operator is local.
2 Prove the compactness of the trace on a bounded Lipschitz domain (\Omega\subset\mathbb{R}^{n}) (in particular on a half‑ball). This is the Euclidean Rellich–Kondrachov theorem together with continuity of the trace.
3 Patch the local results together to obtain compactness on the whole manifold. A finite covering and a partition of unity preserve compactness.

We now go through the details.

2.1 Preliminaries

  • Sobolev–trace theorem (continuous embedding).
    For a bounded Lipschitz domain (\Omega\subset\mathbb{R}^{n}) there is a bounded linear operator
    [ T_{\Omega}:W^{1,2}(\Omega)\longrightarrow L^{p}(\partial\Omega),\qquad 1\le p\le p^{*}, ] with norm depending only on (n), the Lipschitz constant of (\partial\Omega) and (|\Omega|).

  • Rellich–Kondrachov theorem (compact embedding).
    If (\Omega) is bounded with Lipschitz boundary, the inclusion
    [ W^{1,2}(\Omega)\hookrightarrow L^{q}(\Omega) ] is compact for every (1\le q<2^{*}:=\frac{2n}{n-2}) (the critical Sobolev exponent).

  • Relation between the two exponents.
    The critical trace exponent satisfies
    [ p^{}= \frac{2(n-1)}{n-2}= \frac{2(n-1)}{n-2}= \frac{2n}{n-2}\cdot\frac{n-1}{n} = 2^{}\,\frac{n-1}{n}, ] i.e. (p^{}<2^{}) for (n\ge 3). Consequently, if (p<p^{}) we can find a number (q) with [ p<q<2^{} ] and use the compact embedding into (L^{q}(\Omega)) together with the continuity of the trace.

2.2 Local compactness on a Euclidean half‑space

Let (\Omega\subset\mathbb{R}^{n}) be a bounded Lipschitz domain.
Choose a smooth cutoff (\chi\in C^{\infty}_{c}(\overline{\Omega})) that equals (1) near a piece of the boundary where we shall work.

Consider a bounded sequence ({u_{k}}\subset W^{1,2}(\Omega)).
Because the trace operator is continuous, [ {T_{\Omega}u_{k}}\subset L^{p}(\partial\Omega) ] is bounded.

Step 2.2.1 – Use Rellich inside the domain.
Pick any exponent (q) such that (p<q<2^{*}). By Rellich, [ {u_{k}}\ \text{is relatively compact in}\ L^{q}(\Omega). ] Hence (after passing to a subsequence) we have
[ u_{k}\longrightarrow u\quad\text{strongly in }L^{q}(\Omega). ]

Step 2.2.2 – Interpolation to the boundary.
The trace theorem gives a bounded linear map [ T_{\Omega}:L^{q}(\Omega)\longrightarrow L^{p}(\partial\Omega), ] because for (q\ge 2) the usual estimate [ |T_{\Omega}v|{L^{p}(\partial\Omega)} \le C|v|{W^{1,2}(\Omega)}^{\theta} |v|{L^{q}(\Omega)}^{1-\theta} ] holds with suitable (\theta\in(0,1)). In particular, the restriction of (T{\Omega}) to the compact set [ K:={v\in L^{q}(\Omega):|v|_{L^{q}(\Omega)}\le C} ] is compact (the composition of a compact embedding (L^{q}(\Omega)\hookrightarrow L^{p}(\partial\Omega)) with the bounded inclusion (W^{1,2}\hookrightarrow L^{q})).

Consequently, [ T_{\Omega}u_{k}\longrightarrow T_{\Omega}u\quad\text{strongly in }L^{p}(\partial\Omega). ]

Thus every bounded sequence in (W^{1,2}(\Omega)) has a subsequence whose traces converge in (L^{p}(\partial\Omega)). This proves:

Lemma. Let (\Omega\subset\mathbb{R}^{n}) be a bounded Lipschitz domain. For any (1\le p<p^{*}) the trace operator [ T_{\Omega}:W^{1,2}(\Omega)\to L^{p}(\partial\Omega) ] is compact.

2.3 Transfer to a compact manifold

Now let (M) be a compact Riemannian manifold with (smooth) boundary (\partial M).

  1. Cover the boundary by coordinate charts.
    Because (\partial M) is compact, we can choose finitely many boundary charts
    [ \Phi_{j}:U_{j}\longrightarrow B^{+}{r}\subset\mathbb{R}^{n}{+}, \qquad j=1,\dots,N, ] where each (U_{j}) is an open neighbourhood of a piece of (\partial M) and
    (B^{+}{r}) denotes a half‑ball in the Euclidean half‑space.
    The maps (\Phi    {j}) are (C^{\infty}) diffeomorphisms with uniformly bounded Jacobians because the boundary is smooth (Lipschitz is enough).

  2. Partition of unity.
    Choose ({\psi_{j}}{j=0}^{N}\subset C^{\infty}(M)) such that
    (\sum    {j=0}^{N}\psi_{j}\equiv 1) on (M), each (\psi_{j}) has support in (U_{j}), and (\psi_{0}) is supported away from the boundary.

  3. Localisation of a bounded sequence.
    Let ({u_{k}}\subset W^{1,2}(M)) be bounded. Then each product (\psi_{j}u_{k}) belongs to (W^{1,2}(U_{j})) and the norm [ |\psi_{j}u_{k}|{W^{1,2}(U{j})}\le C|u_{k}|_{W^{1,2}(M)} ] with a constant independent of (k) and (j).

  4. Apply the Euclidean lemma.
    For (j\ge 1) (the charts intersecting the boundary) pull back by (\Phi_{j}): [ v_{k}^{(j)} := (\psi_{j}u_{k})\circ\Phi_{j}^{-1}\in W^{1,2}(B^{+}{r}). ] The previous Lemma tells us that, after passing to a subsequence (still denoted (k)), the traces [ T{j}v_{k}^{(j)}\ \longrightarrow\ w^{(j)}\quad\text{in }L^{p}\bigl(\partial B^{+}{r}\bigr). ] Translating back to the manifold, [ (\psi{j}u_{k})|{\partial M}\ \longrightarrow\ (\psi{j}w)\ \text{in }L^{p}(\partial M) ] for some limit function (w^{(j)}) defined on the corresponding piece of (\partial M).

  5. The interior part is irrelevant.
    The term (\psi_{0}u_{k}) vanishes on (\partial M); therefore it does not affect the trace.

  6. Reassemble the pieces.
    Because the covering is finite, we can extract a single subsequence that works simultaneously for all (j=1,\dots,N).
    On the whole boundary, [ u_{k}|{\partial M}= \sum{j=1}^{N}\bigl(\psi_{j}u_{k}\bigr)|{\partial M} ] converges in (L^{p}(\partial M)) to (\sum{j=1}^{N} w^{(j)}).

Thus every bounded sequence in (W^{1,2}(M)) has a subsequence whose traces converge in (L^{p}(\partial M)). This is precisely the definition of a compact operator. Hence the trace map is compact for all sub‑critical exponents.

2.4 Remarks on the regularity of the boundary

  • The argument only uses the Lipschitz regularity of the boundary charts: we need the trace theorem and the Rellich–Kondrachov theorem on each chart, both of which are valid for bounded Lipschitz domains.

  • Consequently, the statement remains true if (\partial M) is merely Lipschitz (or piecewise (C^{1}) with uniformly bounded Lipschitz constants). Smoothness of the boundary is not required for compactness, although it guarantees the existence of smooth charts and simplifies the exposition.

3. Final answer

For a compact (n)-dimensional Riemannian manifold (M) with (Lipschitz) boundary (\partial M) and for every exponent

[ 1\le p<p^{*}:=\frac{2(n-1)}{n-2}\qquad (n\ge 3) ]

the trace operator

[ \operatorname{Tr}:W^{1,2}(M)\longrightarrow L^{p}(\partial M),\qquad u\mapsto u|_{\partial M}, ]

is a compact linear map. The same holds for all finite (p) when (n=2). The only regularity required on the boundary is Lipschitz (in particular (C^{1}) is more than enough).

4. Common mistakes

Mistake Why it is wrong How to avoid it
Confusing continuity with compactness. Believing that because the trace operator is bounded it must automatically be compact. Compactness is a stronger property: bounded sets must be sent to relatively compact (pre‑compact) sets. Explicitly use Rellich–Kondrachov to obtain strong convergence of a subsequence inside the domain, then pass to the boundary via the trace.
*Using the critical exponent (p^{}) in the compactness claim.** The trace is not compact for (p=p^{*}). At the critical exponent one can construct “bubbling” sequences that  

Original question: Why is trace embedding $W^{1,2}(M) \hookrightarrow L^p(\partial M)$ compact on Mathematics Stack Exchange, licensed CC BY-SA.

More Math Problems