Ngô Quốc Anh

May 6, 2012

A note on the Sobolev trace inequality

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 11:38

The purpose of this note is to talk about the following so-called Sobolev trace inequality

\displaystyle {\left( {\int_{\partial M} {|u{|^{\frac{{2(n - 1)}}{{n - 2}}}}d{s_g}} } \right)^{\frac{{n - 2}}{{n - 1}}}} \leqslant (S + \varepsilon )\int_M {|\nabla u{|^2}d{v_g}} + A(\varepsilon )\int_{\partial M} {{u^2}d{s_g}}

where (M,g) is a smooth n-dimensional, compact, Riemannian manifold with a smooth boundary \partial M with n \geqslant 3 and \varepsilon >0. The constant S appearing from the above inequality is called the best constant.

In fact, this is just a weak type of the true Sobolev trace inequality, which can be stated as follows

\displaystyle {\left( {\int_{\partial M} {|u{|^{\frac{{2(n - 1)}}{{n - 2}}}}d{s_g}} } \right)^{\frac{{n - 2}}{{n - 1}}}} \leqslant S\int_M {|\nabla u{|^2}d{v_g}} + A\int_{\partial M} {{u^2}d{s_g}}

where A and S are positive constant. It is know that in order to prove the above Sobolev inequality, a weaker version is needed.

We now talk about its motivation. First, we start with the standard Sobolev inequality appearing when we talk about the the following embedding

\displaystyle H^1(M) \hookrightarrow L^\frac{2n}{n-2}(M).

More precise, the following

\displaystyle {\left( {\int_M {|u{|^{\frac{{2n}}{{n - 2}}}}d{v_g}} } \right)^{\frac{{n - 2}}{{n }}}} \leqslant S\int_M {|\nabla u{|^2}d{v_g}} + B\int_M {{u^2}d{v_g}}

holds. The idea of the Sobolev inequality is the following: if u \in H^1(M), by definition, it is clear to see that u \in L^2(M). One now wishes to find the biggest constant p>2 such that u \in L^p(M). The Sobolev embedding (or the Sobolev inequality) can answer this question. Indeed, for n \geqslant 3, the biggest p is nothing but \frac{2n}{n-2} known as the critical exponent in the Sobolev embedding. It is important to note that in the case n=2, the critical exponent \frac{2n}{n-2} is no longer available, in this case, we have the so-called Trudinger inequality.

Since the embedding H^1(M) \hookrightarrow L^\frac{2n}{n-2}(M) is continuous, by definition, there exists a constant C>0 such that

\displaystyle {\left\| u \right\|_{{L^{\frac{{2n}}{{n - 2}}}}(M)}} \leqslant C{\left\| u \right\|_{{H^1}(M)}}.

By squaring both sides and using the standard norms, we get the above form for the Sobolev inequality.

In the theory of the Sobolev spaces, the trace of any u \in H^1(M) is defined to be Tu \in L^2(\partial M) where T : H^1(M) \to L^2(\partial M) is the unique linear bounded operator independent of u. Again, we wish to find the biggest number p such that u \in L^p(\partial M). Once this procedure is done, by the continuous embedding, the corresponding inequality is called the Sobolev trace inequality. In other words, we wish to find the following

\displaystyle {\left( {\int_{\partial M} {|u{|^p}d{s_g}} } \right)^{\frac{2}{p}}} \leqslant S\int_M {|\nabla u{|^2}d{v_g}} + C\int_M {{u^2}d{v_g}}.

Clearly, in order to find the above Sobolev trace inequality, the following weak version is needed

\displaystyle {\left( {\int_{\partial M} {|u{|^p}d{s_g}} } \right)^{\frac{2}{p}}} \leqslant (S + \varepsilon )\int_M {|\nabla u{|^2}d{v_g}} + C(\varepsilon )\int_M {{u^2}d{v_g}}.

This is equivalent to showing that the integral on the left hand side of the Sobolev inequality can be reduced to the integral on the boundary \partial M. Indeed, it was shown by P. L. Lions that

\displaystyle\frac{1}{S} = \inf \left\{ {\frac{{\displaystyle\int_{\mathbb{R}_ + ^n} {|\nabla u{|^2}d{v_g}} }}{{{{\left( {\displaystyle\int_{\partial \mathbb{R}_ + ^n} {|u{|^{\frac{{2(n - 1)}}{{n - 2}}}}d{s_g}} } \right)}^{\frac{{n - 2}}{{n - 1}}}}}} \quad \bigg| \quad \nabla u \in {L^2}(\mathbb{R}_ + ^n), \quad u \in {L^{\frac{{2(n - 1)}}{{n - 2}}}}(\partial \mathbb{R}_ + ^n)\backslash \{ 0\} } \right\}

is achieved. But this is clear by partition of unity. Moreover, by the work of Lions, we find that p =\frac{2(n-1)}{n-2}.

In the last part of this note, we make use of this weak version to show that one can change the term

\displaystyle C(\varepsilon )\int_M {{u^2}d{v_g}}

to the term

\displaystyle A(\varepsilon )\int_{\partial M} {{u^2}d{s_g}}

in order to get the following weak version

\displaystyle {\left( {\int_{\partial M} {|u{|^{\frac{{2(n - 1)}}{{n - 2}}}}d{s_g}} } \right)^{\frac{{n - 2}}{{n - 1}}}} \leqslant (S + \varepsilon )\int_M {|\nabla u{|^2}d{v_g}} + A(\varepsilon )\int_{\partial M} {{u^2}d{s_g}}.

Indeed, by an argument by contradiction, there exists some \delta >0 such that for all \alpha>1, there holds

\displaystyle {\xi _\alpha } = \mathop {\inf }\limits_{{H^1}(M)\backslash \{ 0\} } \left\{ {\frac{{\displaystyle\int_M {|\nabla u{|^2}d{v_g}} + \alpha \int_{\partial M} {{u^2}d{v_g}} }}{{{{\left( {\displaystyle\int_{\partial M} {|u{|^{\frac{{2(n - 1)}}{{n - 2}}}}d{s_g}} } \right)}^{\frac{{n - 2}}{{n - 1}}}}}}} \right\} \leqslant \frac{1}{S} - \delta .

Next, we can show that \xi_\alpha is achieved by some non-negative function u_\alpha, and by normalizing, there holds

\displaystyle {\xi _\alpha } = \int_M {|\nabla {u_\alpha }{|^2}d{v_g}} + \alpha \int_{\partial M} {u_\alpha ^2d{s_g}} , \quad \int_{\partial M} {|{u_\alpha }{|^{\frac{{2(n - 1)}}{{n - 2}}}}d{s_g}} = 1.

For a proof, we refer the reader to the reference below. Once we have this, we can see that the sequence \{\|u_\alpha\|_{H^1(M)}\}_\alpha is bounded by some constant independent of \alpha. Therefore, we have some weak limit \overline u \in H^1(M). This leads to

\displaystyle\int_M {|{u_\alpha } - \overline u {|^2}d{v_g}} + \int_{\partial M} {|{u_\alpha } - \overline u {|^2}d{s_g}} = o(1).

Using the constraint for u_\alpha and the strong convergence in L^2(\partial M), that is,

\displaystyle 0\leqslant \int_{\partial M} {u_\alpha ^2d{v_g}} \leqslant \frac{{{\xi _\alpha }}}{\alpha } \leqslant \frac{1}{\alpha }\left( {\frac{1}{S} - \delta } \right),

there holds \overline u \equiv 0 on \partial M by sending \alpha \to \infty. In particular,

\displaystyle\int_{\partial M} {|{u_\alpha } - \overline u {|^{\frac{{2(n - 1)}}{{n - 2}}}}d{v_g}} = 1.

On the other hand, by the Brezis-Lieb lemma, there holds

\displaystyle\int_M {|\nabla {u_\alpha }{|^2}d{v_g}} = \int_M {|\nabla ({u_\alpha } - \overline u ){|^2}d{v_g}} + \int_M {|\nabla \overline u {|^2}d{v_g}} + o(1),

which implies that, for each \varepsilon_0>0 and \alpha large,

\displaystyle\begin{gathered} {\xi _\alpha }= \int_M {|\nabla ({u_\alpha } - \overline u ){|^2}d{v_g}} + \int_M {|\nabla \overline u {|^2}d{v_g}} + \alpha \int_{\partial M} {u_\alpha ^2d{v_g}} + o(1) \hfill \\ \qquad\geqslant \int_M {|\nabla ({u_\alpha } - \overline u ){|^2}d{v_g}} + o(1) \hfill \\ \qquad\geqslant \frac{1}{{S + {\varepsilon _0}}}{\left( {\int_{\partial M} {|{u_\alpha } - \overline u {|^{\frac{{2(n - 1)}}{{n - 2}}}}d{s_g}} } \right)^{\frac{{n - 2}}{{n - 1}}}} + o(1) \hfill \\ \qquad= \frac{1}{{S + {\varepsilon _0}}} + o(1), \hfill \\ \end{gathered}

where we have used the weak form of the Sobolev inequality since, due to the strong convergence in L^2(M),

\displaystyle\int_M {|{u_\alpha } - \overline u {|^2}d{v_g}} = o(1).

Sending \alpha to \infty we get

\displaystyle\frac{1}{S} - \delta \geqslant \frac{1}{{S + {\varepsilon _0}}}.

Sending \varepsilon_0 to 0 we get a contradiction.

Regarding to the Sobolev trace inequality for the embedding \displaystyle H^p(M) \hookrightarrow L^p(\partial M), with p<n, the exponent \frac{2(n-1)}{n-2} will be changed to \frac{p(n-1)}{n-p}, that is,

\displaystyle {\left\| u \right\|_{{L^{\frac{{p(n - 1)}}{{n - p}}}}(\partial M)}} \leqslant C{\left\| u \right\|_{{H^p}(M)}}.

It is worth recalling that the critical exponent for the embedding \displaystyle H^p(M) \hookrightarrow L^p(M) with p<n is \frac{pn}{n-p}. The only difference is that the dimensional constant appearing in the denominator, n-p, represents the preimage space H^p(M), while the dimensional constant appearing in the numerator, either pn or p(n-1), represents the image space, either L^p(M) or L^p(\partial M). In this manner, we should remember these inequalities in the following way

\displaystyle {\left\| u \right\|_{{L^{\frac{{p\dim (N)}}{{\dim (M) - p}}}}(N)}} \leqslant C{\left\| u \right\|_{{H^p}(M)}}.

If you want a good reference, the following is a good one: Sharp Sobolev trace inequalities on Riemannian manifolds with boundaries.


  1. Nice post. Keep going Quốc Anh. How is your thesis going?

    Comment by doanchi — May 11, 2012 @ 9:36

    • Thanks doanchi for your interest in my post. The first draft of my thesis had been done, thanks God, I still have almost three months to go before the submission.

      Comment by Ngô Quốc Anh — May 11, 2012 @ 12:16

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