# 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, 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 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.

## 2 Comments »

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

This site uses Akismet to reduce spam. Learn how your comment data is processed.