Week 6: Poincaré inequality, difference quotients and the trace operator

Poincaré inequality

The following inequality is an important consequence of the Rellich-Kondrachov inequality. Given $f\in L^{1}(\Omega)$, write $(f)_{\Omega}:=\frac{1}{\mu(\Omega)}\int_{\Omega}f$.

Theorem 6.1 (Poincaré inequality). Let $\Omega$ be a connected bounded domain of class $C^{1}$. For any $p\in[1,\infty]$ there exists a constant $C>0$ depending only on $n$, $p$ and $\Omega$ such that whenever $u\in W^{1,p}(\Omega)$,

Proof. Suppose this is not the case so that for all $k\in\mathbb{N}$, there exists a $u_{k}\in W^{1,p}(\Omega)$ such that

$||u_{k}-(u_{k})_{\Omega}||_{p}>k\cdot ||\D u_{k}||_{p}.$

Define $v_{k}:=\frac{u_{k}-(u_{k})_{\Omega}}{||u_{k}-(u_{k})_{\Omega}||_{p}}$ for each $k\in \mathbb{N}$. Clearly $||v_{k}||_{p}=1$, $(v_{k})_{\Omega}=0$, and

$||\D v_{k}||_{\Omega}=\frac{1}{||u_{k}-(u_{k})_{\Omega}||_{p}}\cdot ||\D u_{k}||_{p}<\frac{1}{k}.$

In particular, $\{v_{k}\}_{k=1}^{\infty}$ is a bounded sequence in $W^{1,p}(\Omega)$ so that by the Rellich-Kondrachov theorem, we may pass to a subsequence convergent in $L^{p}(\Omega)$, which we again denote by $\{v_{k}\}$. Set $v=L^{p}\lim_{k\rightarrow\infty}v_{k}$. We immediately obtain that $||v||_{p}=1$ and $(v)_{\Omega}=0$. On the other hand, if $\varphi\in C_{0}^{\infty}(\Omega)$,

$\left|\int_{\Omega}\partial_{i}\varphi\cdot v_{k} \right|=\left|\int_{\Omega}\varphi\cdot \partial_{i}v_{k}\right|\leq \frac{1}{k}\cdot \int_{\Omega}|\varphi|\xrightarrow{k\rightarrow\infty}0,$

i.e. $\int_{\Omega}\partial_{i}\varphi\cdot v=0$ so that $v\in W^{1,p}(\Omega)$ with $\D v\equiv 0$. Since $\Omega$ is connected (exercise 3 of tutorial sheet 1), we deduce that there exists a constant $C_{0}$ such that $v\equiv C_{0}$. Since $(v)_{\Omega}=0$, we must have $C_{0}=0$ so that $v\equiv 0$, but then this contradicts the statement that $||v||_{p}=1$!

Remark 6.2. In the case of a ball $B(x,r)$, we can get a clearer picture of how the constant in Poincaré's inequality depends on $r$ using scaling techniques as follows: If $\Omega=B(0,1)$, then we know that there is a constant $C(n,p)>0$ such that for all $u\in W^{1,p}(B(0,1))$,

Now suppose $v\in W^{1,p}(B(x,r))$ and define $u:B(0,1)\rightarrow\mathbb{K}$ such that

It is not hard to see that $u\in W^{1,p}(B(0,1))$, and its weak derivative is given by $\D u(z)=r\cdot \D v(x+rz)$ for almost all $z\in B(0,1)$. Now, the Poincaré inequality (*) holds for $u$, and it is easily seen by changing variables in the various integrals that $(u)_{B(0,1)}=(v)_{B(x,r)}$, $||u-(u)_{B(0,1)}||_{p,B(0,1)}=r^{-n/p}||v-(v)_{B(x,r)}||_{p,B(x,r)}$ and $||\D u||_{p,B(0,1)}=r^{1-\frac{n}{p}}||\D v||_{p,B(x,r)}$ so that (*) reduces to

i.e. the constant in Poincaré's inequality linearly depends on $r$.

Difference quotients

Suppose $u\in C^{1}(\Omega)$ and recall that for each $x\in \Omega$ and $i\in \{1,\dots,n\}$,

and for any $U\Subset \Omega$, the right-hand limit exists uniformly in $x\in U$. More generally, given $u\in \lloc^{1}(\Omega)$, $i\in\{1,\dots,n\}$ and $h\in\mathbb{R}$, we define the difference quotient $\delta_{i}^{h}u:\Omega\rightarrow \mathbb{K}$ by

It turns out that if $u\in H^{1,p}(\Omega)$, then $\delta_{i}^{h}u\xrightarrow{h\rightarrow 0}\partial_{i}u$ in $\lloc^{p}(\Omega)$. This is a consequence of the following lemma.

Lemma 6.3. Suppose $u\in H^{k,p}(\Omega)$ and $U\Subset\Omega$. The inequality

holds whenever $0<|h|<\dist(U,\partial\Omega)$, and $\lim_{h\rightarrow 0}||\delta_{i}^{h}u-\partial_{i}u||_{k-1,p,U}=0$, i.e. $\delta_{i}^{h}u\xrightarrow{h\rightarrow 0}\partial_{i}u$ in $W^{k-1,p}_{\textup{loc}}(\Omega)$.

Proof. First suppose $u\in C^{k,p}(\Omega)$ and note that for any $\alpha$, $\D^{\alpha}\delta_{i}^{h}u\equiv \delta_{i}^{h}\D^{\alpha}u$ on $U$. Moreover, whenever $v\in C^{1,p}(\Omega)$, we compute that

$|\delta_{i}^{h}v|^{p}=\left|\frac{1}{h}\int_{0}^{1}\frac{\partial}{\partial t}v(x+the_{i})dt\right|^{p}\leq \left|\int_{0}^{1}|\partial_{i}v|(x+the_{i})dt\right|^{p}\leq \int_{0}^{1}|\partial_{i}v|^{p}(x+the_{i})dt,$

where in the last step we used Hölder's inequality. Integrating over $U$ and applying Tonelli's theorem,

$\int_{U}|\delta_{i}^{h}v|^{p}\leq \int_{0}^{1}\int_{U}|\partial_{i}v|^{p}(x+the_{i})dx\ dt= \int_{0}^{1}\left(\int_{U+the_{i}}|\partial_{i}v|^{p}\right)dt\leq \int_{\Omega}|\partial_{i}u|^{p}.$

Therefore, combining the above, we deduce that

$||\delta_{i}^{h}u||_{k-1,p,U}=\left(\sum_{|\alpha|\leq k-1}\int_{U}|\D^{\alpha}\delta_{i}^{h}u|^{p} \right)^{1/p}\leq \left(\sum_{|\alpha|\leq k-1}\int_{\Omega}|\partial_{i}\D^{\alpha}u|^{p} \right)^{1/p}\leq ||u||_{k,p,\Omega}.\tag{\#}$

Now suppose $u\in H^{k,p}(\Omega)$ and $\{u_{m}\}\subset C^{k,p}(\Omega)$ is a sequence having $u$ as its $W^{k,p}$-limit. Applying (#) to $\{u_{m}\}$ and taking the limit $m\rightarrow\infty$ implies the first claim. To establish the second claim, we use the triangle inequality to deduce that

$||\delta_{i}^{h}u-\partial_{i}u||_{k-1,p,U}\leq ||\delta_{i}^{h}(u-u_{m})||_{k-1,p,U} + ||\delta_{i}^{h}u_{m}-\partial_{i}u_{m}||_{k-1,p,U} + ||\partial_{i}(u_{m}-u)||_{k-1,p,U}\leq 2 ||u-u_{m}||_{k,p,\Omega} + ||\delta_{i}^{h}u_{m}-\partial_{i}u_{m}||_{k-1,p,U}.$

If we choose $m$ large enough so that $||u- u_{m}||_{k,p,\Omega}\leq \frac{\eps}{2}$, we see that for such $m$,

$||\delta_{i}^{h}u-\partial_{i}u||_{k-1,p,U}\leq \eps + ||\delta_{i}^{h}u_{m}-\partial_{i}u_{m}||_{k-1,p,U}\xrightarrow{h\rightarrow 0}\eps.$

Hence, $\limsup_{h\rightarrow 0}||\delta_{i}^{h}u-\partial_{i}u||_{k-1,p,U}\leq\eps$ for all $\eps>0$, i.e. $\limsup_{h\rightarrow 0}||\delta_{i}^{h}u-\partial_{i}u||_{k-1,p,U}=0$, establishing the claim.

Note that a direct consequence of this lemma is the following: If $u\in H^{k,p}(\Omega)$, then for any $U\Subset\Omega$, $||\delta_{i}^{h}u||_{k-1,p,U}$ is bounded from above, and this bound is independent of $h$ and $U$. We shall show that the converse is true in the case where $p\in\left]1,\infty\right[$. First, we recall an important theorem from functional analysis and one of its consequences.

Theorem 6.4 (Banach-Alaoǧlu). Suppose $X$ is a reflexive Banach space, i.e. the mapping

is an isomorphism. If $\{x_{n}\}\subset X$ is a bounded sequence with $||x_{n}||\leq C$, then it has a weakly convergent subsequence $\{x_{n_{k}}\}_{k=1}^{\infty}$. Moreover, the weak limit $x$ of $\{x_{n_{k}}\}$ satisfies the inequality $||x||\leq C$.

Examples of reflexive Banach spaces include all finite-dimensional spaces and $L^{p}(\Omega)$ for $p\in\left]1,\infty\right[$. We now apply this theorem to deduce some information on bounded sequences in Sobolev spaces.

Lemma 6.5. Suppose $p\in\left]1,\infty\right[$ and $\{u_{m}\}\subset W^{k,p}(\Omega)$ is a bounded sequence (i.e. $||u_{m}||_{k,p}\leq C$ for some $C>0$ independent of $m$) such that $u_{m}\xrightharpoondown{m\rightarrow\infty}u$ in $L^{p}(\Omega)$. Then $u\in W^{k,p}(\Omega)$ and for all $|\alpha|\leq k$, $\D^{\alpha}u_{m}\xrightharpoondown{m\rightarrow\infty}\D^{\alpha}u$ in $L^{p}(\Omega)$.

Proof. Fix $\alpha$ with $|\alpha|\leq k$. The boundedness of $\{u_{m}\}$ in $W^{k,p}(\Omega)$ implies that for each multiïndex $\alpha$ with $|\alpha|\leq k$, $\D^{\alpha}u_{m}$ is bounded in $L^{p}(\Omega)$. By applying Banach-Alaoǧlu, we may pass to a subsequence $\{u_{m_{k}}\}$ such that $\D^{\alpha}u_{m_{k}}\xrightharpoondown{k\rightarrow\infty}v_{\alpha}$ in $L^{p}(\Omega)$. Now, since for any $\varphi\in \cs(\Omega)$

$\int_{\Omega}u_{m_{k}}\cdot \D^{\alpha}\varphi= (-1)^{|\alpha|}\int_{\Omega}\D^{\alpha}u_{m_{k}}\cdot \varphi,$

and the left-hand side (resp. right-hand side) is a bounded linear function of $u_{m_{k}}$ (resp. $\D^{\alpha}u_{m_{k}}$), we may take limits to deduce that

$\int_{\Omega}u\cdot\D^{\alpha}\varphi = (-1)^{|\alpha|}\int_{\Omega}v_{\alpha}\cdot\varphi,$

i.e. $\D^{\alpha}u$ exists and is equal to $v_{\alpha}\in L^{p}(\Omega)$. Since $\alpha$ was arbitrary, we see that $u\in H^{k,p}(\Omega)$. To see that $\D^{\alpha}u_{m}\xrightharpoondown{m\rightarrow\infty}\D^{\alpha}u$ in $L^{p}(\Omega)$, we first note that for $\varphi\in \cs(\Omega)$,

$\int_{\Omega}\D^{\alpha}u_{m}\cdot \varphi =(-1)^{|\alpha|}\int_{\Omega}u_{m}\cdot \D^{\alpha}\varphi\xrightarrow{m\rightarrow\infty}(-1)^{|\alpha|}\int_{\Omega}u\cdot \D^{\alpha}\varphi = \int_{\Omega}\D^{\alpha}u\cdot\varphi.$

Recall that $f\in (L^{p}(\Omega))^{\ast}$ may be written in the form

$f(u)=\int_{\Omega}u\cdot v$

for $v\in L^{q}(\Omega)$. In particular, by the triangle inequality and Hölder's inequality,

$\begin{aligned}\left|\int_{\Omega}\D^{\alpha}u_{m}\cdot v - \int_{\Omega}\D^{\alpha}u\cdot v \right|&\leq \int_{\Omega}|\D^{\alpha}u_{m}|\cdot |v-\varphi| + \left|\int_{\Omega}\D^{\alpha}u_{m}\cdot \varphi - \int_{\Omega}\D^{\alpha}u\cdot \varphi \right| + \int_{\Omega}|\D^{\alpha}u|\cdot |\varphi - v|\\ &\leq 2C||v-\varphi||_{q} + \left|\int_{\Omega}\D^{\alpha}u_{m}\cdot \varphi - \int_{\Omega}\D^{\alpha}u\cdot \varphi \right|.\end{aligned}$

For fixed $\eps>0$, we may choose $\varphi$ such that $||v-\varphi||\leq \frac{\eps}{2C}.$ Taking the $\limsup$ of both sides as $m\rightarrow\infty$ then $\eps\searrow 0$ implies $\D^{\alpha}u_{m}\xrightharpoondown{m\rightarrow\infty}\D^{\alpha}u$ in $L^{p}(\Omega)$.

Corollary 6.6. Let $p\in\left]1,\infty\right[$, $k\in\mathbb{N}$ and $u\in W^{k-1,p}(\Omega)$. If there is a constant $C>0$ such that for all $U\Subset\Omega$, $i\in \{1,\dots,n\}$, $h$ with $0<|h|<\dist(U,\partial\Omega)$

then $u\in W^{k,p}(\Omega)$.

Proof. Let $U\Subset \Omega$. By the Banach-Alaoǧlu theorem, for each $i\in\{1,\dots,n\}$ there exists a sequence $h_{m}\rightarrow 0$ such that $\delta_{i}^{h_{m}}u\xrightharpoondown{m\rightarrow\infty}v_{i}$ in $L^{p}(U)$, and by the preceding lemma $v_{i}\in W^{k-1,p}(U)$ with $\D^{\alpha}\delta_{i}^{h_{m}}\rightharpoondown\D^{\alpha}v_{i}$ in $L^{p}(U)$ so that, in particular, for $\varphi\in C_{0}^{\infty}(U)$ and $|\alpha|\leq k-1$,

$\int_{\mathbb{R}^{n}}\D^{\alpha}v_{i}\cdot \varphi=(-1)^{|\alpha|}\lim_{m\rightarrow\infty}\int_{\mathbb{R}^{n}}\delta_{i}^{h_{m}}u\cdot \D^{\alpha}\varphi=(-1)^{|\alpha|}\lim_{m\rightarrow\infty}\int_{\mathbb{R}^{n}}u\cdot (-\delta_{i}^{-h_{m}}\D^{\alpha}\varphi),$

where in the last step follows from a change of variables. Now, since $\varphi$ is smooth and $u$ locally summable, we may interchange limit and integral to deduce that

$\int_{\mathbb{R}^{n}}\D^{\alpha}v_{i}\cdot \varphi = (-1)^{|\alpha|+1}\int_{\mathbb{R}^{n}}u\cdot \partial_{i}\D^{\alpha}\varphi,$

i.e. $\D^{\alpha+e_{i}}u$ exists weakly on $U$ and is equal to $\D^{\alpha}v_{i}\in L^{p}(U)$. This implies in particular that $u\in W^{k,p}(U)$. Finally, let $\Omega_{\delta}=\{x\in\Omega:\dist(x,\partial\Omega)>\delta\}\cap B(0,\delta^{-1})$ and $|\alpha|=k-1$. The bound

$\int_{\Omega_{\delta}}|\delta_{i}^{h}\D^{\alpha}u|^{p}\leq C^{p}$

implies that $\int_{\Omega_{\delta}}|\partial_{i}\D^{\alpha}u|^{p}\leq C^{p}$, i.e. for all $\alpha$ with $|\alpha|=k$, $\int_{\Omega_{\delta}}|\D^{\alpha}u|^{p}\leq C^{p}$. Taking the limit $\delta\searrow 0$ implies that $||\D^{\alpha}u||_{p,\Omega}\leq C$ so that $u\in W^{k,p}(\Omega)$.

Trace operator

We now turn our attention to the question of `boundary values' of a function $u\in W^{1,p}(\Omega)$ in the case where $\Omega$ is a bounded domain of class $C^{1}$. Since the boundary $\partial\Omega$ is a set of measure zero, the quantity $\left.u\right|_{\partial\Omega}$ is a priori not well defined. We can however make sense of this function using the approximation techniques we have been using.

The following lemma is our starting point and tells us how the $W^{1,p}$ norm of a compactly supported $C^{1}$ function on $\overline{\mathbb{R}^{n}_{+}}$ controls its $L^{p}$ norm on $\partial\mathbb{R}^{n}_{+}$.

Lemma 6.7. Let $u\in C^{1}_{0}(\overline{\mathbb{R}^{n}_{+}})$. The following inequality holds:

Proof. We compute, applying the divergence theorem and Young's inequality:

$\int_{\partial\mathbb{R}^{n}_{+}}|u|^{p} = \int_{\mathbb{R}^{n}_{+}}\partial_{n}|u|^{p}\leq p\cdot \int_{\mathbb{R}^{n}_{+}}|u|^{p-1}|\partial_{n}u|\leq p\cdot \int_{\mathbb{R}^{n}_{+}}(1-\frac{1}{p})|u|^{p} + \frac{1}{p}|\partial_{n}u|^{p}\leq p\cdot \int_{\mathbb{R}^{n}_{+}}|u|^{p} + \sum_{i=1}^{n}|\partial_{i}u|^{p}.$

Remark 6.8. Note that this lemma says that the mapping $C_{0}^{1}(\overline{\mathbb{R}^{n}_{+}})\subset H^{1,p}(\mathbb{R}^{n}_{+})\rightarrow L^{p}(\partial\mathbb{R}^{n}_{+})$ given by $u\mapsto \left.u\right|_{\partial\Omega}$ is continuous. Using the same techniques as Theorem 2.11, it may be shown that $H^{1,p}(\mathbb{R}^{n}_{+})=\overline{C_{0}^{1}(\overline{\mathbb{R}^{n}_{+}})}$ so that by Lemma 4.5, it extends uniquely to a continuous mapping $H^{1,p}(\mathbb{R}^{n}_{+})\rightarrow L^{p}(\partial\mathbb{R}^{n}_{+})$, which we call the trace operator. In particular, if $u\in H_{0}^{1,p}(\mathbb{R}^{n})$, then $u=\lim_{m\rightarrow\infty}u_{m}$ in $W^{1,p}(\mathbb{R}^{n})$ for a sequence $\{u_{m}\}_{m=1}^{\infty}\subset \cs(\mathbb{R}^{n}_{+})$, and by Lemma 4.5,

which is consistent with our intuition.