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From Evan's Partial Differential Equations in Ch. 5, there is what seems to be a relatively straightforward exercise (5.7) which is a proof of the trace inequality. Here is the statement of the problem:

Assume that $U$ is bounded and there exists a smooth vector field $\alpha$ such that $\alpha\cdot \nu \ge 1$ along $\partial U$, where $\nu$ as usual denotes the outward unit normal. Assume $1\le p <\infty$. Apply the Gauss-Green Theorem to $\int_{\partial U}|u|^{p}\alpha\cdot \nu dS$, to derive a new proof of the trace inequality $$\int_{\partial U}|u|^{p} dS\leq C\int_{U}|Du|^{p}+|u|^{p}dx$$ for all $u\in C^{1}(\bar{U})$.

I can handle the proof relatively easily up until the end. By assumption, since $\alpha\cdot \nu \geq 1$, we have

$$\int_{\partial U}|u|^{p}\alpha\cdot\nu dS \geq \int_{\partial U}|u|^{p}dS$$

Now applying the Gauss-Green Theorem to the left-side of this inequality we obtain

$$\int_{\partial U}|u|^{p}\alpha\cdot\nu dS=\int_{U}div(|u|^{p}\alpha)dx=\int_{U} div(\alpha)|u|^{p}+D|u|^{p}\cdot \alpha dx$$

I am not entirely sure what to do next with this, how does the smoothness of $\alpha$, for example, fit into this proof? Any suggestions on how to proceed would be incredibly helpful.

MrStormy83
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1 Answers1

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Suppose first $1<p<\infty$ so that $|u|^p$ is in $C^1(\overline{U})$. Smoothness of $\alpha$ (which can be defined on all of $\Bbb{R}^n$) allows you to justify the use of the divergence theorem, and since $\alpha$ is a fixed smooth vector field, you have uniform bounds on it and all its first derivatives. So, \begin{align} \int_{\partial U}|u|^p\,dS&\leq\int_{\partial U}|u|^p\alpha\cdot\nu\,dS\\ &=\int_U[\text{div}(\alpha)|u|^p+p|u|^{p-1}\text{sgn}(u)Du\cdot \alpha]\,dV\\ &\leq C(\|\alpha\|_{\infty},\|D\alpha\|_{\infty})\int_U\left(|u|^p+p|u|^{p-1}|Du|\right)\,dV\\ &\leq C(\|\alpha\|_{\infty},\|D\alpha\|_{\infty})\int_{U}\left[|u|^p+p\left(\frac{p-1}{p}|u|^p+\frac{1}{p}|Du|^p\right)\right]\,dV\tag{$*$}\\ &\leq C(\|\alpha\|_{\infty},\|D\alpha\|_{\infty})\int_U\left(p|u|^p+|Du|^p\right)\,dV, \end{align} which completes the proof for $1<p<\infty$. If $p=1$, then you can work through the above derivation with the divergence theorem being applied in a weak sense, or you can (why?) simply take the limit $p\to 1^+$ above… this is why I kept the explicit lack of dependence of the constant $C$ on $p$ (why is this important?). Another way of arguing for $p=1$ is to apply the inequality above for $p=1$ with a ‘regularized’ function $u_{\epsilon}=\sqrt{|u|^2+\epsilon}$, and then taking the limit $\epsilon\to 0^+$. There are several analysis tricks here for dealing with the edge case $p=1$, and you should gain familiarity with as many of them as you can.

For $(*)$, first recall Young’s inequality $ab\leq \frac{a^p}{p}+\frac{b^q}{q}$, where $a,b\geq 0$ and $p,q\in (1,\infty)$ are Holder-conjugate exponents (so $\frac{1}{p}+\frac{1}{q}=1$). Now for any $\xi,\eta\in [0,\infty)$, apply this with $a=\eta$ and $b=\xi^{p-1}$ to get \begin{align} \xi^{p-1}\eta\leq \frac{\eta^p}{p}+\frac{(\xi^{p-1})^q}{q}=\frac{\eta^p}{p}+\frac{p-1}{p}\xi^p. \end{align}

Note that if the Young’s inequality stuff is confusing, then to keep the gist of the argument in mind, focus on $p=2$, then the inequality simply says $ab\leq \frac{a^2+b^2}{2}$ (which is equivalent to the obviously true inequality $(a-b)^2\geq 0$). This means in the above derivation, we simply use $2|u||Du|\leq |u|^2+|Du|^2$.

Having said that, Young’s inequality, and Holder’s inequality are basic inequalities fundamental for real analysis, and as such they should always come to mind as suitable things to try (even if it takes you a while to figure out the right constants/exponents).

peek-a-boo
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