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Let $\Omega\subset\mathbb{R}^3$ be a bounded Lipschitz-domain. We then have, for $s\in[1,6)$ the compact embedding $H^1(\Omega)\stackrel{c}{\hookrightarrow}L^s(\Omega)$ ensuring the existence of a $C>0$ such that

$$\|u\|_{L^s(\Omega)}\leq C \|u\|_{H^1(\Omega)}\leq C\left(\|u\|_{L^2(\Omega)}+\|\nabla u\|_{L^2(\Omega)^3}\right)$$

for all $u\in H^1(\Omega)$.

I came across a different conclusion in this paper (in the middle of p. 8). This conclusion being: For all $\alpha>0$ there exists $C(\alpha,\Omega)>0$ such that

$$\|u\|_{L^4(\Omega)}\leq \alpha\|\nabla u\|_{L^2(\Omega)^3}+C(\alpha,\Omega)\|u\|_{L^2(\Omega)}$$

for all $u\in H^1(\Omega)$.

My questions: 1) Why is this statement true? I fear that this is trivial, but I fail to come up with a justification. Maybe a hint or a good reference would already do the trick for me.

2) Beyond that: Is it possible to determine the order of (an optimal) $C(\alpha,\Omega)$ w.r.t. $\alpha$? (Possibly something like $C(\alpha,\Omega)= \mathcal{O}(\alpha^{-1})$ when fixing $\Omega$).

Muschkopp
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1 Answers1

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  1. It's a lemma of J.-L. Lions, see An application of J.-L. Lion's Lemma where a book reference is given (Brezis, Functional Analysis...).

  2. You don't get quantitative control of $C$ from the compactness argument by which the lemma is usually proved. Instead, we can use the Sobolev embedding with the Peter-Paul trick as here. Below I write $\|u\|_p$ instead of $\|u\|_{L^p}$ to improve readability. First some preliminaries:

Fact 1. Given exponents $1\le p<q<r \le \infty$, let $\lambda \in (0,1)$ be such that $$\frac{1}{q} = \frac{\lambda }{p}+\frac{1-\lambda }{r}$$ Then for every measurable $f$ we have $$\|f\|_q\le \|f\|_p^\lambda \|f\|_r^{1-\lambda}$$

Source: Folland's Real Analysis, Prop. 6.10. This is immediate from Hölder's inequality.

Fact 2. If $p^{-1}+q^{-1}=1$ and $\epsilon>0$, then for any positive $a,b$ we have $$ab\le \epsilon \frac{a^p}{p} + \epsilon^{-q/p}\frac{b^q}{q}$$

Indeed, this is just Young's inequality applied to $\epsilon^{1/p}a$ and $\epsilon^{-1/p}b$.

Fact 3. [Sobolev inequality] $\|u\|_{2^*} \le C (\|u\|_2 + \|\nabla u\|_2)$ where $2^*=2N/(N-2)$ and $N$ is the dimension.

Now to the proof. Given $s\in (2,2^*)$, let $\lambda = 1-N\dfrac{s-2}{2s}$, so that Fact 1 applies: $$\|u\|_s\le \|u\|_2^\lambda \|u\|_{2^*}^{1-\lambda}$$ From Fact 2, $$\|u\|_s\le \epsilon (1-\lambda)\|u\|_{2^*} + \epsilon^{(\lambda-1)/\lambda} \lambda \|u\|_2 $$ From Fact 3, it follows that
$$\|u\|_s\le C\epsilon (1-\lambda)\|\nabla u\|_{2} + (C \epsilon (1-\lambda) + \epsilon^{(\lambda-1)/\lambda} \lambda) \|u\|_2 \tag{1}$$

With $N=3$ and $s=4$, we have $\lambda=1/4$. Hence, (1) yields the estimate $$\|u\|_{4}\leq \alpha \|\nabla u\|_{2}+C(\alpha,\Omega)\|u\|_{2}$$ with $C(\alpha,\Omega)=O(\alpha^{-3})$.

Danilo Gregorin Afonso
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user127096
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