0

I want to prove that $n>kp$ implies $W^{k,p}(\mathbb R^n)\subset L^q(\mathbb R^n)$ continuously for all $p\leq q\leq \infty$.

Idea of proof:

According to Wikipedia, if $n>kp$ then $W^{k,p}(\mathbb R^n)\subset C^{r,\alpha}(\mathbb R^n)$ for some $\alpha,r$ such that $\frac{1}{p}-\frac{k}{n}=-\frac{r+\alpha}{n}$ with $\alpha\in (0,1)$.

Can we say that if $f\in W^{k,p}(\mathbb R^n)\subset C^{r,\alpha}(\mathbb R^n)$ then $f\in L^\infty(\mathbb R^n)$? That is, do we have $\lVert f\rVert_{L^\infty(\mathbb R^n)}\leq c\lVert f\rVert_{C^{r,\alpha}(\mathbb R^n)}$ for some $c>0$?

Then we could apply Holder's inequality to prove that $\lVert f\rVert_{L^q(\mathbb R^n)}\leq \lVert f\rVert^\theta_{L^p(\mathbb R^n)}\lVert f\rVert_{L^\infty(\mathbb R^n)}^{1-\theta}\leq C\lVert f\rVert_{W^{k,p}(\mathbb R^n)}^\theta c\lVert f\rVert_{C^{r,\alpha}(\mathbb R^n)}^{1-\theta}\leq cC\lVert f\rVert_{W^{k,p}(\mathbb R^n)}$.

schrodingerscat
  • 421

1 Answers1

1

The embedding you are looking for does NOT hold for the full range of parameters $q$.

In your recent question on the case $n = kp$, I had given you a reference (Demengel's book) which states and proves all cases in Theorem 2.31, including the case of this new question for $n > kp$. I encourage you to read it again.

When $n>kp$, the embedding $W^{k,p}(\mathbb{R}^n)\hookrightarrow L^q(\mathbb{R}^n)$ is only true for $p \leq q \leq \frac{np}{n-kp}$. For example, with $n = 2$, $k = 1$, $p = 1$, you only have $W^{1,1}(\mathbb{R}^2) \hookrightarrow L^q(\mathbb{R}^2)$ for $1 \leq q \leq 2$, but not above. This is easily seen by scaling arguments (considering functions of the form $f(\lambda x)$ for a family of $\lambda \to 0$ or $\lambda \to \infty$). You can read Remark 2.34 in the same reference which explains this more thoroughly.

cs89
  • 3,341