Sobolev embedding into $L^\infty$

Question

I heard that $W^{n,1}(\mathbb R^n)\hookrightarrow L^\infty(\mathbb R^n)$.

I can only prove that $W^{1,1}(\mathbb R)\hookrightarrow L^\infty(\mathbb R)$ by Newton-Lebniz formula, how to prove for general $n$?

thanks!

Answer 1

If $u\in C_0^\infty(\mathbb{R}^n)$, then $$u(x_1,\cdots,x_n)=\int_{-\infty}^{x_1}\cdots \int_{-\infty}^{x_n} \frac{\partial ^n u(y_1,\cdots,y_n)}{\partial y_n\cdots \partial y_1}dy_n\cdots dy_1,$$

which implies that

$$\tag{1}\|u\|_\infty\le \|u\|_{n,1}.$$

Now, for any $u\in W^{n,1}(\mathbb{R}^n)$, take a sequence $u_i\in C_0^\infty(\mathbb{R}^n)$, which converges to $u$ in $W^{n,1}(\mathbb{R}^n)$. From $(1)$, we have that $$\|u_i-u_j\|_\infty\le \|u_i-u_j\|_{n,1},$$

hence, $u_i$ converge to $u$ in $L^\infty(\mathbb{R}^n)$ and $$\|u\|_\infty\le \|u\|_{n,1}.$$