Three-dimensional Sobolev inequality

Statement

For $u\in H^1(\mathbb R^3)=W^{1,2}(\mathbb R^3)$ ,

\|u\|_6\le C\|\nabla u\|_2.

The norms are whole-space Lebesgue norms. The same estimate holds for a vector field, with a dimensional constant.

An elementary route

For real $f\in C_c^1(\mathbb R^3)$ , the fundamental theorem of calculus gives $|f(x,y,z)|\le A_1(y,z)=\int|\partial_1f(s,y,z)|\,ds$ , and analogously $A_2(x,z),A_3(x,y)$ . Hence $|f|^{3/2}\le(A_1A_2A_3)^{1/2}$ . Integrating first in $x$ and using Cauchy–Schwarz, then in $(y,z)$ and using it again, yields

\|f\|_{3/2}\le\prod_{j=1}^3\|\partial_jf\|_1^{1/3}\le C\|\nabla f\|_1.

Apply this to $f=|u|^4$ . Hölder gives $\|u\|_6^4\le C\|u\|_6^3\|\nabla u\|_2$ ; divide if $u\ne0$ . Smooth compact-support approximation extends the inequality to $H^1$ .

References

Hunter, Sobolev Spaces, §§3.5–3.7.