Sobolev conjugate

Sobolev conjugate

The Sobolev conjugate of 1leq p is: p^*=frac{pn}{n-p}>pThis is an important parameter in the Sobolev inequalities.

Motivation

A question arises whether "u" from the Sobolev space W^{1,p}(R^n) belongs to L^q(R^n) for some "q">"p". More specifically, when does |Du|_{L^p(R^n)} control |u|_{L^q(R^n)}? It is easy to check that the following inequality:|u|_{L^q(R^n)}leq C(p,q)|Du|_{L^p(R^n)} (*)can not be true for arbitrary "q". Consider u(x)in C^infty_c(R^n), infinitely differentiable function with compact support. Introduce u_lambda(x):=u(lambda x). We have that:|u_lambda|_{L^q(R^n)}^q=int_{R^n}|u(lambda x)|^qdx=frac{1}{lambda^n}int_{R^n}|u(y)|^qdy=lambda^{-n}|u|_{L^q(R^n)}^q:|Du_lambda|_{L^p(R^n)}^p=int_{R^n}|lambda Du(lambda x)|^pdx=frac{lambda^p}{lambda^n}int_{R^n}|Du(y)|^pdy=lambda^{p-n}|Du|_{L^p(R^n)}^pThe inequality (*) for u_lambda results in the following inequality for u:|u|_{L^q(R^n)}leq lambda^{1-p/n+q/n}C(p,q)|Du|_{L^p(R^n)}If 1-n/p+n/q ot = 0, then by letting lambda going to zero or infinity we obtain a contradiction. Thus the inequality (*) could only be true for:q=frac{pn}{n-p},which is the Sobolev conjugate.

ee also

*Sergei Lvovich Sobolev


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