e/Sobolev conjugate

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has gloss	eng: The Sobolev conjugate of p for 1\leq p <n, where n is space dimensionality, is : p^=\fracpn}n-p}>p This 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=\frac1}\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)}^p The inequality (*) for u_\lambda results in the following inequality for u :\\|u\\|_L^q(R^n)}\leq \lambda^1-n/p+n/q}C(p,q)\\|Du\\|_L^p(R^n)} If 1-n/p+n/q\not = 0, then by letting \lambda going to zero or infinity we obtain a contradiction.
lexicalization	eng: Sobolev conjugate
instance of	e/Sobolev space