I see that there are a couple of topics on this question, but neither of them has the answer I am looking for.

In general the goal is to prove the Morrey's inequality $\|u\|_{C^{0,\gamma}(\mathbb{R}^n)}\leq C\|u\|_{W^{1,p}(\mathbb{R}^n)}$, $\gamma=1-\frac{n}{p}$ which is equivalent to $\sup_{{\mathbb{R}}^n}|u|+\sup_{x\neq y}\left\{ \frac{|u(x)-u(y)|}{|x-y|^{1-\frac{n}{p}}}\right\}\leq C\|u\|_{W^{1,p}(\mathbb{R}^n)}$. And the only part I don't understand is the following:

The proof comes to the conclusions that:

(1) $\sup_{{\mathbb{R}}^n}|u|\leq C\|u\|_{W^{1,p}(\mathbb{R}^n)}$


(2) $[u]_{C^{0,1-\frac{n}{p}}(\mathbb{R}^n)}=\sup_{x\neq y}\left\{ \frac{|u(x)-u(y)|}{|x-y|^{1-\frac{n}{p}}}\right\}\leq C\|Du\|_{{L^p}(\mathbb{R}^n)}$.

And then it says that (1) and (2) lead to the desired result. I assume that (1) and (2) are summed, but in this case, we would obtain

$\sup_{{\mathbb{R}}^n}|u|+\sup_{x\neq y}\left\{ \frac{|u(x)-u(y)|}{|x-y|^{1-\frac{n}{p}}}\right\}\leq C\|u\|_{W^{1,p}(\mathbb{R}^n)}+C\|Du\|_{{L^p}(\mathbb{R}^n)}$, which is not what the Morrey's inequality is.

Could anyone please give me a hint! Thank you.

  • $\begingroup$ $\|Du\|_{L^p(\mathbb R^n)}$ is a part of $\|u\|_{W^{1,p}(\mathbb R^n)}$..... $\endgroup$ – user99914 Feb 25 '16 at 7:58
  • $\begingroup$ Yes, that's true. In this case the right hand side of (1) is the same as the right side of the Morrey's inequality and summation of (1) and (2) will yield a right hand side larger than in Morrey's inequality. As one could infer from the book, this is probably the easiest part of the proof, but I just can't see it :( $\endgroup$ – Candidate Feb 25 '16 at 9:56
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    $\begingroup$ That $C$ in the equation is just any constant.... then you will have $ 2C \|u\|_{W^{1,p}(\mathbb R^n)}$ and call this $2C$ your new $C$. $\endgroup$ – user99914 Feb 25 '16 at 11:15
  • $\begingroup$ OMG, now I realise what a stupid question I asked! Thanks a lot, John Ma :) I promise not to ask questions arising at 1 AM or at least to take a sleep and reconsider them before posting! Thank you again! :) $\endgroup$ – Candidate Feb 25 '16 at 11:43

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