Why the constant C in this inequality depends on $q$ but not depend on $u$.
In the proof $C= (\int_{\mathbb{R}^n} |u|^{(\gamma -1)\frac{p}{p-1}}dx )^\frac{p}{p-1}$.
Can anyone explain why it does not depend on $|u|$?
Why the constant C in this inequality depends on $q$ but not depend on $u$.
In the proof $C= (\int_{\mathbb{R}^n} |u|^{(\gamma -1)\frac{p}{p-1}}dx )^\frac{p}{p-1}$.
Can anyone explain why it does not depend on $|u|$?
That is not what the proof tells you. It is derived $$\left(\int |u|^{\frac{\gamma n}{n-1}}\right)^\frac{n-1}{n} \leq \gamma \left( \int |u|^{(\gamma-1)\frac{p}{p-1}} \right)^\frac{p-1}{p} \|Du\|_{L^p}^p.$$ But the proof does not stop here. Now, $\gamma>1$ is chosen s.t. $\frac{\gamma n}{n-1}=(\gamma-1)\frac{p}{p-1}=p^*$, hence it implies $$\left(\int |u|^{p^*}\right)^\frac{n-1}{n} \leq \gamma \left( \int |u|^{p^*} \right)^\frac{p-1}{p} \|Du\|_{L^p}^p.$$ Hence $$\left(\int |u|^{p^*}\right)^{\frac{n-1}{n}-\frac{p-1}{p}} \leq \gamma \|Du\|_{L^p}^p.$$ Notice $\frac{n-1}{n}-\frac{p-1}{p}=\frac{1}{p^*}$, i.e., $C=\gamma$.