Let $u\in H^{1}(\mathbb R).$

Is Gagliardo–Nirenberg interpolation inequality valid for the $p=3, q=r=2, m=1, 0< \alpha < 1$ ; and $j=0$ ? That is, is it true that,$\|u\|_{L^{3}(\mathbb R)} \leq C \|Du\|_{L^{2}(\mathbb R)}^{\alpha} \|u\|_{L^{2}(\mathbb R)}^{1-\alpha}$ ?

[My confusion is in the above link if they take $j=0,$ then they are assuming that $q=\infty$, is it necessarry]


1 Answer 1


Well, I don't think that $j=0$ implies $q=\infty$. In any case, you can do interpolation in $L^p$ spaces and then Sobolev embedding

$$ \|u\|_{L^3}^3\leq \|u\|_{L^2}^2\|u\|_{L^\infty}\leq C\|u\|_{L^2}^{2.5}\|\partial_x u\|_{L^2}^{0.5}. $$

You can also do Sobolev embedding and then interpolation between Sobolev spaces:

$$ \|u\|_{L^3}\leq C\|u\|_{H^{1/6}}\leq C\|u\|_{L^2}^{5/6}\|u\|_{H^1}^{1/6} $$

  • $\begingroup$ thanks a lot; I am curious to know how does $\|u\|^{3}_{L^{3}} \leq C \|u\|_{L^{2}}^{2.5}\|Du\|_{L^{2}}^{0.5}$ follows ? Would you please tell me bit more or reference(if it is long)? $\endgroup$ Mar 10, 2015 at 6:08
  • $\begingroup$ The step $\|u\|_{L^{3}}^{3} \leq \|u\|_{L^{2}}^{2} \|u\|_{L^{\infty}}$ is clear to me; after this I don't follow. $\endgroup$ Mar 10, 2015 at 6:19
  • 1
    $\begingroup$ In one dimension you have $$ \|u\|_{L^\infty}^2\leq C\|u\|_{L^2}\|D u\|_{L^2} $$ $\endgroup$
    – guacho
    Mar 10, 2015 at 15:56

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