# $\|u\|_{L^{3}(\mathbb R)} \leq C \|Du\|_{L^{2}(\mathbb R)}^{\alpha} \|u\|_{L^{2}(\mathbb R)}^{1-\alpha}$?

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]

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
$$\|u\|_{L^3}^3\leq \|u\|_{L^2}^2\|u\|_{L^\infty}\leq C\|u\|_{L^2}^{2.5}\|\partial_x u\|_{L^2}^{0.5}.$$
$$\|u\|_{L^3}\leq C\|u\|_{H^{1/6}}\leq C\|u\|_{L^2}^{5/6}\|u\|_{H^1}^{1/6}$$
• 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)? Mar 10, 2015 at 6:08
• The step $\|u\|_{L^{3}}^{3} \leq \|u\|_{L^{2}}^{2} \|u\|_{L^{\infty}}$ is clear to me; after this I don't follow. Mar 10, 2015 at 6:19
• In one dimension you have $$\|u\|_{L^\infty}^2\leq C\|u\|_{L^2}\|D u\|_{L^2}$$ Mar 10, 2015 at 15:56