Generalization of Gagliardo-Nirenberg Inequality

The standard Gagliardo-Nirenberg Inequality is $$\Vert u\Vert_{L^{\frac{n}{n-1}}(\mathbb R^n)}\le C_n \Vert \nabla u\Vert_{L^{1}(\mathbb R^n)}, \tag{\ast}$$ and constitutes a key step to proving Sobolev Injection Theorems. Of course we may assume that $$u$$ is smooth and compactly supported, the important point is that the constant $$C_n$$ is dimensional and independent of $$u$$.

Question 1. Are there some improvements of ($$\ast$$) where the rhs is replaced by a quasi-norm of $$\nabla u$$ in weak $$L^1$$?

Question 2. Are there some improvements of ($$\ast$$) where the lhs is replaced by a norm of $$u$$ in the Lorentz space $$L^{\frac{n}{n-1}, 1}$$ or $$L^{\frac{n}{n-1}, p}$$ with $$p?

Question 3. Probably linked to the previous questions: is the following inequality true ? For $$n\ge 3$$, $$\Vert u\Vert_{L^{\frac{n}{n-2}}(\mathbb R^n)}\le C_n \Vert \Delta u\Vert_{L^{1}(\mathbb R^n)}.$$

Question 1 I think the short answer is no. Try testing functions of the form $$|x|^{-n + 1}$$ around $$0$$. Its gradient would be like $$|x|^{-n}$$ and thus in $$L^{1,\infty}$$ but the function is not in the required $$L_p$$ space. The Sobolev inequalities for $$p=1$$ are connected with the isoperimetric inequality, see [S] and are thus very tight. I recall a result similar to what you want but at the other end of the range, i.e. if $$\nabla u \in L^{n,\infty}$$ then $$u$$ is in BMO.

Question 2 Try expressing $$v = \nabla u$$ as $$u = D_n \ast v$$ for some $$\mathbf{R}^n$$-valued distribution $$D_n$$. Then check in which Lorentz spaces $$D_n$$ is and apply the Young's inequalities in Lorentz spaces, as in [N].

Question 3 This would be immediately correct with weak $$n/n-2$$ as a consequence of Hardy-Littlewood-Sobolev inequalities, which hold for every fractional power of $$(-\Delta)$$ i.e.: $$\| \, u \, \|_{L^{\frac{n}{n - \alpha},\infty}} \lesssim \| \, (-\Delta)^{\frac{\alpha}{2}} u \, \|_{L^1}$$ see [V]. When $$\alpha = 2$$, the inequality would still be true without weak bounds after changing $$\Delta$$ by the double gradient $$\nabla^2$$ by iteration of the Gagliardo-Niremberg inequality. I do not think that you can change the $$\nabla^2$$ by a $$\Delta$$ unless the iterated Riesz transforms $$R_i R_j$$ were bounded in $$L^1$$, which they are not. To produce an explicit counterexample i would first show that the inequality if tight for $$\nabla^2$$ ans then use that $$\|\nabla^2 u\|_1$$ and $$\|\Delta u\|_1$$ are not comparable.

[N] Nursultanov, Erlan; Tikhonov, Sergey, Convolution inequalities in Lorentz spaces, J. Fourier Anal. Appl. 17, No. 3, 486-505 (2011). ZBL1235.44012.https://link.springer.com/article/10.1007/s00041-010-9159-9

[S] Saloff-Coste, Laurent, Aspects of Sobolev-type inequalities, London Mathematical Society Lecture Note Series. 289. Cambridge: Cambridge University Press. x, 190 p. (2002). ZBL0991.35002.

[V] Varopoulos, N. Th.; Saloff-Coste, L.; Coulhon, T., Analysis and geometry on groups, Cambridge Tracts in Mathematics 100. Cambridge: Cambridge University Press (ISBN 978-0-521-08801-5/pbk). xii, 156 p. (2008). ZBL1179.22009.> Blockquote

• To answer Q3 (in the negative, as expected): let $f$ be a compactly supported, positive function. Then $u = \triangle^{-1} f$ can be evaluated by convolving against the Newton potential, which is signed when $n \geq 3$. In particular, $u$ behaves asymptotically like $|x|^{2-n}$ and thus fails to be in $L^{n/(n-2)}$, while $f = \triangle u \in L^1$. Note that the $u$ constructed is smooth and decays like $|x|^{2-n}$ so is in weak $L^{n/(n-2)}$ as guaranteed by HLS. – Willie Wong Jul 5 at 19:36