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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 $p<n/(n-1)$?

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

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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

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    $\begingroup$ 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. $\endgroup$ – Willie Wong Jul 5 at 19:36

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