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 $n\ge 3$, $$ \Vert u\Vert_{L^{\frac{n}{n-2}}(\mathbb R^n)}\le C_n \Vert \Delta u\Vert_{L^{1}(\mathbb R^n)}. $$


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

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.