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It is a classical result due to Gagliardo and Nirenberg that there exists a constant C such that it holds $$ ||\nabla \psi|| _{L ^\infty (\mathbb{R}^2)} ^2 \le ||D ^2 \psi|| _{L ^\infty (\mathbb{R}^2)} ||\psi|| _{L ^\infty (\mathbb{R}^2)}, $$ for any $ \psi \in C ^2 (\mathbb{R} ^2) $.

My question would be if anyone knows if it is possible to obtain the same result or a similar one with the laplacian instead of the second derivative, namely if it holds that $$ ||\nabla \psi|| _{L ^\infty (\mathbb{R}^2)} ^2 \le ||\Delta \psi|| _{L ^\infty (\mathbb{R}^2)} ||\psi|| _{L ^\infty (\mathbb{R}^2)}. $$

Any help or reference will be greatly appreciated!

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By the classical regularity theory for the Poisson equation, you have $$ \Vert \nabla \psi \Vert_{L^\infty (B_1)} \le C \bigl(\Vert \Delta \psi \Vert_{L^\infty (B_2)} + \Vert \psi \Vert_{L^\infty (B_2)} \bigr). $$ See for example Gilbarg and Trudinger, Elliptic partial differential equations of second order, 1983, theorem 3.9, where it is proved by the maximum principle. (An alternative would be to use a Green representation formula of the solution in a ball.) By scaling this gives $$ R\Vert \nabla \psi \Vert_{L^\infty (B_R)} \le C \bigl(R^2 \Vert \Delta \psi \Vert_{L^\infty (B_{2 R})} + \Vert \psi \Vert_{L^\infty (B_{2 R})} \bigr). $$ By translating the estimate, you obtain thus for every $R > 0$, $$ R\Vert \nabla \psi \Vert_{L^\infty (\mathbb{R}^n)} \le C \bigl(R^2 \Vert \Delta \psi \Vert_{L^\infty (\mathbb{R}^n)} + \Vert \psi \Vert_{L^\infty (\mathbb{R}^n)} \bigr). $$ By taking $R = \sqrt{\Vert \psi \Vert_{L^\infty (\mathbb{R}^n)} / \Vert \Delta \psi \Vert_{L^\infty (\mathbb{R}^n)}}$, we conclude that $$ \Vert \nabla \psi \Vert_{L^\infty (\mathbb{R}^n)} \le 2C \Vert \Delta \psi \Vert_{L^\infty (\mathbb{R}^n)}^{1/2} \Vert \psi \Vert_{L^\infty (\mathbb{R}^n)}^{1/2}. $$

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