Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My question is rather simple. Given $f$ a $L^2(\mathbb{R}^2)$ function with zero mean but supported in the whole plane, is there a bound of the form $$ \|\nabla(\Delta)^{-1} f\|_{L^2(\mathbb{R}^2)}\leq C\|f\|_{L^2(\mathbb{R}^2)}?. $$ Any comment or reference is very welcome!

share|cite|improve this question
may be using Fourier Transform... – Pocho la pantera Sep 3 '13 at 21:16
Consider $f_r(x)=f(rx)$ then $\| f_r\|_{L^2}=r^{-n/2}\| f\|_{L^2}$ and $\| \nabla (\Delta^{-1})f_r\|_{L^2}=r^{-n/2-1}\| \nabla(\Delta^{-1})f\|_{L^2}$. – Jose27 Sep 3 '13 at 21:23
But how can you bound the singularity of the multiplier at zero? Using the zero mean you have $$ \frac{i \xi}{|\xi|^2}\left(\hat{f}(\xi)-\hat{f}(0)\right)= \int_0^1\nabla_\xi\hat{f}(\lambda\xi)d\lambda $$ – guacho Sep 3 '13 at 21:24
Is $f$ compactly supported? – Shuhao Cao Sep 3 '13 at 21:31
@Rafa: $\Delta^{-1} f=u$ is the solution of $\Delta u=f$ so that $u_r=r^{-2}u(rx)=\Delta^{-1}f_r$. Take the gradient and change variables. – Jose27 Sep 3 '13 at 21:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.