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Suppose that $\Delta f\in L^1(\mathbb{R}^n)$, where $\Delta$ denotes the Laplacian. Then I’m asking whether or not we have the following estimates $$ \|\nabla|f|^{\frac12}\|_{L^2}\leq C\|\Delta f\|_{L^1}^{\frac12}. $$ i.e., does the mapping $f\to |f|^{\frac12}$ send functions from $\dot{W^{2,1}}$ to $\dot{W^{1,2}}$ ?

It seems that a paper of W. Sickel (Boundedness properties of the mapping $f\to |f|^{\mu}$, $0 <\mu< 1$ in the framework of Besov spaces) has shown that the inhomogeneous version of the inequality above is indeed true. However I'm not able to get that paper for now, and more importantly, I'm not sure whether the method there will also work for the case I'm concerning.

Any comment is welcome.

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  • $\begingroup$ I don't think this can be true. Take $f$ harmonic (and bounded, say) on a large subset of $\mathbb R^n$. $\endgroup$
    – Christian Remling
    Commented Mar 1, 2017 at 16:16
  • $\begingroup$ @ChristianRemling, Thanks, But the inhomogeneous version is true according to Sickel's paper I mentioned, which sounds a little bit strange. $\endgroup$
    – Tomas
    Commented Mar 2, 2017 at 2:59
  • $\begingroup$ @anonymous, I don't think so. I think both sides scale like $\lambda$ $\endgroup$
    – Tomas
    Commented Mar 2, 2017 at 5:57
  • $\begingroup$ In your case, the right hand side also has $\lambda^{\frac12}$. I thought you mean the fowllowing scaling: $f(x)\to f(\lambda x)$, then both sides scale like $\lambda$ $\endgroup$
    – Tomas
    Commented Mar 2, 2017 at 8:23
  • $\begingroup$ (I've removed my superfluous comments): I claimed earlier that the two sides scaled differently but they do not; @Hannes was kind enough to point out my error to me, thanks for that! $\endgroup$
    – anonymous
    Commented Mar 2, 2017 at 16:03

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