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Assume $\Omega \subset \mathbb{R}^N$, $ N>4 $ is open.

There is a well-known picone identity that says

Let $u,v \in C^2(\Omega)$ satisfy $v>0$ and $-\Delta v \geq 0$ in $\Omega$. The following functionals, $$L(u,v)= \Big ( \Delta u - \frac{u}{v} \Delta v \Big)^2,$$ $$R(u,v)=|\Delta u|^2 - \Delta v \, \Delta (\frac{u^2}{v} ),$$ verify that $R(u,v) \geq L(u,v) \geq 0 $ in $\Omega$.

Proof comes from simple calculations.

Now using the above identity and by integration by parts we can deduce that

Let $ v \in C^2(\Omega) $ be such that $ v \geq \delta >0 $ in $\Omega$. Then for all $ u \in C_0^{\infty}(\Omega) $,

$$ \int_{\Omega} |\Delta u |^2 \, \mathrm{d}x \geq \int_{\Omega} \frac{\Delta ^2 v}{v} u^2 \, \mathrm{d}x . $$

Now my Question is this: In this article (at pages 10, 11 ) authors improve the above inequality and generalize it not only for $ v \in C^2(\Omega)$ but also for $ v \in W^{2,2}(\Omega) $ (see lemma 3.8). Then By using this improved inequality they prove the the general picone inequality as follow

Theorem 3.9 (Picone inequality) Consider $u,v \in W^{2,2} \cap W_0^{1,2}(\Omega) $ such that $ \Delta ^2 v = \mu $, where $\mu$ is a positive Radon measure, $v= \Delta v = 0 $ on $ \partial \Omega $ and $ v >0 $ in $\Omega$. Then,

$$ \int_{\Omega} |\Delta u |^2 \, \mathrm{d}x \geq \int_{\Omega} \frac{\Delta ^2 v}{v} u^2 \, \mathrm{d}x . $$

Now in the above inequality I can not understand how to interpret the RHS of the inequality. What is the definition of the RHS. Why positive Radon measure comes to it? (why it is has not assumed to be Lebesgue measure). If $ \Delta^2 v $ is not positive Radon measure what happens? $v$ is only two times differentiable in the weak sense but $ \Delta ^2 $ is fourth-order differential operator?

Thanks.

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