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In page 101, corollary 4.3.3., from Armitage and Gardiner's book on potential theory, the authors prove that any subharmonic function, can be identified with a positive measure (Riesz measure).

In doing so, they consider the functional $L_s(\phi)=\int_\Omega s(x)\Delta \phi(x)$, $\Omega$ a open set, $\phi\in C_0^\infty(\Omega)$ and $s$ the subharmonic function.

For any $\phi\in C_0(\Omega)$, he consider the mollified sequence $\phi_n$, which converges to $\phi$ uniformly and claim that $L_s(\phi_n)$ is Cauchy. Maybe, as he says, this is evident, however I fail to see it.

If, for instance the regularized sequence $s_n$, is such that, $\Delta s_n$ is bounded uniformly in $L^1_{loc}(\Omega)$ then, the above is true, however, I also fail to see it. Any help is appreciated.

Remark: I know how to extend $L_s$ to $C_0(\Omega)$, but with another approach, although the extension is the same, because of unicity.

Update: For $x\in \Omega$, fix some $r>0$ such that $\overline{B(x,r)}\subset \Omega$.The regularization $s_n$ of $s$, is a sequence of subharmonic functions in $B(x,r)$ which converges decreasingly to $u$ in $B(x,r)$. Note that (Green's theorem)

\begin{eqnarray} \int_{B(x,r)}|\Delta s_n| &=& \int_{B(x,r)} \Delta s_n \nonumber \\ &=& \int_{\partial B(x,r)} \frac{\partial s_n}{\partial\eta} \nonumber \\ &=& r^{N-1}\frac{\partial}{\partial r}\left(\frac{1}{r^{N-1}}\int_{\partial B(x,r)}s_n\right) \end{eqnarray}

Does anyone knows how to bound the right hand side uniformly in $n$? Also, note that the right hand side, can be written in terms of the average integral of $s_n$.

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This is the "continuity from positivity" trick. First, recall that the authors use $C_0$ to denote the space of compactly supported functions, not those vanishing on the boundary. Since $\phi$ is compactly supported, so are $\phi_n$ for large $n$, and we can take a compact set $K\subset \Omega$ such that all large $n$ the functions $\phi_n$ and $\phi$ vanish outside of $K$.

Let $\chi\in C_0^\infty(\Omega)$ be a nonnegative function such that $\chi\equiv 1$ on $K$. Let $\delta_{nm}=\sup|\phi_n-\phi_m|$. Then $$- \delta_{nm} \chi \le \phi_n-\phi_m\le \delta_{nm} \chi$$ everywhere in $\Omega$. Since $L_s$ is a positive functional, it respects pointwise inequalities; thus $$-\delta_{nm} L_s(\chi) \le L_s(\phi_n)-L_s(\phi_m)\le \delta_{nm} L_s(\chi)$$ Here $L_s(\chi)$ is a fixed number, independent of $m,n$. Since $\delta_{nm}\to 0$ as $m,n\to\infty$, the conclusion follows.

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  • $\begingroup$ I was going so far to prove a simple thing... $\endgroup$ – Tomás Jul 28 '14 at 22:09
  • $\begingroup$ Tomás. I can answer you old question: how to bound the integral. Since the sequence $(s_{n})$ is decreasing, $s_{n}\leq s_{0}$. But how did you go from the 2end equality to the third one? $\endgroup$ – M. Rahmat Jul 1 at 23:55

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