Let $\phi$ be upper semi-continuous function defined on $\Omega\subset \mathbb C^2$.

Let $\Omega_n= \{z \in \Omega: d(z,\partial \Omega) > \frac1n \}$.

Let $\chi\in C_c^\infty$ of $|z_1|$, $|z_2|$ with support contained in $|z|<1$ satisfying $\chi \geq 0$ and $\int \chi d\zeta\wedge d\bar{\zeta} =1$, and set $$\phi_n(z)= \int_{\triangle\times \triangle} \phi\left(z-\frac{\zeta}{n}\right)\chi(\zeta)d\zeta\wedge d\bar{\zeta}$$

Notation is index notation: That is $\zeta= (\zeta_1,\zeta_2)$ and same way $z-\zeta$ is also in index notation.

If i understood correctly the below reference, I need to show that $\phi_n$ is $C^\infty $ in $\Omega_n$.

Reference: Giuseppe Zampieri; Complex analysis and CR geometry, Page number 31. You can see the above things clicking the link.. First line in page 31.

  • $\begingroup$ Something is odd here, should the $\phi$ in the integral defining $\phi_n$ be $f$? $\endgroup$ – mrf May 5 '12 at 12:23
  • $\begingroup$ @mrf what do you mean by $f$. Can you please click on the link and go to the page 31. Thanks a lot. $\endgroup$ – zapkm May 5 '12 at 13:17
  • $\begingroup$ page 31 is not available for me. But your first line says: "Let $f$ be an usc function...", and then $f$ doesn't show up after that. $\endgroup$ – mrf May 5 '12 at 13:20
  • $\begingroup$ oh sorry.. i edited it.. $\endgroup$ – zapkm May 5 '12 at 13:20
  • $\begingroup$ Thanks for looking the question, I edited it.. $\endgroup$ – zapkm May 5 '12 at 13:35

If all you need to know is that $\phi_n$ is $C^\infty$, it follows by differentiating under the integral sign. The general technique of constructing smooth approxations via convolutions with regularizing kernels is very common, and definitely worth mastering.

Assume $\psi \in C_0^\infty$ with support in the ball $\| z \| < 1$ and $f$ almost any function, for example $f \in L^1_{\text{loc}}(\Omega)$. Then $$ f\star \psi(z) = \int f(z-\zeta) \psi(\zeta) \,dV(\zeta)$$ is well defined for $z \in \Omega$ were $\operatorname{dist}(z,\partial\Omega) > 1$. Also $D(f \star \psi) = (Df) \star \psi = f \star (D\psi)$, where $D$ is any partial derivative, in particular, since $\psi$ is smooth, $f \star \psi$ is also smooth.

Furthermore, if $\psi_\varepsilon(\zeta) = \varepsilon^{-n}\psi(\zeta/\varepsilon)$, then $\psi_\varepsilon \to \delta$ in the sense of distributions, and $f\star \psi_\varepsilon \to f$ in $L^1_\text{loc}$ and pointwise almost everywhere. If $\psi$ is chosen positive and radial and $f$ is (pluri-)subharmonic, then the convergence is even monotone.

(See for example: Wikipedia: Mollifier

  • $\begingroup$ Thanks for ur kind reply... I will try to understand it.. and get back to u, if i need.. $\endgroup$ – zapkm May 6 '12 at 3:40

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