respectively. The Dyson equation for $F_L$ is UV divergent if for $p/\mu >> 1$, and $d(p) \ge \log^{1/2}(p^2)$. This divergence can be removed by the Coulomb operator renormalization constant. The renormalized equation is obtained from the once-subtracted equation $F_\L(\l) - F_{\L_0}(\l_0)$. For example, if the subtraction is chosen at $|\l_0|  = \mu$ and $\L_0 = {\bf 0}$, the renormalized coupling $F_0(\mu)$ can be fixed from the Coulomb potential. After integrating [Eq. (\[qq\])]{} (over $\x$) one obtains $\delta(\y_1 -
   \y_2)$ multiplying $E(\z_1, \cdots, \y_2)$. Therefore, it follows from [Eq. (\[ft\])]{} that $V_C(\R)$ is determined by $F_0(\l)$ and $F_0(\l) = F_0(l) = f(l)$ with $f$ defined in [Eq. (\[ff\])]{}.

In [Eq. (\[F\])]{} $\L$ is a parameter, [*i.e.*]{} the Dyson equation does not involve self-consistency in $\L$. We have just shown that as $\L \to {\bf 0}$, $F_\L(\l)$ has a finite limit: it is given by $f$. For large $L=|\L|$ ($L/\mu >> 1$), due to asymptotic freedom, $F_\L$ is expected to vanish logarithmically, $F_\L \to d^2(L) \propto 1/\log(L^2)$. We do not attempt here to solve [Eq. (\[F\])]{}, instead we use a simple interpolation formula between the $L=0$ and $L\to \infty$ limits,

$$F_\L(\l) = 
 f(\l) \theta(\mu - |\l|) \theta(\mu -|{\L\over 2}|) + \left[ 1 