= Solution
A four-dimensional vacuum Einstein solution with the fixed <cosmological constant> $\Lambda$ has $R_{ab}=\Lambda g_{ab}$ and $R=4\Lambda$, constant. Thus every derivative of $f'(R)$ vanishes and
$$
E_{ab}=\left(\Lambda f'(4\Lambda)-\frac12f(4\Lambda)\right)g_{ab}.
$$
The <metric tensor> is nondegenerate, so this is zero precisely when
$$
\boxed{f(4\Lambda)=2\Lambda f'(4\Lambda).}
$$
This is the necessary and sufficient <Einstein metric condition in f(R) gravity> for the specified $\Lambda$, and works for every such Einstein metric, even when its <Weyl tensor> is nonzero. There is no need to divide by $f'(4\Lambda)$; the degenerate case where both $f$ and $f'$ vanish at that curvature is included. At $\Lambda=0$, the condition is simply $f(0)=0$.
If the intention is to demand the inclusion for every real $\Lambda$ simultaneously, the stronger functional condition is $Rf'(R)=2f(R)$ for all $R$. On each nonzero half-line it integrates to $f(R)=CR^2$; smoothness across zero makes the constants equal. Thus the all-$\Lambda$ version gives \b[$f(R)=CR^2$], including $C=0$. This stronger reading is distinct from fixing one cosmological constant.
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