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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 $f^{\prime }(R)$ vanishes andThe metric tensor is nondegenerate, so this is zero precisely whenThis 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 $f^{\prime }(4\Lambda )$; the degenerate case where both $f$ and $f^{\prime }$ 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 $R{f}^{\prime }(R)=2f(R)$ for all $R$. On each nonzero half-line it integrates to $f(R)=C{R}^2$; smoothness across zero makes the constants equal. Thus the all-$\Lambda$ version gives $f(R)=C{R}^2$, including $C=0$. This stronger reading is distinct from fixing one cosmological constant.