PomeLemma 16.28.18.50B3044AA630 For $0<\alpha <\beta $, one has on $\mathbb C^ n$
11BD756B4F41
16.28.19
\begin{equation} (\beta -\alpha )V_ K^*\leq F_\beta -F_\alpha \leq \frac{\beta -\alpha }{\alpha }(F_\alpha -m). \end{equation}
Lemma 16.28.1850B3044AA630
Proof. If $u$ is a competitor for $V_\alpha =V_{\alpha ,\psi }$ and $v$ is a competitor for $V_ K=V_{1,0}$, then $u+(\beta -\alpha )v$ is a competitor for $V_\beta $. Taking the suprema over $u$ and $v$ independently gives
\[ V_\beta \geq V_\alpha +(\beta -\alpha )V_ K. \]
By proposition 16.28.10, the raw envelopes agree almost everywhere with their regularizations, so this inequality holds almost everywhere for the regularizations, and hence everywhere by lemma 5.24.3. This is the first inequality. The same argument applied to (16.28.15) gives the second. $\square$
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