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‎we have: A function ‎$‎f:I‎‎\rightarrow‎‎\mathbb{R}‎^+‎$ ‎is ‎log-convex ‎if ‎and ‎only ‎if‎ ‎\begin{align*}‎‎ ‎f(‎\lambda‎ x + u y)\leq f^{‎\lambda‎}(x) f^u (y) ‎\end{align*}‎‎ ‎for ‎$‎x, y\in I‎$ ‎and ‎‎$‎‎\lambda‎, u>0‎$ ‎with ‎‎$‎‎\lambda +‎ u‎ =‎ ‎1‎$‎.‎

‎Now, my question is:‎‎ why? The reformulation of log-convexity implied by above inequality is equivalent to the following working definition: the function ‎$‎f:I‎‎\rightarrow‎‎\mathbb{R}‎$ ‎is ‎log-convex ‎if ‎and ‎only ‎if ‎for ‎all ‎‎$‎x, y, z\in I‎$ ‎with ‎‎$‎x<y<z‎$‎‎ ‎\begin{align*}‎‎ ‎f^{z-x}(y)\leq f^{z-y}(x) f^{y-x}(z). ‎\end{align*}

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The two are equivalent. If $x <y<z$ we can write $y=\lambda z+(1-\lambda)x$ where $\lambda=\frac {y-x} {z-x}$. Applying the first definition to this we get the second one. [ We get $f(y) \leq f(z)^{\frac {y-x} {z-x}} f(x)^{(1-\frac {y-x} {z-x})}$. Rise to power $z-x$]. Conversely, give, $x<z$ and $\lambda ,u >0$ with $\lambda +u =1$ take $y=\lambda z+(1-\lambda)x$. The $x <y<z$ and we can use the second definition to get the first (with $z$ in place of $y$).

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  • $\begingroup$ Kavi Rama Murthy. Thanks a lot. Excusme, how can I if $f$ is log-convex, then $f$ satisfies ‎‎ ‎\begin{align*}‎‎ ‎f^{z-x}(y)\leq f^{z-y}(x) f^{y-x}(z) ‎\end{align*}‎‎for ‎all ‎‎$‎x, y, z\in I‎$ ‎with ‎‎$‎x<y<z‎$. $\endgroup$ Apr 23 '19 at 8:46
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    $\begingroup$ @koohyareslami With the way I defined $\lambda$ verify that $y=\lambda z+(1-\lambda)x$. Apply the definition and rise both sides to power $z-y$. $\endgroup$ Apr 23 '19 at 8:48
  • $\begingroup$ Kavi Rama Murthy. Thanks $\endgroup$ Apr 23 '19 at 8:53

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