Equivalence of logarithm-convex.

‎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 ‎\begin{align*}‎‎ ‎f^{z-x}(y)\leq f^{z-y}(x) f^{y-x}(z). ‎\end{align*}

The two are equivalent. If $$x we can write $$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 and $$\lambda ,u >0$$ with $$\lambda +u =1$$ take $$y=\lambda z+(1-\lambda)x$$. The $$x and we can use the second definition to get the first (with $$z$$ in place of $$y$$).
• 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 ‎‎$‎x<y<z‎$. Apr 23 '19 at 8:46
• @koohyareslami With the way I defined $\lambda$ verify that $y=\lambda z+(1-\lambda)x$. Apply the definition and rise both sides to power $z-y$. Apr 23 '19 at 8:48