Let $I\subset\mathbb{R}$. A function $f:I\rightarrow\mathbb{R}$, is said to be log-convex if $\log f$ is convex or equivalently for all $x,y\in I$ and $\alpha\in [0,1]$ $$f(\alpha x+(1-\alpha)y)\leq [f(x)]^\alpha [f(y)]^{1-\alpha}.$$ Let $X$ be a vector lattice algebra and $F:X\rightarrow X$. Can we define the log-convexity of $F$ in a similar way? If yes, can someone give me a reference to such definition?
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$\begingroup$ What would the exact statement be? $\|F(\alpha x+(1-\alpha)y)\|\leq\|F(x)\|^\alpha\|F(y)\|^{1-\alpha}$? This means that $N\circ F:X\to[0,\infty)$ is log-convex, where $N(x)=\|x\|$, so it reduces to log-convexity of real valued functions. If you mean something else, what kind of order and powers do you have in mind? $\endgroup$– Joonas IlmavirtaOct 16, 2014 at 12:56
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$\begingroup$ It should be $$F(\alpha x+(1-\alpha)y)\leq [F(x)]^\alpha [F(y)]^{(1-\alpha)},\quad x,y\in X$$ $\endgroup$– user786Oct 16, 2014 at 18:38
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$\begingroup$ I don't quite understand what you are trying to state, because the usual structure of a normed space does not allow taking powers and comparing elements. For $z\in X$, what does $z^\alpha$ mean? If $z,w\in X$, what do $zw$ and $z\leq w$ mean? It would help if you could elaborate on the meaning of your statement. $\endgroup$– Joonas IlmavirtaOct 16, 2014 at 18:42
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$\begingroup$ It would definitely make sense if I replace the normed space with $\textbf{Vector Lattice Algebra}$. $\endgroup$– user786Oct 16, 2014 at 18:49
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$\begingroup$ Thank you prof. Joonas Ilmavirta for correction. I have edited the question too. $\endgroup$– user786Oct 16, 2014 at 19:04
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