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On the Functional Inequality $$f(x)f(y)-f(xy)\le f(x)+f(y)-f(x+y)$$ f ( x ) f ( y ) - f ( x y ) ≤ f ( x ) + f ( y ) - f ( x + y )
Computational Methods and Function Theory ( IF 2.1 ) Pub Date : 2020-07-04 , DOI: 10.1007/s40315-020-00327-8
Horst Alzer , Luis Salinas

We prove that all solutions \(f: {\mathbb {R}} \rightarrow {\mathbb {R}}\) of the functional inequality

$$\begin{aligned} (*) \quad f(x)f(y)-f(xy)\le f(x)+f(y)-f(x+y), \end{aligned}$$

which are convex or concave on \({\mathbb {R}}\) and differentiable at 0 are given by

$$\begin{aligned} f(x)=x\quad \text{ and } \quad f(x)\equiv c, \quad \text{ where } \quad 0\le c\le 2. \end{aligned}$$

Moreover, we show that the only non-constant solution \(f: {\mathbb {R}} \rightarrow {\mathbb {R}}\) of \((*)\), which is continuous on \({\mathbb {R}}\) and differentiable at 0 with \(f(0)=0\) is \(f(x)=x\).



中文翻译:

关于泛函不等式$$ f(x)f(y)-f(xy)\ le f(x)+ f(y)-f(x + y)$$ f(x)f(y)-f( xy)≤f(x)+ f(y)-f(x + y)

我们证明函数不等式的所有解\(f:{\ mathbb {R}} \ rightarrow {\ mathbb {R}} \)

$$ \ begin {aligned}(*)\ quad f(x)f(y)-f(xy)\ le f(x)+ f(y)-f(x + y),\ end {aligned} $ $

\({\ mathbb {R}} \)上凸或凹并且在0处可微分

$$ \ begin {aligned} f(x)= x \ quad \ text {和} \ quad f(x)\ equiv c,\ quad \ text {其中} \ quad 0 \ le c \ le2。\ end {已对齐} $$

而且,我们证明了\((*)\)的唯一非恒定解\(f:{\ mathbb {R}} \ rightarrow {\ mathbb {R}} \)\({ mathbb {R}} \)且可与\(f(0)= 0 \)在0处微分的是\(f(x)= x \)

更新日期:2020-07-05
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