Given, $f(x)$ is an increasing real valued function on $(0,1)$.
Also, $f(x)\gt 1$ on $(0,1)$.
Then, what condition one need to show for any $0<x<y<1$, $f(y)^x\lt f(x)^y$
I am totally blind here that what is the way to show it.
please help me with anything.
On
To simplify the problem we can set $f(x) = e^{x h(x)}$.
Then $f(x)>1$ means $h(x)>0$ for $x\in (0,1)$. Also, the condition that $f$ is increasing, means that $xh(x)$ is increasing.
For the final condition $0<x<y<1 \Rightarrow f(y)^x<f(x)^y$, note that $$f(y)^x<f(x)^y \Longleftrightarrow f(y)^{1/y}<f(x)^{1/x} \Longleftrightarrow e^{yh(y)/y}<e^{xh(x)/x}=e^{h(y)}<e^{h(x)}$$ Which means that $h$ is a decreasing function.
A typical example of such a function $h$, is $h(x)= x^{-\alpha}$ where $\alpha\in(0,1)$.
Moreover, the function $h$ (and hence $f$) does not have to be continuous.
We can set $h(x)= x^{-\alpha}$ for $0<x<1/2$, and $h(x)= x^{-\beta}$ for $1/2\leq x<1$, provided $0<\beta<\alpha<1$.
(However, monotone functions are differentiable almost everywhere, and hence continuous ”almost everywhere” (a theorem of Lebesgue). Which means that it can not be discontinuous in a ”thick” set.)
You want $x \ln f(y)<y\ln f(x)$ or $\frac {\ln f(y)} y<\frac {\ln f(x)} x$ for $x<y$. This means $\frac {\ln f(x)} x$ is a decreasing function. To get a simple usable condition I will assume that $f$ is differentiable.
The derivative of this function is $(x\frac {f'(x)} {f(x)} -\ln f(x))/x^{2}$ so you need $x\frac {f'(x)} {f(x)} < \ln f(x)$ or $f'(x)<\frac { f(x)\ln f(x)} x$. I hope this helps.