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Indeed, then $f(x)f(x)/f(xx)\to0$ as $x\downarrow0$, so that, for any real $x_+$, there is no real $\beta>0$ such that $f(x)f(y)\le\beta f(xy)$$f(x)f(y)\ge\beta f(xy)$ for all $x,y$ in $(0,x_+]$
A counterexample is given by
$$f(x)=e^{1/x}.$$
Indeed, then $f(x)f(x)/f(xx)\to0$ as $x\downarrow0$, so that, for any real $x_+$, there is no real $\beta>0$ such that $f(x)f(y)\le\beta f(xy)$ for all $x,y$ in $(0,x_+]$
A counterexample is given by
$$f(x)=e^{1/x}.$$
Indeed, then $f(x)f(x)/f(xx)\to0$ as $x\downarrow0$, so that, for any real $x_+$, there is no real $\beta>0$ such that $f(x)f(y)\ge\beta f(xy)$ for all $x,y$ in $(0,x_+]$
Indeed, then $f(x)f(x)/f(xx)\to0$ as $x\downarrow0$, so that, for any real $x_+$, there is no real $\beta>0$ such that $f(x)f(y)\le\beta f(xy)$ for all $x,y$ in $(0,x_+]$
