Question

A function $f\left(x\right)$ satisfies the following property: $f(x\cdot y)=f\left(x\right)f\left(y\right)$. Show that the function $f\left(x\right)$ is continuous for all values of x if it is continuous at $x=1$.

Answer 1

$f(x+y)=f\left(x\right)f\left(y\right)$

$f(0\to 0)=f{\left(0\right)}^2$ $\therefore f\left(0\right)=1$

$\Rightarrow f\left(1\right).f\left(1\right).f\left(0\right)$ $\therefore f\left(1\right)=a$ ( say )

$f\left(2\right)=a^2,f\left(b\right)=f\left(1\right).f\left(2\right)=a^3$

Hence $f\left(n\right)=a^n$ ( $\forall n\in$ positive integers )

$\Rightarrow f(1-1)=f\left(1\right).f(-1)$

$\Rightarrow f(-1)=a^{-1}$

$\therefore f\left(n\right)=a^{v}n$ ( for all $n$ )

and $a^n$ is continuous for all $n$

$\therefore f\left(x\right)$ is continuous for all values of $x.$

Hence, solved.