Let $f$:$\mathbb{R} \rightarrow \mathbb{R}$ be a continous function such that for all$x \in \mathbb{R}$ and all $t\geq 0$
choose any 2 arbitary nos $x,y$ using the functional relationship prove that $f(x)=f(y)$
when $x,y$ are of opposite signs then show that $$f(x)=f(\frac{x}{2})=f(\frac{x}{4})\dots$$
use continuity to show that $f(x)=f(0)$
Let us take any $2$ real nos $x$ and $y$.
$x$ and $y$ are of same sign . WLG $0<x<y$
Then$\frac{y}{x}>1$
so there is a no $t\geq 0$ such that
$\frac{y}{x}=k^t$
$f(y)=f(k^tx)=f(x)$ [using$f(x)=f(k^tx)$]
$x,y$ are of opposite sign. WLG $x<0<y$
Then $f(x)=f(k^tx)$
$\Rightarrow f(k^tx)=f(k^t2\frac{1}{2}x)$
$\Rightarrow f(k^t2\frac{1}{2}x)=f(k^tk^{log_k2}\frac{x}{2})$
$\Rightarrow f(k^tk^{log_k2}\frac{x}{2})=f(k^{t+log_k2}\frac{x}{2})$
$\Rightarrow f(k^{t+log_k2}\frac{x}{2})=f(\frac{x}{2})$
Using this logic repeatedly we get
$f(x)=f(\frac{x}{2})=f(\frac{x}{4})\dots =f(\frac{x}{2^n})$
Now $\frac{x}{2^n}\rightarrow0$ and $f$ is a continous function hence $\lim_{n\to\infty}f(\frac{x}{2^n})=f(0)$.
[Because we know if $f$ is a continous function and $x_n$ is a sequence that converges to $x$ then $\lim_{n\to\infty}f(x_n)=f(x)$]
using similar logic we can show that $f(y)=f(0)$ so $f(x)=f(y)$ for any $x,y\in \mathbb{R}$
Let $f$:$\mathbb{R} \rightarrow \mathbb{R}$ be a continous function such that for all$x \in \mathbb{R}$ and all $t\geq 0$
choose any 2 arbitary nos $x,y$ using the functional relationship prove that $f(x)=f(y)$
when $x,y$ are of opposite signs then show that $$f(x)=f(\frac{x}{2})=f(\frac{x}{4})\dots$$
use continuity to show that $f(x)=f(0)$
Let us take any $2$ real nos $x$ and $y$.
$x$ and $y$ are of same sign . WLG $0<x<y$
Then$\frac{y}{x}>1$
so there is a no $t\geq 0$ such that
$\frac{y}{x}=k^t$
$f(y)=f(k^tx)=f(x)$ [using$f(x)=f(k^tx)$]
$x,y$ are of opposite sign. WLG $x<0<y$
Then $f(x)=f(k^tx)$
$\Rightarrow f(k^tx)=f(k^t2\frac{1}{2}x)$
$\Rightarrow f(k^t2\frac{1}{2}x)=f(k^tk^{log_k2}\frac{x}{2})$
$\Rightarrow f(k^tk^{log_k2}\frac{x}{2})=f(k^{t+log_k2}\frac{x}{2})$
$\Rightarrow f(k^{t+log_k2}\frac{x}{2})=f(\frac{x}{2})$
Using this logic repeatedly we get
$f(x)=f(\frac{x}{2})=f(\frac{x}{4})\dots =f(\frac{x}{2^n})$
Now $\frac{x}{2^n}\rightarrow0$ and $f$ is a continous function hence $\lim_{n\to\infty}f(\frac{x}{2^n})=f(0)$.
[Because we know if $f$ is a continous function and $x_n$ is a sequence that converges to $x$ then $\lim_{n\to\infty}f(x_n)=f(x)$]
using similar logic we can show that $f(y)=f(0)$ so $f(x)=f(y)$ for any $x,y\in \mathbb{R}$
