Let $f$ be a differentiable function satisfying $\log_2(f(3x))=x+\log_2(3f(x))\;\forall x\in\mathbb R$ and $f'(0)=1$. Find the value of $[f(3)]$, where $[k]$ denotes greatest integer less than or equal to $k$.
Solution:
$\log_2(f(3x))=x+\log_2(3f(x))$
$\frac{f(3x)}{3f(x)}=2^x$
$\frac{f(x)}{3f(\frac x3)}\cdot\frac{f(\frac x3)}{3f(\frac x{3^2})}\cdot\frac{f(\frac x{3^2})}{3f(\frac x{3^3})}\cdot\cdot\cdot\frac{f(\frac x{3^{n-1}})}{3f(\frac x{3^n})}=2^{\frac x3}\cdot2^{\frac x{3^2}}\cdot\cdot\cdot2^{\frac x{3^n}}$
$\frac{f(x)}{3^n\cdot f(\frac x{3^n})}=2^{\left(\frac x3+\frac x{3^2}+…+\frac x{3^n}\right)}$
$\lim_{n\to\infty}\frac{f(x)\cdot\frac x{3^n}}{x\cdot f(\frac x{3^n})}=\lim_{n\to\infty}2^{\left(\frac x3+\frac x{3^2}+…+\frac x{3^n}\right)}$
$\frac{f(x)}{x\cdot f'(0)}=2^{\left(\frac {\frac x3}{1-\frac13}\right)}$
$f(x)=x\cdot2^{\frac x2}$
$\therefore f(3)=3\cdot2^{\frac32}=6√2≈8.4$
$\implies[f(3)]=8$
My doubt:
They seem to have written $\lim_{n\to\infty}\frac{f(\frac x{3^n})}{\frac x{3^n}}=f'(0)$
I don't understand this step.
My Attempt:
$f'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}$
Replacing $h$ by $\frac x{3^n}$
$f'(0)=\lim_{n\to\infty}\frac{f(\frac x{3^n})-f(0)}{\frac x{3^n}}$
I don't think $f(0)$ is zero because that's not in the range.
Best Answer
Here is a proof of absence of Such a function for $x\in \mathbb{R}$
Assume A differentiable function $f(x)$ such that $\log_2(f(3x))=x+\log_2(3f(x))$ $\forall x\in \mathbb{R}$,
Given $f'(0)=1$
Noticing the domain of This functional equation, $f(x)\neq 0$ $\forall x\in \mathbb{R}$
taking Antilogarithm with base $2$ on both sides of functional equation
It follows
$f(3x)=2^x\cdot3f(x) \tag 1$
As the function $f(x)$ is differentiable ,
differentiate equation $1$ on both sides
$3f'(3x)=2^x\ln2(3f(x)) +2^x\cdot3f'(x) \tag 2$
Substituting $x=0$ in equation $2$
$3f'(0)=\ln2\cdot(3f(0)) +3f'(0) \tag 3$
substituting $f'(0)=1$ in equation $3$
it follows
$1=\ln2\cdot f(0)+1$
yielding
$f(0)=0$
But this contradicts the statement $f(x)\neq0$ which was a direct consequence of domain of functional equation, Hence by contradiction, Such function does not exist $\forall x\in \mathbb{R}$