Real analysis- Finding all the possible, $f:\mathbb{R}\rightarrow \mathbb{R}$

Question

Given, $f:\mathbb{R}\rightarrow \mathbb{R}$, which satisfies $f\left(\dfrac{x}{y}\right)=\dfrac{f(x)}{f(y)}$, where f is continuous and differentiable $\forall x,y\in\mathbb{R},f(y)\neq0$, $f'(1)=2$ and $f(x)\neq0$,$\forall x\neq0$. Find all such possible fuctions. We can easily figure it out that $f(0)=0$,

If we put $x=0$, then we get, $f(y)f(0)=f(0)$, and since $f'(1)\neq 0\Rightarrow f(0)=0$.

Thanks in anticipation.

There are missing assumptions. Where is it assumed to be differentiable, where is it assumed to be continuous? — Keen-ameteur, Jun 14 '18 at 06:08
The equation does not even make sense since $0$ is in the domain. If you take the domain as $(0,\infty)$ then we can show that there is no solution. — Kavi Rama Murthy, Jun 14 '18 at 07:20
@ChandramauliChakraborty I wrote an answer, but I also have a question, you wrote $f(y)\neq 0$ for all $y\in\mathbb{R}$ but how can it be that $f(0)=0$? — zxcvber, Jun 14 '18 at 07:27
There is nothing wrong with the question. $f(0)=0$, can be easily proved. — Chandramauli Chakraborty, Jun 14 '18 at 07:34
The point is that the function is defined to be non - zero $\forall x\in \mathbb{R}$, which leads to a contradiction if $f(0)=0$. — , Jun 14 '18 at 07:37
If $f(x) \neq 0$ for all $x$, you can actually cancel out $f(0)$ from $f(y)f(0)=f(0)$, and get $f(y)=1$ for all $y$. (Well this doesn't satisfy $f'(1)=2$ but...) Is this supposed to happen? I'm confused. It seems that the problem statement is unclear. — zxcvber, Jun 14 '18 at 07:44
I guess defining $f(x)\neq 0\ \forall x\neq 0$ would be better? — , Jun 14 '18 at 07:47
Similar older question: If $f(\frac{x}{y})=\frac{f(x)}{f(y)} , , f(y),y \neq 0$ and $f'(1)=2$ then $f(x)=$?. — Martin Sleziak, Jun 18 '18 at 18:09

score 4 · Accepted Answer · edited Jun 14 '18 at 07:38

First, plug in $x=y=1$ to see that $f(1) = 1$. Now, since the function is differentiable, we compute it's derivative. By definition, (assume $x\neq 0)$ $$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \rightarrow 0} f(x)\cdot \frac{\frac{f(x+h)}{f(x)}-1}{h} = \lim_{h \rightarrow 0} \frac{f(x)}{x}\cdot \frac{{f(1+h/x)}-f(1)}{h/x}$$ As $h\rightarrow 0$, $h/x \rightarrow 0$. Thus we have $$f'(x) = \frac{f(x)}{x}\cdot f'(1) = \frac{2}{x}f(x)$$

Now just solve the differential equation. Multiply $1/x^2$ on both sides to get $$\frac{1}{x^2}f'(x) - \frac{2}{x^3}f(x)=0$$ This is equivalent to $$\frac{d}{dx}\left(\frac{1}{x^2}f(x)\right) = 0$$ Thus we see that $f(x)=Cx^2 (x\neq 0)$, where $C$ is a constant.

Note: I assumed that $x$ is never 0. So I will have to find the correct value for $x=0$ but I don't see how I can calculate it from $f(x/y)=f(x)/f(y)$. Perhaps the domain is not $\mathbb{R}$?

Edit: The question was edited, so that $f(0)=0$. Then $f(x)=Cx^2$ would certainly work.

Edit: Using the fact $f'(1)=2$, we get $c=1$, hence, $f(x)=x^2$.

How silly of me to forget to plug in the initial conditions ! Thanks! — zxcvber, Jun 14 '18 at 07:39

Real analysis- Finding all the possible, $f:\mathbb{R}\rightarrow \mathbb{R}$

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