Functional equation $f(\frac xy)=f(x)/f(y)$ and $f'(1)=2$

Question

Find $f(x)$ that verifies:

• $f\left(\dfrac{x}{y}\right)=\dfrac{f(x)}{f(y)}$ for all $x,y$ such that
• $f(y)\neq 0$
• $f'(1)=2$

How to solve such a question ?

Answer 1

As $f'(1)$ exists and is non-zero, there is a neighbourhood of $1$ where $f$ is not identically zero. In particlar, there exists $a>0$ with $f(a)\ne 0$. Then $$\tag1 f(1)=f(\tfrac aa)=\frac{f(a)}{f(a)}=1.$$ If $x>0$ and $f(\sqrt x)\ne 0$, we see from $f(\sqrt x)=f(\frac x{\sqrt x})=\frac{f(x)}{f(\sqrt x)}$ that $$\tag{$\star$} f(x)>0\qquad \text{if }x\ge0\text{ and }f(\sqrt x)\ne0. $$

By continuity at $x=1$, we have $f(x)>0$ in a neighbourhood of $1$. Let $u=\sup\{\,x<1\mid f(x)\le 0\,\}$ and $v=\inf\{\,x>1\mid f(x)\le 0\,\}$. If $v<\infty$, then certainly $v>1$ and so for $v\le x< v^2$, we have $1<\sqrt x< v$ and from $(\star)$ conclude that $f(x)>0$, contradicting the definition of $v$. Similarly, if $u>0$, we find $f(x)>0$ for $u^2<x\le u$, contradicting the definition of $u$. We conclude that $v=+\infty$ and $u\le 0$, i.e., $$ f(x)>0\qquad\text{for }x>0. $$

Define $g(x)=\ln f(e^x)$ (allowed as $f(e^x)>0$). Then $g(x+y-y)=g(x+y)-g(y)$, and so $$\tag2 g(x+y)=g(x)+g(y)$$ for all $x,y\in\Bbb R$. As $f$ is differentiable at $1$, it must in particular be continuous there, and that means that $g$ is continuous at $0$. It is well-known that the only solutions to $(2)$ that are continuous at at least one point are of the form $$g(x)=cx$$ for some constant $c$. It follows that, $$ f(x)=f(e^{\ln x})=e^{g(\ln x)}=e^{c\ln x}=x^c\qquad\text{for x>0}.$$

By taking derivatives at $x=1$, we find $c=2$, i.e., $$\tag3 f(x)=x^2\qquad\text{for x>0}.$$

From $$ f(0)= f(\tfrac 02)=\frac{f(0)}{f(2)}=\frac{f(0)}4$$ we find $$\tag4 f(0)=0.$$ Assume there exists $b<0$ with $f(b)\ne 0$. Then $$ \tag5f(x)=f(\tfrac{bx}b)=\frac{f(bx)}{f(b)}=\frac{b^2x^2}{f(b)}\qquad\text{for x<0}.$$ In particular $f(-1)=\frac{b^2}{f(b)}\ne 0$ and from $$ f(-1)=f(\tfrac1{-1})=\frac{f(1)}{f(-1)}$$ we obtain $$ f(-1)=\pm 1.$$ Plugging this into $(5)$ we find that either $$\tag{6a}f(x)=x^2 \qquad\text{for x<0}$$ or $$\tag{6b}f(x)=-x^2 \qquad\text{for x<0}.$$ Together with $(3)$ and $(4)$, we find that the only candidate solutions are $$\fbox{$ f(x)=x^2$}$$ and $$ \fbox{$f(x)=|x|\cdot x=\begin{cases}x^2&x\ge 0\\-x^2&x\le 0\end{cases}$}$$ as well as the solution where no $b<0$ with $f(b)\ne 0$ exists, i.e., $$ \fbox{$f(x)=\begin{cases}x^2&x\ge 0\\0&x\le 0\end{cases}$}.$$ One immediately verifies that indeed all three candidates have the desired properties.

Remark: As the results show, it turns out that $f$ must be differentiable, even though we were given only differentiability at $x=1$ and did not even have to assume continuity at any other point.

Answer 2

First I'd start by thinking of potential families of functions that would satisfy the condition.

I know that $f(x) = x^k$ would satisfy the condition $f(x/y) = f(x)/f(y)$.

After that I just need to find one such $k$ and I am done.

To do this, I will differentiate both sides with respect to $x$ to get:

$\frac{1}{y}f'(x/y) = f'(x)/f(y)$.

Substitute $x = 1$:

$\frac{1}{y}f'(1/y)f(y) = 2$.

Next use $f(y) = y^k$.

We have $k=2$.

By subsitution $f(x) = x^2$ solves this.

A trial an error type effort with $f(x)=x^k$ for small $k$ would work too.

Answer 3

If $f(\dfrac{x}{y}) =\dfrac{f(x)}{f(y)} $ then $f(\dfrac{x}{y})f(y) =f(x) $.

Setting $\dfrac{x}{y} = z$, this is $f(z)f(y) = f(zy)$.

Differentiable solutions to this are $f(x) =x^a $ with $ a =f'(1) $ (shown below).

Since $f'(1) = 2$, we have $a = 2$ so $f(x) = x^2$.

As a check, $(x^2)' = 2x$ so $(x^2)'(1) = 2$.

Show:

If $f(z)f(y) = f(zy)$, then, setting $z=1$, $f(1) = 1$ and $f(x+h) =f(x(1+h/x)) =f(x)f(1+h/x) $ so

$\begin{array}\\ f(x+h)-f(x) &=f(x)f(1+h/x)-f(x)\\ &=f(x)(f(1+h/x)-1)\\ &=f(x)(f(1+h/x)-f(1))\\ \text{so}\\ \dfrac{f(x+h)-f(x)}{h} &=f(x)\dfrac{f(1+h/x)-f(1)}{h}\\ &=\dfrac{f(x)}{x}\dfrac{f(1+h/x)-f(1)}{h/x}\\ \end{array} $

Letting $h \to 0$, $f'(x) =\dfrac{f(x)}{x}f'(1) $ or $\dfrac{f'(x)}{f(x)} =\dfrac{f'(1)}{x} $.

Integrating, $\ln(f(x)) =f'(1)\ln(x)+c =\ln(Cx^{f'(1)}) $ so $f(x) =Cx^{f'(1)} $.

Setting $x=1$, $1 = f(1) =C $ so $f(x) =x^{f'(1)} $.