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If $f(x)$ is a continuous and differentiable function which satisfies the function equation If $$2f (x) = f(xy) + f\left(\frac xy\right)\quad \forall x,y \in \mathbb{R}^{+}$$ and $f'(1)=1$ then find $f(x)$

I can see that $f(x)=\ln(x)$ is one function which satisfies all the properties but how can it be proved? I tried using first principle of differentiation but wasn't able to obtain the function. Could someone help me with this?

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    $\begingroup$ Interestingly, if you assume that $f$ is defined and continuous at $0$, then such a function cannot exist. For any $x > 0$ and $y = 2$ rearranging of the equation gives: $$f(2x) - f(x) = f(x) - f\left(\frac{x}2\right) = \ldots = f\left(\frac{x}{2^n}\right) - f\left(\frac{x}{2^{n+1}}\right) \xrightarrow{n\to\infty} f(0) - f(0) = 0$$ So $$f(2x) = f(x) = f\left(\frac{x}2\right) = \ldots = f\left(\frac{x}{2^n}\right) \xrightarrow{n\to\infty} f(0)$$ Hence, $f = f(0)$ so $f' = 0$, which contradicts $f'(1) = 1$. $\endgroup$ – mechanodroid Nov 18 '17 at 20:19
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Let's differentiate your equation with respect to $y$

$$0=xf^{\prime}\left(xy\right)-\dfrac{x}{y^{2}}f^{\prime}\left(\dfrac{x}{y}\right) \tag{1}$$

This is allowed since $f$ is differentiable. Now set $x=y$ and use $f^{\prime}(1)=1$ to have

$$0=xf^{\prime}\left(x^{2}\right)-\dfrac{1}{x} \tag{2}$$

or if we define $u=x^{2}$

$$f^{\prime}(u)=\dfrac{1}{u} \tag{3}$$

Thus $f\left(u\right)=\ln(u)+C$ for $C\in\mathbb{R}$.

Edit 1: Following the answer by @Bumblebee, we can solve this equation in general by not specifying $C_{1}=f^{\prime}(1)$. In this case, Eq. $(3)$ is in fact

$$f^{\prime}(u)=\dfrac{C_{1}}{u} \tag{4}$$

with a solution

$$f(u)=C_{1}\ln(u)+C_{2} \tag{5}$$

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Lets find all $f:\Bbb{R}^+\to\Bbb{R}$ continuous functions with $f(xy)=2f(x)-f\left(\dfrac{x}y\right).$

When $x=y,$ we have $f(x^2)=2f(x)-f(1).$ Now assume for any $r\in\Bbb{N}\cup\{0\},$ $$f(x^r)=r(f(x)-f(1))+f(1).$$ Now $x\to x^r$ and $y\to x$ gives us $f(x^{r+1})=2f(x^r)-f(x^{r-1}).$ Therefore we can proof this claim by strong induction. Using the fact that $f(x^{-r})=-f(x^r)+2f(1),$ we can extend our result to negative integer $r$ values as $$f(x^{-r})=-r(f(x)-f(1))+f(1)$$ Moreover $x\to x^{1/r}$ gives us $$f(x^{1/r})=\dfrac1r(f(x)-f(1))+f(1).$$ Hence the result establishes for rational (positive and negative) $r$ values. Also due to the continuity, we have the result for any $r\in\Bbb{R}.$
Let $x\to e$ and $r\to\log_e{x},$ then $$f(x)=a\ln x+b\,\,\,\,\,\,\forall x\in\Bbb{R}^+$$ where $a=f(e)-f(1)$ and $b=f(1)$ are some constants.

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  • $\begingroup$ You can add that $a=1$ since $f^{\prime}(1)=1$. $\endgroup$ – eranreches Nov 18 '17 at 21:17
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    $\begingroup$ @eranreches: Yes, of course. But we can solve this functional equation without the differentiability condition. So I thought this answer is more general. $\endgroup$ – Bumblebee Nov 18 '17 at 21:20
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Problem:   Find all continuous functions $f:\Bbb{R}^+\to\Bbb{R}$ which satisfy $$ \forall_{x\ y>0}\quad 2\cdot f(x) \,\ =\,\ f(x\cdot y)\ +\ f\left(\dfrac{x}y\right).$$

Solution:

Let $\ s\ :=\ x\cdot y\ $ and $\ t\ :=\ \frac xy;\ \ $ i.e. $\ \ x\ =\ \sqrt{s\cdot t}\ $ and $\ y\ =\ \sqrt{\frac st}.\ $ Then the above equation is equivalent to:

$$ f\left(\sqrt{s\cdot t}\right)\,\ =\,\ \frac {f(s)+f(t)}2 $$

Define function

$$ g:=f\circ exp: \mathbb R\rightarrow\mathbb R $$

Then:

$$ \forall_{\sigma\ \tau\in\mathbb R}\quad g\left(\frac{\sigma+\tau}2\right)\ =\ \frac{g(\sigma)+g(\tau)}2 $$

It's known that such $\ g\ $ is a linear function (meaning, a polynomial of degree $1$), i.e. there exist $\ A\ B\in\mathbb R\ $ such that:

$$ g(\xi)\ =\ A\cdot\xi+B $$ hence $$ f(x)\ =\ (f\circ\exp\circ\log)(x)\ =\ (g\circ\log)(x)\ =\ A\cdot\log(x)+B $$

Solution:   The set of required functions $\ f\ $ is $$ \{ A\!\cdot\!\log\, +\, B\ :\,\ A\,\ B\in\mathbb R\} $$

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