Function-identities

[evinda]

Hi guys!!!I have a question..How can I show that the function that has the following identities:


$\displaystyle \bullet$ $\displaystyle f(x)\neq 0$ ,$\displaystyle x\in\mathbb{R}$.
$\displaystyle \bullet$ $\displaystyle f(0)=f\left(\dfrac{2}{3}\right)$.
$\displaystyle \int_{0}^{x}\frac{f(t)f(x-t)}{f(\frac{2+t}{3})f(\frac{2+x-t}{3})}=\int_{0}^{x}\frac{f^2{(t)}}{f^2{(\frac{2+t}{3}})}$
is f(x)=c..??
That's what I did:


=>$\displaystyle \int_{0}^{x}\frac{f(t)f(x-t)}{f(\frac{2+t}{3})f(\frac{2+x-t}{3})}=\int_{0}^{x}\frac{f^2{(t)}}{f^2{(\frac{2+t}{3}})}$


$\displaystyle \frac{f(x)f(0)}{f(\frac{2+x}{3})f(\frac{2}{3})}$=$\displaystyle \frac{f^2{(x)}}{f^2(\frac{2+x}{3})}$ .


Because $\displaystyle f(0)=f(\frac{2}{3})$


$\displaystyle \frac{f(x)}{f(\frac{2+x}{3})}=\frac{f^{2}(x)}{f^2(\frac{2+x}{3}){}}$


=> $\displaystyle f^2{(\frac{2+x}{3})}f(x)=f^2{(x)}f{(\frac{2+x}{3})}$


=>$\displaystyle f{(\frac{2+x}{3})}=f(x)$ , $\displaystyle f(x) \neq 0$


=>f(x)=c


Can I find the derivative $\displaystyle \int_{0}^{x}\frac{f(t)f(x-t)}{f(\frac{2+t}{3})f(\frac{2+x-t}{3})}=\int_{0}^{x}\frac{f^2{(t)}}{f^2{(\frac{2+t}{3}})}$ without using any theorem??
Also,how can I prove that $\displaystyle f{(\frac{2+x}{3})}=f(x)$=>f(x)=c??

I hope you can help me...Thanks in advance!

 

[daon2]

Of course f(x) is not necessarily constant. In fact here's a counter-example:

\(\displaystyle f(x) = \left\{ \begin{array}{lr}
1; \,\, x = 0, \dfrac{2}{3}\\
-1; \,\,\text{ otherwise}
\end{array}
\right.
\)

Is there some other information you may not be including?

 

[evinda]

A ok...No,there is no other imformation given...But can I find the derivative of the equation $\displaystyle \int_{0}^{x}\frac{f(t)f(x-t)}{f(\frac{2+t}{3})f(\frac{2+x-t}{3})}=\int_{0}^{x}\frac{f^2{(t)}}{f^2{(\frac{2+t}{3}})}$ without using any theorem???

 

[daon2]

A ok...No,there is no other imformation given...But can I find the derivative of the equation $\displaystyle \int_{0}^{x}\frac{f(t)f(x-t)}{f(\frac{2+t}{3})f(\frac{2+x-t}{3})}=\int_{0}^{x}\frac{f^2{(t)}}{f^2{(\frac{2+t}{3}})}$ without using any theorem???

"The derivative of an equation" doesn't make sense. But if you mean "get rid of the integrals", you may use the fundamental theorem of calculus.

Assume:


$\displaystyle \displaystyle F(x)=\int_0^x f(t) dt = \int_0^x g(t) dt = G(x)$

Then

$\displaystyle \displaystyle F'(x)= f(x) = g(x) = G'(x)$

 

[evinda]

Nice..Thanks I have also an other question...from the equation f(x)=f((2+x)/3),how can I find the value of f??

 

[daon2]

Nice..Thanks I have also an other question...from the equation f(x)=f((2+x)/3),how can I find the value of f??

Why are you assuming you know that equality is true? I was under the impression you are supposed to SHOW the integrals are equal.

Anyway that equality tells us that

$\displaystyle f(x) = f((x-1)/3^n + 1)$ for all $\displaystyle n$.

So, assuming f is nice enough, what can you do?

 

[evinda]

I got stuck..How can I ind the value of f???

Why are you assuming you know that equality is true? I was under the impression you are supposed to SHOW the integrals are equal.

Anyway that equality tells us that

$\displaystyle f(x) = f((x-1)/3^n + 1)$ for all $\displaystyle n$.

So, assuming f is nice enough, what can you do?

 

[daon2]

I fot stuck..How can I ind the value of f???

You cant. But you can show it is constant IF it is continuous, but you said there were no other assumptions.

 

[evinda]

I understood..Thank you!!!

I fot stuck..How can I ind the value of f???