Determining the Function with Identities

[evinda]

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

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

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

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

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

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

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

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

=>f(x)=c

Is this right?

 

[I like Serena]

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

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

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

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

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

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

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

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

=>f(x)=c

Is this right?

Hi evinda!

It all looks good except for the last step.
(Oh, and you may want to add $dt$ to both integrals. ;))

For instance, since you did not specify that f is continuous, another solution would be:
$$f(x)=\begin{cases}
1 & \text{ if } \exists k \in \mathbb Z \text{ such that } x=1-3^k \\
2 & \text{ otherwise}
\end{cases}$$

With the additional condition that f is continuous, it does follow that $f(x)=C$.
This can be proven by picking a point with a different value and consider what that means for the limit where $x \to 1$ (proof by contradiction).

 

[evinda]

Nice!Thank you veery much! :p

 

[johng1]

Hi,
I can't quite figure out the hint. But here's a direct proof:
Let a and b be any reals with -1<a<1 and f a continuous function with f(ax+b)=f(x) for all x. Then f is a constant function.
Proof.
Let x be any real. Define the sequence x_n inductively by x₀ = x and for for n>0, x_n=ax_n-1+b. Then an easy induction shows x_n=aⁿx+b(aⁿ-1)/(a-1). By hypothesis, f(x)=f(x_n) for all n. So f(x)=limit as n approaches infinity f(x_n)=f(-b/(a-1)); i.e. f is constant.

 

[evinda]

Great..Thank you!