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If $f(x)$ is a continuous function such that $f(x) = f(2x)$ and $f(1) = 3\;,$ Then $\displaystyle \int_{-1}^{1}f(f(x))\,dx$

$\bf{My\; Try::}$ Here $-\infty <x<\infty$ and Given $f(x) = f(2x)$

So Using Recursively $$\displaystyle f(x) = f(2^1x)=f(2^2x)=..........=\lim_{n\rightarrow \infty}f\left(2^{n-1}x\right)$$

OR we can write it as $$f(x) = f\left(\frac{x}{2}\right)=f\left(\frac{x}{2^2}\right)=........=\lim_{n\rightarrow \infty}f\left(\frac{x}{2^{n-1}}\right)$$

Now How can i prove that $f(x)$ is a constant function.

Help me

Thanks

StubbornAtom
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juantheron
  • 53,015

1 Answers1

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$f(x) = f\left(\frac{x}{2}\right)=f\left(\frac{x}{2^2}\right)=........=\lim_{n\rightarrow \infty}f\left(\frac{x}{2^{n-1}}\right)=f(0)$ by continuity.

So, $f(x)=f(0)$ for all $x$, and in particular, $f(x)=f(1)=3$ for all $x$.

Then, $\int_{-1}^{1}f(f(x))dx=\int_{-1}^{1}f(3)dx=\int_{-1}^{1}3dx=6$

πr8
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