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I came across this question in an answer given to a question here on MSE.

Given that $f(a)=f(a^2)$ for all $a\in [0,1]$,$f$ continuous, it is easy to prove by induction that $p(n): f(a) = f(a^{2^{n}})$ for all $a\in [0,1]$ and for all $n\in \mathbb{N}$.

I am having a hard time proving it. Here's my attempt though, and I'll appreciate any help I can get.

For $n=1$, $p(n)$ is true.
Now , suppose it is true for $n=k$. i.e. $f(a)=f\left(a^{2^{k}}\right)$ and consider it for for $n=k+1$. So we would have $$ f\left(a^{2^{k+1}}\right)=f\left(a^{2^k}\cdot a^2\right)$$

Now this is where I get stuck.


Okay, so thanks to the quick answers below, I see where I went wrong. I make some modification to the $k+1$ step. $$ f\left(a^{2^{k+1}}\right)=f\left(a^{2^k}\cdot a^{2^k}\right)=f\left(\left(a^{2^k}\right)^2\right)=f(a)~~\text{by the induction hypothesis}.$$

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2 Answers 2

Note that $$a^{2^k}a^2 = a^{2^k+2}\neq a^{2^{k+1}}$$ (unless $k=1$).

What you want is $$a^{2^k}a^{2^k} = a^{2^k+2^k} = a^{2(2^k)} = a^{2^{k+1}}.$$

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Thanks for your response. I've modified my attempt. Is it better now? –  Jon Nov 14 '11 at 21:04
    
I would add the intermediate step $f((a^{2^k})^2) = f(a^{2^k})$ (by the given property of $f$), and then $f(a^{2^k}) = f(a)$ by the induction hypothesis. Otherwise, it's fine. –  Arturo Magidin Nov 14 '11 at 21:08
    
Right!...Thanks very much. –  Jon Nov 14 '11 at 21:10

HINT : Your mistake is with your index laws. It is not true that $ a^{2^{k+1} } = a^{2^k} \cdot a^2 . $ The second term is actually $ a^{2^k +2}$, since $ a^m \cdot a^n = a^{m+n}.$ Instead, the correct manipulation is $ a^{2^{k+1} } = ( a^{2^k} )^2 $ because of the general index law $ (a^m)^n = a^{mn}.$

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Thanks for your swift response. –  Jon Nov 14 '11 at 21:04

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