If $f,\varphi$ are two even periodic functions with $T=2$. Suppose $$f(x) = x(2-x),\quad \varphi(x) = x,\quad x\in [0,1]$$ prove that: $$f(x)=\sum\limits_{n=0}^\infty\frac{1}{2^{2n}}\varphi(2^nx),\quad\forall x\in\mathbb{R}.$$ I wonder if I need to make use of the fourier expansion of $f$ and $\varphi$, can anyone offer me some help?
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Have you verified that they are periodic and even? – Weierstraß Ramirez May 22 '23 at 08:02
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The expression given above is only in $[0,1]$, and that's enough to uniquely determine $f$ and $\varphi$ since they are even and periodic. – Exjudger May 23 '23 at 02:31
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Do you mean f(x)=f(x+2n) for $n\in \mathbb{N}$, and the respective for the second function? – Weierstraß Ramirez May 23 '23 at 06:35
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Yes. The 2 functions are therefore determined. – Exjudger May 23 '23 at 08:53
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It seems that I should make use of the fourier expansion. – Exjudger May 25 '23 at 12:16
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Have you tried? – Weierstraß Ramirez May 25 '23 at 12:50