Functional Integrations

Calculus Level 3

Let f : R → R f : \mathbb {R \to R} be a continuous function such that f ( x ) = f ( 2 x ) f(x) = f(2x) is true ∀ x ∈ R \forall \ x \in \mathbb R and f ( 1 ) = 3 f(1)=3 . Find ∫ − 1 1 f ( f ( f ( x ) ) ) d x \displaystyle \int_{-1}^1 f(f(f(x))) \, dx .

Notation : R \mathbb R denotes the set of real numbers .


The answer is 6.

Dharani Chinta by Dharani Chinta , 22, India

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1 solution

Note that the functional equation implies that f ( 1 2 n x ) = f ( x ) , ∀ x ∈ R , n ∈ N f\left(\frac{1}{2^n} x\right)=f(x),\ \forall x\in \mathbb{R},\ n\in \mathbb{N} . From the continuity of f f , it thus follows that f ( x ) = lim ⁡ n → ∞ f ( 1 2 n x ) = f ( lim ⁡ n → ∞ 1 2 n x ) = f ( 0 ) f(x)=\lim_{n\to \infty}f\left(\frac{1}{2^n} x\right)=f\left(\lim_{n\to \infty}\frac{1}{2^n} x\right)=f(0) . Thus, f f is a constant function with value 3 3 , which implies that the result is 6 \boxed{6} .

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