Perfect Harmony III

Calculus Level 4

Let F \mathcal F be the collection of all continuous increasing functions ψ \psi defined on [ 0 , 1 ] . \left[0, 1\right]. For any ψ ∈ F , \psi\in\mathcal F, consider the expression s ( ψ ) = ∫ 0 1 x ψ ( x ) d x ∫ 0 1 ψ ( x ) d x . s(\psi)=\frac {\displaystyle\int_0^1 x\psi(x)\, dx}{\displaystyle\int_0^1 \psi(x)\, dx}. Find the smallest possible value of s ( ψ ) s(\psi) as ψ \psi ranges over F . \mathcal F.

Bonus: Compare ∫ 0 s ψ ( x ) d x \displaystyle\int_0^s \psi(x)\, dx and ∫ s 1 ψ ( x ) d x . \displaystyle\int_s^1 \psi(x)\, dx.


The answer is 0.5.

Brian Lie by Brian Lie , 26, China

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1 solution

Abhishek Sinha
Jan 1, 2020

Since the functions ψ : [ 0 , 1 ] → R \psi : [0,1] \to \mathbb{R} and x : [ 0 , 1 ] → R x: [0,1] \to \mathbb{R} are non-decreasing, using the Hardy-Littlewood inequality (a generalization of the well-known Rearrangement Inequality ) on the numerator, we have s ( ψ ) ≥ ∫ 0 1 ( 1 − x ) ψ ( x ) d x ∫ 0 1 ψ ( x ) d x = 1 − s ( ψ ) . s(\psi) \geq \frac{\int_{0}^{1} (1-x) \psi(x) dx}{\int_{0}^{1} \psi(x) dx} = 1- s(\psi). This yields s ( ψ ) ≥ 1 2 . s(\psi) \geq \frac{1}{2}. The minimum value is achieved by taking, e.g., ψ ( x ) = 1 , ∀ x ∈ [ 0 , 1 ] . \psi(x)=1, \forall x \in [0,1].

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