max (I(f)-J(f))

[Cezar Lupu]

Iata o problema data la Putnam in 2006, cam usurica dupa umila-mi parere.
Enuntul problemei este urmatorul:

Fie \( f:[0,1]\to\mathbb{R} \) o functie continua. Consideram \( I(f)=\int_0^1 x^2 f(x)dx \) si \( J(x)=\int_0^1 xf^2(x)dx \). Sa se gaseasca
\( \max(I(f)-J(f)) \) si pentru ce functie \( f \) se realizeaza acest maxim.

[Doru Popovici]

f:[0,1]->R
\( I(f)=\int_{0}^{1} x^2f(x)dx \)
\( J(f)=\int_{0}^{1} xf^2(x)dx \)
=> \( I(f)-J(f)=\int_{0}^{1} xf(x)(x-f(x)) \)
dar \( f(x)(x-f(x))\leq x^2/4 \), deoarece:
\( 4xf(x) - 4f^2(x)\leq x^2 \), trece tot in partea dreapta
=>\( 4f^2(x)-4xf(x)+x^2\geq 0 \)
=>\( (2f(x) - x)^2\geq 0 \) adevarat

=> \( I(f)-J(f)=\int_{0}^{1} xf(x)(x-f(x)) \leq \int_{0}^{1}x*x^2/4= (\int_{0}^{1}x^3)/4=((x^4/4)|_{0}^{1})/4=1/16 \)
=>max(I(f)-J(f))=1/16 si are loc agalitatea daca f(x)=x/2

QED