I was wondering if: $\int_0^1x(t)\int_0^tx(s)ds\ dt$ is positive for a general $x\in L_2[0,1]$ . Can you help me with this?
Is this function positive?
4
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integration
hilbert-spaces
1 Answers
6
By Fubini's theorem, the integral equals $ I=\int_{0\leq s\leq t\leq 1}x(s)x(t)\,dsdt. $ By symmetry, $ I=\int_{0\leq t\leq s\leq 1}x(s)x(t)\,dsdt. $ Adding the two equations, $ 2I = \int_{0\leq s\leq t\leq 1}x(s)x(t)\,dsdt+\int_{0\leq t\leq s\leq 1}x(s)x(t)\,dsdt, $ whence $ 2I = \int_{0\leq s,t\leq 1} x(s)x(t)\,dsdt. $ By Fubini's theorem again, $ 2I = \left(\int_0^1 x(s)\,ds\right)\left(\int_0^1 x(t)\,dt\right) = \left(\int_0^1 x(r)\,dr\right)^2.$ The right hand side is nonnegative, hence so is $I$.
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1@HasanhasanHasan: As long as $x(t)$ is real valued, this is exactly how I was going to prove this. – 2012-12-16