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?
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$\begingroup$
integration
hilbert-spaces
1 Answers
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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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0I am convinced and this sounds good for me. If anybody thinks that this is not true please let me know. Thanks @GH at MO – 2012-12-16
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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