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Suppose that $f, g$ are real, integrable functions, and their product is also integrable. I am considering this expression:

$$\displaystyle\left( \int_{a}^{b} f(x)g(x) \ \mathrm{d}x \right)^2,$$

and I would like to say that

$$\displaystyle \left( \int_{a}^{b} f(x)g(x) \ \mathrm{d}x \right)^2 = \left| \int_{a}^{b} f(x)g(x) \ \mathrm{d}x \right|^2 \leqslant \left| \int_{a}^{b} f(x)|g(x)| \ \mathrm{d}x \right|^2.$$

Is this valid? Without the square, I know that this would definitely not be true. The reason I am asking is because I am dealing with an integral of a product of two Bessel functions, and for the purposes of estimating the integral, I'd like to bound only one of them using the bound $|J_{\nu}(z)| \leqslant 1$, and leave the other one. Am I allowed to do this? I cannot think of a counter-example.

1 Answers 1

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Let $f=g$ defined by $f(x)=1$ on $[a, (a+b)/2]$, and $f(x)=-1$ on $[(a+b)/2,b]$. The LHS is $(b-a)^2$, the RHS 0