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A few hours ago I posted this solution : https://math.stackexchange.com/a/3579886/629594 It seems to be all right at a first glance, given that I actually obtain the same constant as the other user, but I have just realised something : my first integral$\left(\int_0^1 \left(\frac{ax^2+bx}{\sqrt{1-x^2}}\right)^2dx\right)$ is actually improper.
In my case, it is true that by sheer luck I actually obtain the same inequality after setting $a+b=0$ (because I may just cancel the $1+x$ after all), but is this really correct? I only know the Cauchy-Schwarz inequality for Riemann integrals, but does it also hold for (convergent) improper integrals?

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  • $\begingroup$ If the integrals are absolutely convergent improper integrals, C-S will work. $\endgroup$
    – zhw.
    Mar 13, 2020 at 23:02

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If both $\int_a^b f^2(x)\, dx$ and $\int_a^b g^2(x)\, dx$ exist as (finite) improper Riemann integrals then for every (sufficiently small) $\delta > 0$, $\epsilon > 0$ $$ \int_{a+\delta}^{b-\epsilon} |f(x) g(x)| \, dx \le \left( \int_{a+\delta}^{b-\epsilon} f(x)^2 \, dx\right)^{1/2}\left( \int_{a+\delta}^{b-\epsilon} g(x)^2 \, dx\right)^{1/2} \\ \le \left( \int_{a}^{b} f(x)^2 \, dx\right)^{1/2}\left( \int_{a}^{b} g(x)^2 \, dx\right)^{1/2} \, . $$ The left-hand side is decreasing in both $\delta$ and $\epsilon$ and bounded, so that the limits for $\delta \to 0^+$ and $\epsilon\to 0^+$ exist. It follows that $\int_a^b |f(x) g(x)| \, dx$ exists as an improper Riemann integral, and $$ \int_{a}^{b} |f(x) g(x)| \, dx \le \left( \int_{a}^{b} f(x)^2 \, dx\right)^{1/2}\left( \int_{a}^{b} g(x)^2 \, dx\right)^{1/2} \, . $$ This implies that $\int_{a}^{b} f(x) g(x) \, dx$ exists as an improper Riemann integral, compare Proof verification: existence of improper integral, given that the integral of the absolute value is finite.. Since $f(x) g(x) \le |f(x)g(x)|$ we can conclude that $$ \int_{a}^{b} f(x) g(x) \, dx \le \left( \int_{a}^{b} f(x)^2 \, dx\right)^{1/2}\left( \int_{a}^{b} g(x)^2 \, dx\right)^{1/2} \, . $$

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  • $\begingroup$ Thank you very much ! So, just to make sure since I am not really experienced with improper integrals, we just need the improper integrals to exist, it doesn't matter whther they are finite or infinite, right? $\endgroup$
    – Alexdanut
    Mar 13, 2020 at 23:54
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    $\begingroup$ @Alexdanut: If the right-hand side is infinite then $\int_{a}^{b} |f(x) g(x)| \, dx \le \left( \int_{a}^{b} f(x)^2 \, dx\right)^{1/2}\left( \int_{a}^{b} g(x)^2 \, dx\right)^{1/2}$ trivially holds. But in that case we can say nothing about the existence of $\int_{a}^{b} f(x) g(x) \, dx$. $\endgroup$
    – Martin R
    Mar 14, 2020 at 2:53

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