# Does the Cauchy-Schwarz integral inequality still hold for convergent improper integrals?

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?

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

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 $$\int_{a}^{b} f(x) g(x) \, dx \le \left( \int_{a}^{b} f(x)^2 \, dx\right)^{1/2}\left( \int_{a}^{b} g(x)^2 \, dx\right)^{1/2} \, .$$
• @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 $\int_{a}^{b} f(x) g(x) \, dx$. Mar 14, 2020 at 2:53