Let \(f, g\) be non-negative measurable functions on \([0, \infty)\) with \[ A &\da \int_0^{\infty } f(y) y^{-1/2} \dy < \infty \\ B &\da \qty{ \int_0^{\infty } \abs{ g(y) } }^2 \dy < \infty .\]
Show that \[ \int_0^{\infty } \qty{ \int_0^{\infty } f(y) \dy } {g(x) \over x} \dx \leq AB .\]
As printed, the inner integral \(\int_0^\infty f(y)\,dy\) does not depend on \(x\), and \(B\) is written with the square outside the integral. The inequality above takes the inner integral over \([0,x]\) and \(B = \|g\|_{L^2}\).
Solution
We prove \(\int_0^\infty \frac{g(x)}{x}\int_0^x f(y)\,dy\,dx \le AB\) with \(A = \int_0^\infty f(y)y^{-1/2}\,dy\) and \(B = \left(\int_0^\infty g(y)^2\,dy\right)^{1/2}\); see the remark below.
1. \(\int_0^\infty \frac{g(x)}{x}\int_0^x f(y)\,dy\,dx = \int_0^\infty f(y) \int_y^\infty \frac{g(x)}{x}\, dx\, dy\).
The integrand \(f(y)g(x)x^{-1}\chi_{\theset{y < x}}\) is nonnegative and measurable on \((0,\infty)^2\), so Tonelli’s theorem allows exchanging the order of integration.
2. For \(y > 0\), \(\int_y^\infty \frac{g(x)}{x}\, dx \le B y^{-1/2}\).
By the Cauchy–Schwarz inequality, \(\int_y^\infty \frac{g(x)}{x}\, dx \le \left(\int_y^\infty g^2\right)^{1/2}\left(\int_y^\infty x^{-2}\,dx\right)^{1/2} \le B\,y^{-1/2}\).
3. Q.E.D.
By steps 1 and 2, the left side is at most \(\int_0^\infty f(y)\,B y^{-1/2}\, dy = AB\).