\[ F(x):=\int_{0}^{x} f(y) d y \quad \text { and } \quad G(x):=\int_{0}^{x} g(y) d y \\ \implies \int_{0}^{1} F(x) g(x) d x=F(1) G(1)-\int_{0}^{1} f(x) G(x) d x .\]
Solution
Let \(f, g \in L^1[0,1]\) and let \(T = \theset{(x,y) : 0 \le y \le x \le 1}\).
1. \(\iint_{T} |f(y)||g(x)|\,dy\,dx \le \|f\|_1\|g\|_1 < \infty\).
By Tonelli’s theorem, \(\iint_T |f(y)||g(x)|\,dy\,dx \le \int_0^1\int_0^1 |f(y)||g(x)|\,dy\,dx = \|f\|_1\|g\|_1\).
2. \(\int_0^1 F(x)g(x)\,dx = \int_0^1 f(y)\big(G(1) - G(y)\big)\,dy\).
By the definition of \(F\), \(\int_0^1 F(x)g(x)\,dx = \int_0^1\int_0^x f(y)g(x)\,dy\,dx\). By step 1, Fubini’s theorem allows integrating over \(T\) in the other order, with \(y \in [0,1]\) and \(x \in [y,1]\), giving \(\int_0^1 f(y)\int_y^1 g(x)\,dx\,dy\). Finally \(\int_y^1 g = G(1) - G(y)\).
3. Q.E.D.
By step 2 and \(F(1) = \int_0^1 f\), \(\int_0^1 Fg = G(1)\int_0^1 f - \int_0^1 fG = F(1)G(1) - \int_0^1 fG\).