convergence of improper integral wrt comparison to improper integrals known to exist

Question

If $\int_{0}^{\infty} (f(x))^2dx$ exists as an improper integral, and $\int_{0}^{\infty} (g(x))^2dx$ exists as an improper integral, we want to prove that $\int_{0}^{\infty} f(x)g(x)dx$ also exists as an improper integral. These functions are real-valued only.

My ideas so far: We know that for all $a, b \in \mathbb{R}$, $2ab \leq a^2 + b^2$, so we may be able to show that, by comparison (which only works if $f(x)g(x) \geq 0$ which is where I'm getting stuck and wondering if this is the right idea), that $\int_{0}^{\infty} 2 \cdot f(x)g(x)dx$ exists as an improper integral, which would imply $\int_{0}^{\infty} f(x)g(x)dx$ exists.

Thanks in advance!

Answer 1

If I haven't misses your points, then I think this is just a special case when we apply Hölder's inequality to the $ f,g \in L^2(0,+\infty) $ , where $ p=2,q=2 $: $$ \int_0^{\infty}fg\leq\int_0^{\infty}|fg|\leq\left(\int_0^{\infty}|f|^2\right)^{\frac12}\left(\int_0^{\infty}|g|^2\right)^{\frac12}<+\infty $$

Answer 2

This is clear: $$\int|fg|\le\left(f^2\right)^{1/2}\left(g^2\right)^{1/2},$$so the integral $\int fg$ actually converges absolutely, and hence it converges.