# Application of Cauchy-Schwarz to improper integrals

I would like to know if I can apply the Cauchy-Schwarz inequality to prove the following claim:

$$\left( \int_0^\infty P(t) e^{-\theta t} t \ dt \right)^2 \leq \left( \int_0^\infty P(t) e^{-\theta t} \ dt \right) \left( \int_0^\infty P(t) e^{-\theta t} t^2 \ dt \right)$$ where $$P(t)$$ is a non-negative real function, $$\theta$$ is a positive scalar, and all integrals involved in the inequality converge.

My idea is to apply the Cauchy-Schwarz inequality (for non-negative integrands) $$\left( \int f g \right)^2 \leq \int f^2 \int g^2$$
Setting $$f= \sqrt{P(t) e^{-\theta t}}$$ and $$g = \sqrt{P(t) e^{-\theta t}} t$$

• Yes, you can... Nov 18, 2021 at 13:46
• great, thx! I just wasn't sure as it involves improper integrals. I would like to find a source of the exact statement of the Cauchy-Schwarz inequality for L^2 spaces. Nov 18, 2021 at 13:52
• You can always write the CS inequality for proper integrals $\int_0^b$ first, and then take the limit $b \to \infty$. Here is a similar question math.stackexchange.com/q/3580202/42969. Nov 18, 2021 at 13:54

Yes, you can. CS is a special case of Holder's inequality, taking $$p = q = 2$$ ($$\in (1,\infty), 1/p+1/q=1$$).
$$\left( \int_0^\infty P(t) e^{-\theta t} t \ dt \right)\leq \left( \int_0^\infty P(t) e^{-\theta t} \ dt \right)^{\frac{1}{2}} \left( \int_0^\infty P(t) e^{-\theta t} t^2 \ dt \right)^{\frac{1}{2}}$$
Let $$f_t,g_t$$ be defined as you have and write $$f,g$$ for simplicity. Let $$R=\mathbb{R}^+\cup \:\{0\}$$. Then you recover Holder's:
$$\int_{R}|fg| = \|fg\|_{L^1(R)}\le\|f\|_{L^p(R)}\|g\|_{L^q(R)}= \left(\int_{R}|f|^p\right)^{\frac{1}{p}} \left(\int_{R}|g|^q\right)^{\frac{1}{q}}$$