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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 $g = \sqrt{P(t) e^{-\theta t}} t$

Cameron Williams
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mkb90
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    Yes, you can... – Martin R Nov 18 '21 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. – mkb90 Nov 18 '21 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 https://math.stackexchange.com/q/3580202/42969. – Martin R Nov 18 '21 at 13:54

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Yes, you can. CS is a special case of Holder's inequality, taking $p = q = 2$ ($\in (1,\infty), 1/p+1/q=1$).

Spelled out:

$$ \left( \int_0^\infty P(t) e^{-\theta t} t \ dt \right)\leq \left( \int_0^\infty P(t) e^{-\theta t} \ dt \right)^{\frac{1}{2}} \left( \int_0^\infty P(t) e^{-\theta t} t^2 \ dt \right)^{\frac{1}{2}} $$

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:

$$ \int_{R}|fg| = \|fg\|_{L^1(R)}\le\|f\|_{L^p(R)}\|g\|_{L^q(R)}= \left(\int_{R}|f|^p\right)^{\frac{1}{p}} \left(\int_{R}|g|^q\right)^{\frac{1}{q}}$$

Martin Geller
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