# Cauchy-Schwarz for integrals

One of the exercises (3.2) of Izenman's Modern Multivariate Statistical Techniques is for $A \subset \mathbb{R}$, $$\left( \int_{A}fg\right)^2 \leq \left(\int_{A}f^2\right) \left(\int_{A}g^2\right)$$ where $f: A \to \mathbb{R}$ and $g: A \to \mathbb{R}$ are such that $f^2$, $g^2$ are integrable.

Now I've already proven this for a general inner product, i.e., over a vector space $V$ where $\mathbf{x}, \mathbf{y} \in V$, $$\langle \mathbf{x}, \mathbf{y} \rangle ^2 \leq \langle \mathbf{x}, \mathbf{x} \rangle\langle \mathbf{y}, \mathbf{y}\rangle\text{.}$$

But if I recall from linear algebra, $\int_{A}fg$ might be an inner product, so all I would need to do is show that it's an inner product.

I have a few questions though:

• Doesn't $fg$ have to be integrable as well?
• Related to the previous question: if $\int_{A}fg$ is an inner product (according to my memory), what vector space is it over? Do $f$ and $g$ have to be continuous? Integrable? I'm pretty sure that it's not as simple as $f$ and $g$ are real-valued and defined over $A$, such that $f^2$ and $g^2$ are integrable. (I would THINK $fg$ would have to be integrable as well, too.)

1) If $f$ and $g$ are square-integrable then $fg$ is automatically integrable, because $$|fg| \leq \frac{|f|^2+|g|^2}{2}.$$
2) The vector space is usually noted $L^2(A)$ and consists of the square integrable functions. You don't need more assumptions.