1. ## Explanation of a Proof about L^2

Reading through Apostol's analysis text on $L^{2}(I)$ spaces, he claims that the inequality $|(f, g)| \leq ||f|| \cdot ||g||$ follows immediately from $\displaystyle \int_{I} \Bigl[ \int_{I} |f(x)g(y) - f(y)g(x)|^{2} dy \Bigr] dx \geq 0$. But all I can see is that $|(f, g)| = \Bigl| \displaystyle \int_{I} f(x)g(x) dx \Bigr| \leq \int_{I} |f(x)g(x)| dx$ and $\displaystyle ||f|| \cdot ||g|| = \Bigl( \bigl[ \int_{I} f^{2}(x) dx \bigr] \bigl[ \int_{I} g^{2}(x) dx \bigr] \Bigr)^{\frac{1}{2}}$, and I don't see how these three facts combine to prove the inequality we want. I've thought about squaring both and comparing them, but that doesn't seem like it's helping me.

2. Originally Posted by ragnar
Reading through Apostol's analysis text on $L^{2}(I)$ spaces, he claims that the inequality $|(f, g)| \leq ||f|| \cdot ||g||$ follows immediately from $\displaystyle \int_{I} \Bigl[ \int_{I} |f(x)g(y) - f(y)g(x)|^{2} dy \Bigr] dx \geq 0$. But all I can see is that $|(f, g)| = \Bigl| \displaystyle \int_{I} f(x)g(x) dx \Bigr| \leq \int_{I} |f(x)g(x)| dx$ and $\displaystyle ||f|| \cdot ||g|| = \Bigl( \bigl[ \int_{I} f^{2}(x) dx \bigr] \bigl[ \int_{I} g^{2}(x) dx \bigr] \Bigr)^{\frac{1}{2}}$, and I don't see how these three facts combine to prove the inequality we want. I've thought about squaring both and comparing them, but that doesn't seem like it's helping me.
Notice that $fg$ is measurable, so it suffices to dominate it by an integrable function to show it's integrable. To do this note $|fg|\leq \max \{|f|,|g|\} ^2 =\max \{ f^2,g^2\} \leq f^2 +g^2$. To prove the inequality just expand the square in the first integral to obtain something like:

$\left( \int_I f^2(x)dx \right) \left( \int_I g^2(y)dy \right) - 2\left( \int_I f(x)g(x)dx \right) \left( \int_I f(y)g(y)dy \right) +\left( \int_I f^2(y)dy \right) \left( \int_I g^2(x)dx \right) \geq 0$

Now just remember that the variable of integration is irrelevant, rename, regroup, cancel, etc.