Show $f, g\in\mathscr{L}^2(\mu)$ imply $f\overline{g}, \overline{f}g\in\mathscr{L}(\mu)$, and $\int f\overline{g} +\overline{f}g\leq2\|f\|\|g\|$. The problem arises from Walter Rudin's Principles of Mathematical Analysis:

My Question:

*

*The equality highlighted in yellow. Indeed we have $\|f+g\|^2=\int|f+g|^2d\mu=\int(|f|^2+f\overline{g}+\overline{f}g+|g|^2)d\mu$; but in order to say $\int(|f|^2+f\overline{g}+\overline{f}g+|g|^2)d\mu=\int|f|^2+\int f\overline{g}+\int\overline{f}g+\int|g|^2$, we need to have $\mathbf{f\overline{g}, \overline{f}g\in\mathscr{L}(\mu)}$ first. To that end, we let $f=u+iv$, and let $g=a+ib$, then $\int|f\overline{g}|d\mu=\int\sqrt{(ua+vb)^2+(va-ub)^2}d\mu$. But then I failed to show that $\int\sqrt{(ua+vb)^2+(va-ub)^2}d\mu<+\infty$.


*The inequality highlighted in green. In order to say the inequality holds, we must have  $\mathbf{\int f\overline{g} +\overline{f}g\leq2\|f\|\|g\|}$, which I again failed to show.
Any hint would be greatly appreciated.
 A: By the Schwarz inequality,
$$\int|fg| \leq \|f\|\|g\| < \infty$$
In particular, as $|\bar{f}g| = |f\bar{g}| = |fg|,$ this shows $\bar{f}g, f\bar{g}, fg \in \mathscr{L}$.  This should resolve your first question.

By the triangle inequality in $\mathbb{C}$,
$$\begin{align*}\|f+g\|^2 &= \left|\|f+g\|^2\right| \\ &= \left|\int|f|^2 + \int f\bar{g} + \int \bar{f}g + \int|g|^2\right|\\ &\leq \int |f|^2 + \left|\int f\bar{g}\right| + \left|\int\bar{f}g\right| + \int |g|^2\end{align*}$$
And by theorem 11.26,
$$\left|\int f\bar{g}\right| \leq \int|fg| \\ \left|\int \bar{f}g\right| \leq \int|fg|$$
Putting it together:
$$\|f+g\|^2 \leq \|f\|^2 + 2\int|fg| + \|g\|^2.$$
Again, by Schwarz, $$\|f+g\|^2 \leq \|f\|^2 + 2\|f\|\|g\| + \|g\|^2$$
A: The $L^2$ inner product is generally defined as
$$\langle f,g \rangle_{L^2(\mu)} = \int_X f\overline{g}d\mu.$$
Then recall that for a general inner product, we have $|\langle u,v \rangle| \leq \|u\|\cdot\|v\|$. I think this is what you were missing. Can you fill in the rest?
