Given $f,g\in L^1(\mathbb{D})$, is it true that $|\int fg|\leq C|\int f|g||$ for some constant $C$. Here, $L^1(\mathbb{D})$ is the set of all complex-valued measurable functions on the open unit disc which are integrable with respect to the area measure on $\mathbb{D}$. I am not used to double integrals, but perhaps this question can be answered using only measure-theoretic notions. However, I am unable to gain much traction on this problem. My question is as follows - given $f,g\in L^1(\mathbb{D})$, is it true that $$\Bigg|\int fg\Bigg|\leq C\Bigg|\int f|g|\Bigg|$$ for some constant $C>0$?
Just as an example, it is easy to show for real-valued integrable functions, say $f,g:[0,1]\rightarrow \mathbb{R}$, that $\Bigg|\int fg\Bigg|\leq 2\Bigg|\int f|g|\Bigg|$ by using the definition of the Lebesgue integral along with basic properties of positive and negative numbers. Unfortunately the same techniques will not work in the setting of complex-valued functions.
Any help is greatly appreciated. Thank you.
 A: This is false (even for real valued functions) on virtually every measure space.
Take disjoint subsets $E_1,E_2 \subset X$ with $0<\mu(E_1)=\mu(E_2)<\infty$ and define $f=g=\chi_{E_1}-\chi_{E_2}$.
Since $f,g$ vanish outside of $Y := E_1\cup E_2$, we get
$$
\int f |g|\, d\mu =\int_Y f \, d\mu =0,
$$
where we used $|g|\equiv 1$ on $Y$.
But
$$
\int fg \, d\mu = \int_Y 1 \, d\mu =\mu(Y)>0.
$$
Hence, your inequality can hold for no $C$.
EDIT: This also fails (under reasonable assumptions, which are fulfilled if we take as our measure space the unit disk $\Bbb{D}$) for continuous functions $f,g$.
To see this, assume towards a contradiction that there is some $C>0$ with
$$
\bigg|\int fg \, d\mu \bigg| \leq C \cdot \bigg|\int f |g| \, d\mu \bigg|
$$
for all continuous $f,g : X \to \Bbb{C}$.
We observe that the argument given above yields $f_0, g_0 \in L^2 (\mu)$ such that $\int f_0 g_0 \, d\mu = \mu(Y) > 0$ and $\int f_0 |g_0| \, d\mu = 0$.
For regular measures (for example the usual measure on the unit circle), the continuous functions are dense in $L^2$. This yields sequences $(f_n)_n, (g_n)_n$ of continuous functions with $f_n \to f_0$ and $g_n \to g_0$ in $L^2$.
It is easy to see that this also implies $|g_n| \to |g_0|$ in $L^2$. All in all, we conclude
\begin{eqnarray*}
\mu(Y) &=& \bigg| \int f_0 g_0 \, d\mu\bigg|\\ &=& \big| \langle f_0, g_0 \rangle_{L^2} \big|\\ &=& \lim_n \big| \langle f_n, g_n \rangle_{L^2} \big| \\ &\leq & \lim_n C \cdot \bigg|\int f_n |g_n| \, d\mu\bigg| \\ &=& \lim_n C \cdot \big|\langle f_n, |g_n|\rangle \big| \\ &=& C \cdot \big| \langle f_0, |g_0|\rangle_{L^2} \big| = 0,
\end{eqnarray*}
contradiction.
