Given a measure space $(\Omega, \mathbb{F},\mu)$ and any two measurable real-valued functions $f,g$ defined on $\Omega$, I was wondering if there is an inequality like $$ \int_\Omega |f*g| d\mu \leq \int_\Omega |f| d\mu * \int_\Omega |g| d\mu? $$
Or what are the correct relations between product and integral?
What if the measure is a probability measure?
Thanks and regards!
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$\begingroup$Below, assume that all integrals/expectations are finite.
If the measure is a probability measure, then your question reads $$ {\rm E}|XY| \le {\rm E}|X|{\rm E}|Y|? $$ (Here $X$ and $Y$ are random variables.) It does hold $$ |{\rm E}(XY)|^2 \le {\rm E}(X^2 ){\rm E}(Y^2 ). $$
Note that $X$ and $Y$ (which are measurable functions from $\Omega$ to $\mathbb{R}$) correspond to $f$ and $g$. That is, the correct inequality is $$ \bigg|\int_\Omega {fg\,d\mu } \bigg|^2 \le \bigg(\int_\Omega {f^2 \,d\mu } \bigg)\bigg(\int_\Omega {g^2 \,d\mu } \bigg) $$ (generalized below), where $\mu$ is the probability measure.
Your inequality, however, is evidently false (in general), for if $X$ is nonnegative, it gives ${\rm E}(X^2 ) \le {\rm E}^2 (X)$, or ${\rm Var}(X) \leq 0$.
EDIT: More (but not most) generally, if $p,q \in (1,\infty)$ satisfy $1/p + 1/q = 1$, then $$ {\rm E}|XY| \le ({\rm E}|X|^p )^{1/p} ({\rm E}|Y|^q)^{1/q}, $$ or $$ \int_\Omega {|fg|\,d\mu } \le \bigg(\int_\Omega {|f|^p \,d\mu } \bigg)^{1/p} \bigg(\int_\Omega {|g|^q \,d\mu } \bigg)^{1/q}. $$ For further details, see here (note the section "Generalization for probability measures").
$\endgroup$ 1 $\begingroup$Edit: I realise that we are apparently talking about the pointwise product of functions. Then the statement is false:
Let $f$ and $g$ be constant with value $1$. Let $\mu(\Omega)=1/2$. Then the left hand side is $1/2$ while the right hand side is $1/4$. That's a counterexample.
First of all: Convolution only makes sense if $\Omega$ has a group structure like for instance $\mathbb R$.
If you assume that, then, yes, your statement is true.
A reformulation of your estimation is that the $L^1$-norm on $L^1(G,\mu)$ is sub-multiplicative, see
$\endgroup$ $\begingroup$Assuming that the $*$ means product we have Cauchy-Schwarz (a special case of Hölder)
$$\int |fg| \, d\mu = \sqrt{\int |f|^2 \, d\mu \int |g|^2 \, d\mu}$$
If it is convolution look up the Youngs inequality for convolutions. In the common setting your RHS doesn't make any sense.
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