Is there such an inequality between product and integral for functions 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!
 A: 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").
A: 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
http://en.wikipedia.org/wiki/Convolution_algebra#The_convolution_algebra_L1.28G.29
A: 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.
