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!

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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").

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Note: the assumption of total measure $1$ and the assumption of finite integrals/expectations is not essential. – Shai Covo Feb 18 '11 at 3:32

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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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

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