Suppose that $f\left(x\right)\geq0$ is continuous on $\left[-\infty,\infty\right]$ and $\int_{-\infty}^{\infty}f\left(x\right)dx=1$. Is it true that $\int_{-\infty}^{\infty}\left|x\right|f\left(x\right)dx\leq\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left|x-y\right|f\left(x\right)f\left(y\right)dxdy$?
Thanks.
As suggested by Robin, we would need to prove that if $X$ and $Y$ are independent random variables with the same distribution, the inequality $E|X|\leq E|X-Y|$ holds.
This is certainly NOT true in general; for example suppose $X$ and $Y$ have some constant non-zero value; then the RHS is 0 and the LHS is positive.
However, if you add the condition that $E(X)=0$ then the result is true. (In your notation, $\int_{-\infty}^{\infty}xf(x)=0$ ).
In that case write $p=P(X\geq 0)$, and write $X_+$ for the positive part of $X$ and $X_-\leq 0$ for the negative part of $X$.
We have $0=EX=E|X_+|-E|X_-|$, so $E|X_+|=E|X_-|$.
Hence also $E|X|=E|X_+|+E|X_-|=2E|X_+|$.
Now $E|X-Y| \geq E(|X-Y|I(Y<0)I(X\geq 0)) + E(|X-Y|I(X<0)I(Y \geq 0))$
$=2 E(|X-Y|I(Y<0)I(X\geq0))$
$=2 E(|X_+|I(Y<0)I(X\geq0)) + 2E(|Y_-|I(Y<0) I(X\geq 0))$
$=2(1-p)E(|X_+|I(X\geq 0)) + 2p E(|Y_-|I(Y<0))$
$=2(1-p)E|X_+| + 2pE|Y_-|$
$=2E|X_+|$
$=E|X|$
(using at various times the facts that $X$ and $Y$ are independent and that $X$ and $Y$ have the same distribution)
This can be rephrased as follows.
Let $X$ and $Y$ be independent random variables with the same, continuous, distribution. Is it true that $E(X)\le E(|X-Y|)$.
Is this likely?
Is it true for a discrete distribution?
If one has a discrete counterexample, can one convert it to a continuous counterexample?
Added I now realise that there was some absolute value signs in the original integral (these come out badly on my screen) and I should have written $E(|X|)\le E(|X-Y|)$. Plus ca change...
As it is mentioned in James Martin's answer, if one does not make additional assumptions, the statement is wrong.
One correct statement is: if $X$ and $Y$ are independent r.v.'s with finite expectations, and $\mathbf{E}Y=0$, then $\mathbf{E}f(X)\le \mathbf{E}f(X+Y)$ for any convex function $f$ such that these expectations are finite.
Proof: The sequence of two r.v.'s $X$ and $X+Y$ is a martingale. Therefore, the sequence $f(X), f(X+Y)$ is a submartingale, and the claim follows.
Of course, $f$ should be taken to be the absolute value function, and $Y$ should be replaced by $-Y$, if we want this statement to look more like the statement in question.
